MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 4
Liberty University
Question 1
Question
Determine if the series ∞
X
n=1
n2+ 1
n4−1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s denote the general term of the series as an=n2+1
n4−1.
Step 1: Find a suitable series to compare with. We will choose the series
∞
X
n=1
1
n2
since it is a well-known convergent p-series with p= 2.
Step 2: Compute the limit
lim
n→∞
an
1/n2= lim
n→∞
n2+ 1
n4−1·n2
1
Simplify the expression to get
= lim
n→∞
n4+n2
n4−1= 1
Step 3: Since the limit is a nonzero finite value, both series either both
converge or both diverge. Since P∞
n=1 1
n2converges, by the Limit Comparison
Test, we conclude that P∞
n=1 n2+1
n4−1also converges.
Question 2
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
Step 1: Apply the ratio test. Let’s consider the limit of the ratio of consec-
utive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
Step 3: Analyze the limit.
lim
n→∞
n
n+ 1
= 1
Step 4: Determine convergence. Since the limit is equal to 1, the ratio test
is inconclusive. Therefore, we need to try a different test.
Step 5: Apply the root test. Let’s consider the root test, which involves
calculating:
lim
n→∞
n
p|an|= lim
n→∞
n
s
n!
nn
Step 6: Simplify and calculate the limit.
lim
n→∞
n
s
n!
nn
= lim
n→∞
n1/n
n
This limit simplifies to:
lim
n→∞
1
n= 0
Step 7: Determine convergence. Since the limit is less than 1, by the root
test, the series P∞
n=1 n!
nnconverges.
2
Question 3
Question
Prove or disprove the convergence of the series P∞
n=1 n2+1
n3+n.
Solution
To determine the convergence of the series P∞
n=1 n2+1
n3+n, we can use the limit com-
parison test. Specifically, we will compare it with the harmonic series P∞
n=1 1
n.
Step 1: Determine a comparison series Consider the series P∞
n=1 1
n.
Let’s simplify the given series to find a suitable comparison.
n2+ 1
n3+n=n2/n3+ 1/n3
n+ 1/n =1/n + 1/n3
1+1/n2
Step 2: Find the limit of the ratio Now, we calculate the limit as n
approaches infinity:
lim
n→∞
(1/n + 1/n3)
(1 + 1/n2)= lim
n→∞
1/n + 1/n3
1+1/n2
Step 3: Simplify the limit
lim
n→∞
1/n + 1/n3
1+1/n2=0+0
1+0 = 0
Step 4: Apply the limit comparison test Since the limit is finite and
positive, we can conclude that the given series P∞
n=1 n2+1
n3+nconverges by the
limit comparison test with P∞
n=1 1
n.
Therefore, the series P∞
n=1 n2+1
n3+nconverges.
Question 4
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
3
Step 2: Simplify the expression.
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Further simplify.
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 4: Analyze the limit. Since the limit is equal to 1, the ratio test is
inconclusive. We will try another test.
Step 5: Apply the root test. Let’s try the root test, computing:
lim
n→∞
n
p|an|= lim
n→∞
n
rn!
nn
Step 6: Simplify and analyze the limit.
= lim
n→∞
n
√n!
n
Since n!→ ∞ faster than nn→ ∞, the limit approaches 0.
Step 7: Conclusion. The root test shows that the series converges. There-
fore, the series P∞
n=1 n!
nnconverges.
Question 5
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Calculate the ratio R. Since the ratio test involves calculating
the limit of the absolute value of the ratio of consecutive terms, we have:
R= lim
n→∞
an+1
an
where an=n!
nn. Calculating an+1 and anseparately, we get:
an+1 =(n+ 1)!
(n+ 1)n+1 and an=n!
nn
4
Therefore, the ratio Rbecomes:
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio R. Simplify the ratio further as follows:
R= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
R= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
R= lim
n→∞
nn
(n+ 1)n
R= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit of R. Taking the limit as napproaches infinity,
we get:
R= lim
n→∞ n
n+ 1n
=1
e
Step 4: Determine the convergence or divergence of the series. If R < 1, the
series converges. Otherwise, it diverges. Since R=1
e<1, the series P∞
n=1 n!
nn
converges by the ratio test.
Question 6
Question
Determine whether the series
∞
X
n=1
n2+ 3n
n4+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
bn=1
n2.
Step 1: Find the limit of the ratio of the two series.
lim
n→∞ n2+3n
n4+1
1
n2= lim
n→∞
n4+ 3n3
n4+ 1 = 1
5
Step 2: State the Limit Comparison Test. If limn→∞
an
bnexists and is
a positive finite number, then either both series P∞
n=1 anand P∞
n=1 bnconverge
or both diverge.
Step 3: Conclusion Since limn→∞
an
bn= 1 (a positive finite number), by
the Limit Comparison Test, the series P∞
n=1 n2+3n
n4+1 has the same convergence
behavior as P∞
n=1 1
n2, which is a convergent p-series with p= 2. Therefore, the
given series converges.
Question 7
Question
Determine whether the series ∞
X
n=1
2n
3n+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
2n
3n+ 1, we can use the Ratio
Test.
Step 1: Find the limit. Let an=2n
3n+1 . We compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
2n+1
3n+1 + 1 ·3n+ 1
2n
= lim
n→∞
2
3 + 1
3n
=2
3.
Step 2: Apply the Ratio Test. - If limn→∞
an+1
an<1, then the se-
ries converges. - If limn→∞
an+1
an>1 or limn→∞
an+1
an= 1, then the series
diverges.
Since limn→∞
an+1
an=2
3<1, by the Ratio Test, the series ∞
X
n=1
2n
3n+ 1
converges.
Question 8
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
6
Solution
Let’s use the ratio test to determine the convergence of the series. Recall that
for a series P∞
n=1 an, if
lim
n→∞
an+1
an
=L,
then the series converges if L < 1, and diverges if L > 1.
Step 1: Find the ratio of consecutive terms We have an=n!
nn. Let’s
find an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=n+ 1
(n+ 1)n+1 ·n=n
(n+ 1)n
Step 2: Determine the limit Now, let’s compute the limit as napproaches
infinity:
lim
n→∞
n
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n= lim
n→∞
n
e=∞
Step 3: Conclusion Since the limit of an+1
anis infinity, the series diverges
by the ratio test. Therefore, the series
∞
X
n=1
n!
nn
diverges.
Question 9
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn, then we consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplifying the above expression by dividing the numerator by the denominator
gives:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!nn
(n+ 1)n+1n!
Step 2: Simplify the expression. Further simplifying the expression, we get:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
7
Step 3: Evaluate the limit. As ntends to infinity, the expression n
n+1 tends
to 1, and hence the limit simplifies to:
lim
n→∞ n
n+ 1n
= lim
n→∞ |1|= 1
Step 4: Apply the Ratio Test. Since the limit is equal to 1, we cannot make
any conclusion about convergence or divergence of the series using the Ratio
Test. We will need to use a different test.
In this case, we can use the Root Test to determine the convergence.
Therefore, the convergence of the series P∞
n=1 n!
nnis inconclusive by the Ratio
Test.
Question 10
Question
Determine whether the series ∞
X
n=1
n3
2n
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
Step 2: Simplify the expression and compute the limit:
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1+3/n + 3/n2+ 1/n3
2
=1
2
8
Step 3: Analyze the limit: Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n3
2n
converges.
Therefore, the series converges.
Question 11
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Using the
ratio test, we find:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Analyze the result. Since 1
e<1, the ratio test tells us that the
series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 12
Question
Determine whether the series
∞
X
n=1
n2+n+ 1
n2+ 2n+ 1
converges or diverges.
9
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series
bn=1
n
Step 1: Find the limit of the ratio of the given series and bnWe
will find the limit of the ratio an
bnas napproaches infinity:
lim
n→∞
n2+n+1
n2+2n+1
1
n
Simplify the expression:
lim
n→∞
n3+n2+n
n2+ 2n+ 1
lim
n→∞
n3+n2+n
n2+ 2n+ 1 = lim
n→∞
n3+n2+n
n2+ 2n+ 1 = lim
n→∞
1 + 1
n+1
n2
1 + 2
n+1
n2
= 1
Step 2: Determine the convergence based on the limit of the ratio
Since the limit is a positive finite number, we can conclude that the given series
has the same convergence behavior as the series P1
n, which is a divergent p-
series with p= 1. Therefore, by the limit comparison test, the given series
∞
X
n=1
n2+n+ 1
n2+ 2n+ 1
diverges.
Question 13
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Let’s evaluate the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression inside the limit:
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
10
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Taking the limit as napproaches infinity:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e
Step 4: Apply the ratio test: If the limit is less than 1, the series converges;
if the limit is greater than 1, the series diverges. Since the limit is 1
e<1, the
series P∞
n=1 n!
nnconverges by the ratio test.
Question 14
Question
Consider the series ∞
X
n=1
1
n3ln2n. Determine if the series converges or diverges.
Solution
To determine the convergence of the series, we will use the Integral Test. Let
f(x) = 1
x3ln2x.
Step 1: Find the integral We will calculate the integral of f(x) over the
interval [2,∞).
Z∞
2
1
x3ln2xdx
Step 2: Evaluate the integral Let u= ln x, then du =1
xdx. The integral
becomes Z1
u2du =−1
u=−1
ln x
∞
2
Applying the Fundamental Theorem of Calculus:
−lim
b→∞
1
ln b+1
ln 2
Step 3: Check for convergence Since the improper integral converges,
by the Integral Test, the series ∞
X
n=1
1
n3ln2nalso converges.
11
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3 .
Solution
To determine the convergence or divergence of the series P∞
n=1 n3+2n+1
n4+3 , we will
use the Limit Comparison Test.
Step 1: Let’s find the limit of the ratio of the given series term to the term
of a known convergent or divergent series.
Consider the series P∞
n=1 1
n, which is a known divergent p-series when p= 1.
Let an=n3+2n+1
n4+3 and bn=1
n. We want to find the limit of an
bnas napproaches
infinity.
lim
n→∞
an
bn
= lim
n→∞
n3+ 2n+ 1
n4+ 3 ·n
1= lim
n→∞
n4+ 2n2+n
n4+ 3 .
Dividing by the highest power of nin the denominator, we get:
= lim
n→∞
1+2/n2+ 1/n3
1+3/n4=1
1= 1.
Step 2: Now, we compare the limit of an
bnto a known series.
Since the limit is finite and non-zero, we apply the Limit Comparison Test.
Since the harmonic series diverges, and P∞
n=1 1
ndiverges, we conclude that
P∞
n=1 n3+2n+1
n4+3 also diverges.
Therefore, the series P∞
n=1 n3+2n+1
n4+3 diverges.
Question 16
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
12
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!nn
n!(n+ 1)n+1 =n+ 1
(1 + n)n=n+ 1
(1 + 1/n)n.
Step 2: Examine the limit of the ratio. Taking the limit as napproaches
infinity,
lim
n→∞
n+ 1
(1 + 1/n)n= lim
n→∞
1+1/n
(1 + 1/n)n= lim
n→∞
1
e=1
e,
where we have used the limit definition of the number e.
Step 3: Apply the ratio test. Since the limit of the ratio is 1
e<1, by the
ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 17
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine whether the series P∞
n=1 n!
nnconverges or diverges, we will use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit of
the ratio an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Evaluate the limit. Let L= limn→∞
nn
(n+1)n. Rewrite this limit in
a suitable form for further evaluation.
L= lim
n→∞
1
(1 + 1
n)n
13
Step 4: Find the limit using the fact that limn→∞(1 + 1
n)n=e(a standard
result from calculus).
L=1
e
Step 5: Analyze the ratio. Since L=1
e<1, the ratio test implies that the
series P∞
n=1 n!
nnis absolutely convergent.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 18
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2
4n.
Solution
To determine the convergence or divergence of the given series, we will first
rewrite it in a more convenient form.
Step 1: Rewrite the series
∞
X
n=1
3n+ 2
4n=∞
X
n=1 3n
4n+2
4n=∞
X
n=1 3
4n
+∞
X
n=1
2
4n.
Now, we have split the series into two separate series, both of which we need
to analyze for convergence.
Step 2: Analyze the first series For the first series,
∞
X
n=1 3
4n
,
we have a geometric series with a= 3/4 and r= 3/4. This series converges if
|r|<1.
Since |3/4|<1, the series P∞
n=1 3
4nconverges.
Step 3: Analyze the second series For the second series,
∞
X
n=1
2
4n,
we observe that this is also a geometric series with a= 2/4 and r= 1/4. This
series converges if |r|<1.
Since |1/4|<1, the series P∞
n=1 2
4nconverges.
Step 4: Conclusion Since both series P∞
n=1 3
4nand P∞
n=1 2
4nconverge,
the original series P∞
n=1 3n+2
4nalso converges.
14
Question 19
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the Ratio Test.
Step 1: Compute the ratio r: Let an=n!
nn. Then we have:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
Step 2: Simplify the expression: We know that limn→∞(1 + 1/n)n=e, so:
r= lim
n→∞
n
e=∞
Step 3: Analyze the limit: Since the limit of the ratio test is ∞, the series
∞
X
n=1
n!
nndiverges.
Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 20
Question
Let {an}be a sequence such that an>0 for all nand limn→∞ an= 0. Determine
the convergence or divergence of the series
∞
X
n=1
an
1 + a1+a2+··· +an
15
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Find a suitable series for comparison. Let’s consider the series
P∞
n=1
an
nfor comparison. Notice that an
1+a1+a2+···+an>an
nsince 1 + a1+a2+
··· +an> n for all n.
Step 2: Compute the limit of the ratio. Let bn=an
n. We will calculate
lim
n→∞
an/n
bn
= lim
n→∞
an
n·n
an
= lim
n→∞ 1=1.
Step 3: Apply the Limit Comparison Test. Since P∞
n=1 bnconverges if
and only if P∞
n=1 an/(1+a1+a2+···+an) converges, and we found the limit of
the ratio to be 1, by the Limit Comparison Test, the series P∞
n=1
an
1+a1+a2+···+an
converges if and only if P∞
n=1
an
nconverges.
Therefore, since P∞
n=1
an
nis a p-series with p= 1, it diverges. Thus, the
given series P∞
n=1
an
1+a1+a2+···+analso diverges.
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the given series, we will use the
ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nnbe the
16
general term of the series. Then, we compute:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e= 0
Step 2: Apply the ratio test. Since the limit of the ratio of consecutive
terms is 1
e= 0, the series P∞
n=1 n!
nndiverges by the ratio test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. We compute
the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!/(n+ 1)n(n+ 1)
n!
= lim
n→∞
1
(1 + 1/n)n
Step 3: Evaluate the limit: We recognize that the limit above is the defini-
tion of the constant e. Therefore,
lim
n→∞
an+1
an
=1
e
Step 4: Apply the ratio test: The series converges if the limit is less than
1 and diverges if the limit is greater than 1. Since 1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
17
Question 23
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify further:
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Apply the ratio test: Since 1
e<1, the series P∞
n=1 n!
nnconverges by
the ratio test.
Question 24
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Let’s apply the ratio test to the series. The ratio test states: If
limn→∞
an+1
an=L, then the series converges if L < 1 and diverges if L > 1.
18
Step 2: Calculate the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify the expression:
an+1
an
=(n+ 1)nn
(n+ 1)n+1 =n
n+ 1n
Step 4: Compute the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 5: Since the limit is 1
ewhich is less than 1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 25
Question
Determine whether the series P∞
n=1 2n2+3
n3+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test with the series P∞
n=1 1
n, which is a known series.
Step 1: Let’s find the limit of the ratio of the two series:
lim
n→∞
2n2+3
n3+1
1
n
= lim
n→∞
2n3+ 3n
n3+ 1 ·n
1= lim
n→∞
2 + 3
n2
1 + 1
n3
= 2
Step 2: Since the limit is a finite positive number (= 2), by the Limit Com-
parison Test, we conclude that P∞
n=1 2n2+3
n3+1 and P∞
n=1 1
neither both converge
or both diverge.
Step 3: Since the harmonic series P∞
n=1 1
ndiverges (as a p-series with
p= 1), we can conclude that the given series P∞
n=1 2n2+3
n3+1 also diverges by the
Limit Comparison Test.
Question 26
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
2n3+5 .
19
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s consider the series P∞
n=1 n2+1
2n3+5 and compare it with
the convergent series P∞
n=1 1
n.
Step 1: Find the limit Let an=n2+1
2n3+5 and bn=1
n. We want to find
limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
2n3+ 5 ·n
1= lim
n→∞
n3+n
2n3+ 5
Step 2: Simplify the expression
= lim
n→∞
n3+n
2n3+ 5 = lim
n→∞
1 + 1
n2
2 + 5
n3
=1
2
Since the limit is a positive finite value, we can apply the Limit Comparison
Test.
Step 3: Apply the Limit Comparison Test Since limn→∞
an
bnexists and
is a positive finite value, we can conclude that either both series Panand Pbn
converge or both diverge. Since Pbn=P1
nis a p-series with p= 1 >0, we
know that Pbnconverges. Therefore, by the Limit Comparison Test, the series
P∞
n=1 n2+1
2n3+5 also converges.
Question 27
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test: Let an=n!
nn. Compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
20
Step 3: Evaluate the limit:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Check the ratio: Since the limit of the ratio is 1
e<1, by the ratio
test, the series P∞
n=1 n!
nnconverges.
Question 28
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
following statement: If limn→∞ an= 0, then the series P∞
n=1 1
nanalso converges.
Solution
To prove or disprove the statement, we will provide a counterexample.
Step 1: Counterexample Consider the series defined by an=1
n2. It’s
clear that limn→∞ an= 0 since limn→∞
1
n2= 0. However, the series P∞
n=1 an=
P∞
n=1 1
n2converges (by the p-series test with p= 2).
Step 2: Series P∞
n=1 1
nanNow let’s consider the series P∞
n=1 1
nan=P∞
n=1 1
n·1
n2
=
P∞
n=1 n. This harmonic series P∞
n=1 ndiverges (since it’s a well-known result).
Step 3: Conclusion As seen in Step 1 and Step 2, we have found a
counterexample that disproves the statement. Therefore, the statement ”If
limn→∞ an= 0, then the series P∞
n=1 1
nanalso converges” is false.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
21
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We evaluate:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1
= lim
n→∞
(n+ 1) n
n+ 1n
= lim
n→∞
(n+ 1) 1
1 + 1
nn
= lim
n→∞(n+ 1) ·lim
n→∞ 1
1 + 1
nn
=∞ · 1
e
=∞.
Step 2: Analyze the limit. Since the limit is greater than 1, the series
P∞
n=1 n!
nndiverges by the Ratio Test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 30
Question
Determine whether the series
∞
X
n=1
n
n3+√n+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Comparison Test.
Step 1: We will find a series that converges and is larger than the given
series. Notice that for all n≥1, we have
n
n3+√n+ 1 ≤n
n3=1
n2
Therefore, we will compare our series to P∞
n=1 1
n2.
Step 2: We will show that the series P∞
n=1 1
n2converges. This is a p-series
with p= 2 >1, so by the p-series test, P∞
n=1 1
n2converges.
Step 3: Using the Comparison Test, since n
n3+√n+1 ≤1
n2and P∞
n=1 1
n2
converges, then by the Comparison Test, P∞
n=1 n
n3+√n+1 also converges.
Therefore, the given series converges by the Comparison Test.
22
Question 31
Question
Let {an}be a sequence such that limn→∞ an= 0 and P∞
n=1 anconverges, but
P∞
n=1 |an|2diverges. Prove that P∞
n=1 anconverges absolutely.
Solution
Step 1: Since limn→∞ an= 0, there exists some N∈Nsuch that |an|<1 for
all n≥N. Therefore, for n≥N, we have |an|≤|an|2.
Step 2: Consider the series P∞
n=1 an. Since P∞
n=1 anconverges, the sequence
{an}must be bounded. Thus, there exists some M > 0 such that |an| ≤ Mfor
all n∈N.
Step 3: For n<N, we have |an| ≤ Mand for n≥N, we have |an|≤|an|2≤
M. Therefore, |an| ≤ Mfor all n∈N.
Step 4: Now, we have |an| ≤ Mfor all n∈N. This implies that the series
P∞
n=1 |an|converges by the Comparison Test.
Step 5: Thus, since P∞
n=1 |an|converges, the series P∞
n=1 anconverges ab-
solutely.
Question 32
Question
Determine the convergence of the series ∞
X
n=1
2n3+ 3n2
n4+ 5 .
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=2n3+ 3n2
n4+ 5 . We will compare this series with a known convergent
or divergent series.
Step 1: Calculate the limit
lim
n→∞
an
1
n2
= lim
n→∞
2n3+3n2
n4+5
1
n2
= lim
n→∞
2n+ 3
n2+5
n2
= lim
n→∞
2
n= 0
Step 2: Interpret the limit Since the limit is finite and positive, and
∞
X
n=1
1
n2converges (this is a p-series with p= 2 >1), by the Limit Comparison
23
Test, the given series ∞
X
n=1
2n3+ 3n2
n4+ 5 also converges.
Question 33
Question
Determine the convergence or divergence of the series P∞
n=1
(−1)n
n+ln(n).
Solution
To determine the convergence or divergence of the series, we will use the Alter-
nating Series Test (AST).
Step 1: Determine if the series satisfies the conditions of the AST
The series P∞
n=1
(−1)n
n+ln(n)is an alternating series in the form (−1)nbn, where
bn=1
n+ln(n).
We need to check the following two conditions for the AST to apply: 1. The
sequence bn=1
n+ln(n)is positive, decreasing, and approaches 0 as napproaches
infinity.
2. The terms of the series are decreasing, i.e., bn+1 ≤bnfor all n.
Step 2: Check if the sequence satisfies the conditions of the AST
Let’s evaluate the limit as napproaches infinity of bn=1
n+ln(n)to verify the
first condition:
lim
n→∞
1
n+ ln(n)= lim
n→∞
1/n
1+1/n = 0
So, the sequence bn=1
n+ln(n)is positive, decreasing, and approaches 0 as n
approaches infinity.
To check the second condition, we calculate bn+1
bn:
bn+1
bn
=1
n+ 1 + ln(n+ 1) ·n+ ln(n)
1=n+ ln(n)
n+ 1 + ln(n+ 1)
Since n+ ln(n)> n ≥0 and n+ 1 + ln(n+ 1) > n + 1 ≥1, we have bn+1
bn<1
for all n. Therefore, the terms of the series are decreasing.
Step 3: Apply the Alternating Series Test Since the series satisfies the
conditions of the AST, we conclude that the series P∞
n=1
(−1)n
n+ln(n)converges.
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n−1
n3+ 2
24
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s compare the given series with the series P∞
n=1 1
n.
Step 1: Find the limit of the ratio Let an=n2+3n−1
n3+2 and bn=1
n. We
will consider the limit
lim
n→∞
an
bn
= lim
n→∞
n2+3n−1
n3+2
1
n
Step 2: Simplify the limit Simplify the limit to get
lim
n→∞
n3+ 3n2−n
n3+ 2 = lim
n→∞
n(n2+ 3n−1)
n3+ 2 = lim
n→∞
n3+ 3n2−n
n3+ 2 = 1
Step 3: Apply the Limit Comparison Test Since the limit is a positive
finite number, by the Limit Comparison Test, we conclude that the series
∞
X
n=1
n2+ 3n−1
n3+ 2
and ∞
X
n=1
1
n
either both converge or both diverge.
Step 4: Conclusion The series P∞
n=1 1
nis a harmonic series which diverges.
Therefore, by the Limit Comparison Test, the series
∞
X
n=1
n2+ 3n−1
n3+ 2
also diverges.
Question 35
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test. Let an=n!
nn. Then, we consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
25
Question 2
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
Step 1: Apply the ratio test. Let’s consider the limit of the ratio of consec-
utive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
Step 3: Analyze the limit.
lim
n→∞
n
n+ 1
= 1
Step 4: Determine convergence. Since the limit is equal to 1, the ratio test
is inconclusive. Therefore, we need to try a different test.
Step 5: Apply the root test. Let’s consider the root test, which involves
calculating:
lim
n→∞
n
p|an|= lim
n→∞
n
s
n!
nn
Step 6: Simplify and calculate the limit.
lim
n→∞
n
s
n!
nn
= lim
n→∞
n1/n
n
This limit simplifies to:
lim
n→∞
1
n= 0
Step 7: Determine convergence. Since the limit is less than 1, by the root
test, the series P∞
n=1 n!
nnconverges.
2
Question 3
Question
Prove or disprove the convergence of the series P∞
n=1 n2+1
n3+n.
Solution
To determine the convergence of the series P∞
n=1 n2+1
n3+n, we can use the limit com-
parison test. Specifically, we will compare it with the harmonic series P∞
n=1 1
n.
Step 1: Determine a comparison series Consider the series P∞
n=1 1
n.
Let’s simplify the given series to find a suitable comparison.
n2+ 1
n3+n=n2/n3+ 1/n3
n+ 1/n =1/n + 1/n3
1+1/n2
Step 2: Find the limit of the ratio Now, we calculate the limit as n
approaches infinity:
lim
n→∞
(1/n + 1/n3)
(1 + 1/n2)= lim
n→∞
1/n + 1/n3
1+1/n2
Step 3: Simplify the limit
lim
n→∞
1/n + 1/n3
1+1/n2=0+0
1+0 = 0
Step 4: Apply the limit comparison test Since the limit is finite and
positive, we can conclude that the given series P∞
n=1 n2+1
n3+nconverges by the
limit comparison test with P∞
n=1 1
n.
Therefore, the series P∞
n=1 n2+1
n3+nconverges.
Question 4
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
3
Step 2: Simplify the expression.
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Further simplify.
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 4: Analyze the limit. Since the limit is equal to 1, the ratio test is
inconclusive. We will try another test.
Step 5: Apply the root test. Let’s try the root test, computing:
lim
n→∞
n
p|an|= lim
n→∞
n
rn!
nn
Step 6: Simplify and analyze the limit.
= lim
n→∞
n
√n!
n
Since n!→ ∞ faster than nn→ ∞, the limit approaches 0.
Step 7: Conclusion. The root test shows that the series converges. There-
fore, the series P∞
n=1 n!
nnconverges.
Question 5
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Calculate the ratio R. Since the ratio test involves calculating
the limit of the absolute value of the ratio of consecutive terms, we have:
R= lim
n→∞
an+1
an
where an=n!
nn. Calculating an+1 and anseparately, we get:
an+1 =(n+ 1)!
(n+ 1)n+1 and an=n!
nn
4
Therefore, the ratio Rbecomes:
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio R. Simplify the ratio further as follows:
R= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
R= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
R= lim
n→∞
nn
(n+ 1)n
R= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit of R. Taking the limit as napproaches infinity,
we get:
R= lim
n→∞ n
n+ 1n
=1
e
Step 4: Determine the convergence or divergence of the series. If R < 1, the
series converges. Otherwise, it diverges. Since R=1
e<1, the series P∞
n=1 n!
nn
converges by the ratio test.
Question 6
Question
Determine whether the series
∞
X
n=1
n2+ 3n
n4+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
bn=1
n2.
Step 1: Find the limit of the ratio of the two series.
lim
n→∞ n2+3n
n4+1
1
n2= lim
n→∞
n4+ 3n3
n4+ 1 = 1
5
Step 2: State the Limit Comparison Test. If limn→∞
an
bnexists and is
a positive finite number, then either both series P∞
n=1 anand P∞
n=1 bnconverge
or both diverge.
Step 3: Conclusion Since limn→∞
an
bn= 1 (a positive finite number), by
the Limit Comparison Test, the series P∞
n=1 n2+3n
n4+1 has the same convergence
behavior as P∞
n=1 1
n2, which is a convergent p-series with p= 2. Therefore, the
given series converges.
Question 7
Question
Determine whether the series ∞
X
n=1
2n
3n+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
2n
3n+ 1, we can use the Ratio
Test.
Step 1: Find the limit. Let an=2n
3n+1 . We compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
2n+1
3n+1 + 1 ·3n+ 1
2n
= lim
n→∞
2
3 + 1
3n
=2
3.
Step 2: Apply the Ratio Test. - If limn→∞
an+1
an<1, then the se-
ries converges. - If limn→∞
an+1
an>1 or limn→∞
an+1
an= 1, then the series
diverges.
Since limn→∞
an+1
an=2
3<1, by the Ratio Test, the series ∞
X
n=1
2n
3n+ 1
converges.
Question 8
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
6
Solution
Let’s use the ratio test to determine the convergence of the series. Recall that
for a series P∞
n=1 an, if
lim
n→∞
an+1
an
=L,
then the series converges if L < 1, and diverges if L > 1.
Step 1: Find the ratio of consecutive terms We have an=n!
nn. Let’s
find an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=n+ 1
(n+ 1)n+1 ·n=n
(n+ 1)n
Step 2: Determine the limit Now, let’s compute the limit as napproaches
infinity:
lim
n→∞
n
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n= lim
n→∞
n
e=∞
Step 3: Conclusion Since the limit of an+1
anis infinity, the series diverges
by the ratio test. Therefore, the series
∞
X
n=1
n!
nn
diverges.
Question 9
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn, then we consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplifying the above expression by dividing the numerator by the denominator
gives:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!nn
(n+ 1)n+1n!
Step 2: Simplify the expression. Further simplifying the expression, we get:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
7
Step 3: Evaluate the limit. As ntends to infinity, the expression n
n+1 tends
to 1, and hence the limit simplifies to:
lim
n→∞ n
n+ 1n
= lim
n→∞ |1|= 1
Step 4: Apply the Ratio Test. Since the limit is equal to 1, we cannot make
any conclusion about convergence or divergence of the series using the Ratio
Test. We will need to use a different test.
In this case, we can use the Root Test to determine the convergence.
Therefore, the convergence of the series P∞
n=1 n!
nnis inconclusive by the Ratio
Test.
Question 10
Question
Determine whether the series ∞
X
n=1
n3
2n
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
Step 2: Simplify the expression and compute the limit:
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1+3/n + 3/n2+ 1/n3
2
=1
2
8
Step 3: Analyze the limit: Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n3
2n
converges.
Therefore, the series converges.
Question 11
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Using the
ratio test, we find:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Analyze the result. Since 1
e<1, the ratio test tells us that the
series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 12
Question
Determine whether the series
∞
X
n=1
n2+n+ 1
n2+ 2n+ 1
converges or diverges.
9
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series
bn=1
n
Step 1: Find the limit of the ratio of the given series and bnWe
will find the limit of the ratio an
bnas napproaches infinity:
lim
n→∞
n2+n+1
n2+2n+1
1
n
Simplify the expression:
lim
n→∞
n3+n2+n
n2+ 2n+ 1
lim
n→∞
n3+n2+n
n2+ 2n+ 1 = lim
n→∞
n3+n2+n
n2+ 2n+ 1 = lim
n→∞
1 + 1
n+1
n2
1 + 2
n+1
n2
= 1
Step 2: Determine the convergence based on the limit of the ratio
Since the limit is a positive finite number, we can conclude that the given series
has the same convergence behavior as the series P1
n, which is a divergent p-
series with p= 1. Therefore, by the limit comparison test, the given series
∞
X
n=1
n2+n+ 1
n2+ 2n+ 1
diverges.
Question 13
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Let’s evaluate the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression inside the limit:
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
10
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Taking the limit as napproaches infinity:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e
Step 4: Apply the ratio test: If the limit is less than 1, the series converges;
if the limit is greater than 1, the series diverges. Since the limit is 1
e<1, the
series P∞
n=1 n!
nnconverges by the ratio test.
Question 14
Question
Consider the series ∞
X
n=1
1
n3ln2n. Determine if the series converges or diverges.
Solution
To determine the convergence of the series, we will use the Integral Test. Let
f(x) = 1
x3ln2x.
Step 1: Find the integral We will calculate the integral of f(x) over the
interval [2,∞).
Z∞
2
1
x3ln2xdx
Step 2: Evaluate the integral Let u= ln x, then du =1
xdx. The integral
becomes Z1
u2du =−1
u=−1
ln x
∞
2
Applying the Fundamental Theorem of Calculus:
−lim
b→∞
1
ln b+1
ln 2
Step 3: Check for convergence Since the improper integral converges,
by the Integral Test, the series ∞
X
n=1
1
n3ln2nalso converges.
11
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3 .
Solution
To determine the convergence or divergence of the series P∞
n=1 n3+2n+1
n4+3 , we will
use the Limit Comparison Test.
Step 1: Let’s find the limit of the ratio of the given series term to the term
of a known convergent or divergent series.
Consider the series P∞
n=1 1
n, which is a known divergent p-series when p= 1.
Let an=n3+2n+1
n4+3 and bn=1
n. We want to find the limit of an
bnas napproaches
infinity.
lim
n→∞
an
bn
= lim
n→∞
n3+ 2n+ 1
n4+ 3 ·n
1= lim
n→∞
n4+ 2n2+n
n4+ 3 .
Dividing by the highest power of nin the denominator, we get:
= lim
n→∞
1+2/n2+ 1/n3
1+3/n4=1
1= 1.
Step 2: Now, we compare the limit of an
bnto a known series.
Since the limit is finite and non-zero, we apply the Limit Comparison Test.
Since the harmonic series diverges, and P∞
n=1 1
ndiverges, we conclude that
P∞
n=1 n3+2n+1
n4+3 also diverges.
Therefore, the series P∞
n=1 n3+2n+1
n4+3 diverges.
Question 16
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
12
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!nn
n!(n+ 1)n+1 =n+ 1
(1 + n)n=n+ 1
(1 + 1/n)n.
Step 2: Examine the limit of the ratio. Taking the limit as napproaches
infinity,
lim
n→∞
n+ 1
(1 + 1/n)n= lim
n→∞
1+1/n
(1 + 1/n)n= lim
n→∞
1
e=1
e,
where we have used the limit definition of the number e.
Step 3: Apply the ratio test. Since the limit of the ratio is 1
e<1, by the
ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 17
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine whether the series P∞
n=1 n!
nnconverges or diverges, we will use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit of
the ratio an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Evaluate the limit. Let L= limn→∞
nn
(n+1)n. Rewrite this limit in
a suitable form for further evaluation.
L= lim
n→∞
1
(1 + 1
n)n
13
Step 4: Find the limit using the fact that limn→∞(1 + 1
n)n=e(a standard
result from calculus).
L=1
e
Step 5: Analyze the ratio. Since L=1
e<1, the ratio test implies that the
series P∞
n=1 n!
nnis absolutely convergent.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 18
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2
4n.
Solution
To determine the convergence or divergence of the given series, we will first
rewrite it in a more convenient form.
Step 1: Rewrite the series
∞
X
n=1
3n+ 2
4n=∞
X
n=1 3n
4n+2
4n=∞
X
n=1 3
4n
+∞
X
n=1
2
4n.
Now, we have split the series into two separate series, both of which we need
to analyze for convergence.
Step 2: Analyze the first series For the first series,
∞
X
n=1 3
4n
,
we have a geometric series with a= 3/4 and r= 3/4. This series converges if
|r|<1.
Since |3/4|<1, the series P∞
n=1 3
4nconverges.
Step 3: Analyze the second series For the second series,
∞
X
n=1
2
4n,
we observe that this is also a geometric series with a= 2/4 and r= 1/4. This
series converges if |r|<1.
Since |1/4|<1, the series P∞
n=1 2
4nconverges.
Step 4: Conclusion Since both series P∞
n=1 3
4nand P∞
n=1 2
4nconverge,
the original series P∞
n=1 3n+2
4nalso converges.
14
Question 19
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the Ratio Test.
Step 1: Compute the ratio r: Let an=n!
nn. Then we have:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
Step 2: Simplify the expression: We know that limn→∞(1 + 1/n)n=e, so:
r= lim
n→∞
n
e=∞
Step 3: Analyze the limit: Since the limit of the ratio test is ∞, the series
∞
X
n=1
n!
nndiverges.
Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 20
Question
Let {an}be a sequence such that an>0 for all nand limn→∞ an= 0. Determine
the convergence or divergence of the series
∞
X
n=1
an
1 + a1+a2+··· +an
15
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Find a suitable series for comparison. Let’s consider the series
P∞
n=1
an
nfor comparison. Notice that an
1+a1+a2+···+an>an
nsince 1 + a1+a2+
··· +an> n for all n.
Step 2: Compute the limit of the ratio. Let bn=an
n. We will calculate
lim
n→∞
an/n
bn
= lim
n→∞
an
n·n
an
= lim
n→∞ 1=1.
Step 3: Apply the Limit Comparison Test. Since P∞
n=1 bnconverges if
and only if P∞
n=1 an/(1+a1+a2+···+an) converges, and we found the limit of
the ratio to be 1, by the Limit Comparison Test, the series P∞
n=1
an
1+a1+a2+···+an
converges if and only if P∞
n=1
an
nconverges.
Therefore, since P∞
n=1
an
nis a p-series with p= 1, it diverges. Thus, the
given series P∞
n=1
an
1+a1+a2+···+analso diverges.
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the given series, we will use the
ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nnbe the
16
general term of the series. Then, we compute:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e= 0
Step 2: Apply the ratio test. Since the limit of the ratio of consecutive
terms is 1
e= 0, the series P∞
n=1 n!
nndiverges by the ratio test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. We compute
the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!/(n+ 1)n(n+ 1)
n!
= lim
n→∞
1
(1 + 1/n)n
Step 3: Evaluate the limit: We recognize that the limit above is the defini-
tion of the constant e. Therefore,
lim
n→∞
an+1
an
=1
e
Step 4: Apply the ratio test: The series converges if the limit is less than
1 and diverges if the limit is greater than 1. Since 1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
17
Question 23
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify further:
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Apply the ratio test: Since 1
e<1, the series P∞
n=1 n!
nnconverges by
the ratio test.
Question 24
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Let’s apply the ratio test to the series. The ratio test states: If
limn→∞
an+1
an=L, then the series converges if L < 1 and diverges if L > 1.
18
Step 2: Calculate the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify the expression:
an+1
an
=(n+ 1)nn
(n+ 1)n+1 =n
n+ 1n
Step 4: Compute the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 5: Since the limit is 1
ewhich is less than 1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 25
Question
Determine whether the series P∞
n=1 2n2+3
n3+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test with the series P∞
n=1 1
n, which is a known series.
Step 1: Let’s find the limit of the ratio of the two series:
lim
n→∞
2n2+3
n3+1
1
n
= lim
n→∞
2n3+ 3n
n3+ 1 ·n
1= lim
n→∞
2 + 3
n2
1 + 1
n3
= 2
Step 2: Since the limit is a finite positive number (= 2), by the Limit Com-
parison Test, we conclude that P∞
n=1 2n2+3
n3+1 and P∞
n=1 1
neither both converge
or both diverge.
Step 3: Since the harmonic series P∞
n=1 1
ndiverges (as a p-series with
p= 1), we can conclude that the given series P∞
n=1 2n2+3
n3+1 also diverges by the
Limit Comparison Test.
Question 26
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
2n3+5 .
19
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s consider the series P∞
n=1 n2+1
2n3+5 and compare it with
the convergent series P∞
n=1 1
n.
Step 1: Find the limit Let an=n2+1
2n3+5 and bn=1
n. We want to find
limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
2n3+ 5 ·n
1= lim
n→∞
n3+n
2n3+ 5
Step 2: Simplify the expression
= lim
n→∞
n3+n
2n3+ 5 = lim
n→∞
1 + 1
n2
2 + 5
n3
=1
2
Since the limit is a positive finite value, we can apply the Limit Comparison
Test.
Step 3: Apply the Limit Comparison Test Since limn→∞
an
bnexists and
is a positive finite value, we can conclude that either both series Panand Pbn
converge or both diverge. Since Pbn=P1
nis a p-series with p= 1 >0, we
know that Pbnconverges. Therefore, by the Limit Comparison Test, the series
P∞
n=1 n2+1
2n3+5 also converges.
Question 27
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test: Let an=n!
nn. Compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
20
Step 3: Evaluate the limit:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Check the ratio: Since the limit of the ratio is 1
e<1, by the ratio
test, the series P∞
n=1 n!
nnconverges.
Question 28
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
following statement: If limn→∞ an= 0, then the series P∞
n=1 1
nanalso converges.
Solution
To prove or disprove the statement, we will provide a counterexample.
Step 1: Counterexample Consider the series defined by an=1
n2. It’s
clear that limn→∞ an= 0 since limn→∞
1
n2= 0. However, the series P∞
n=1 an=
P∞
n=1 1
n2converges (by the p-series test with p= 2).
Step 2: Series P∞
n=1 1
nanNow let’s consider the series P∞
n=1 1
nan=P∞
n=1 1
n·1
n2
=
P∞
n=1 n. This harmonic series P∞
n=1 ndiverges (since it’s a well-known result).
Step 3: Conclusion As seen in Step 1 and Step 2, we have found a
counterexample that disproves the statement. Therefore, the statement ”If
limn→∞ an= 0, then the series P∞
n=1 1
nanalso converges” is false.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
21
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We evaluate:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1
= lim
n→∞
(n+ 1) n
n+ 1n
= lim
n→∞
(n+ 1) 1
1 + 1
nn
= lim
n→∞(n+ 1) ·lim
n→∞ 1
1 + 1
nn
=∞ · 1
e
=∞.
Step 2: Analyze the limit. Since the limit is greater than 1, the series
P∞
n=1 n!
nndiverges by the Ratio Test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 30
Question
Determine whether the series
∞
X
n=1
n
n3+√n+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Comparison Test.
Step 1: We will find a series that converges and is larger than the given
series. Notice that for all n≥1, we have
n
n3+√n+ 1 ≤n
n3=1
n2
Therefore, we will compare our series to P∞
n=1 1
n2.
Step 2: We will show that the series P∞
n=1 1
n2converges. This is a p-series
with p= 2 >1, so by the p-series test, P∞
n=1 1
n2converges.
Step 3: Using the Comparison Test, since n
n3+√n+1 ≤1
n2and P∞
n=1 1
n2
converges, then by the Comparison Test, P∞
n=1 n
n3+√n+1 also converges.
Therefore, the given series converges by the Comparison Test.
22
Question 31
Question
Let {an}be a sequence such that limn→∞ an= 0 and P∞
n=1 anconverges, but
P∞
n=1 |an|2diverges. Prove that P∞
n=1 anconverges absolutely.
Solution
Step 1: Since limn→∞ an= 0, there exists some N∈Nsuch that |an|<1 for
all n≥N. Therefore, for n≥N, we have |an|≤|an|2.
Step 2: Consider the series P∞
n=1 an. Since P∞
n=1 anconverges, the sequence
{an}must be bounded. Thus, there exists some M > 0 such that |an| ≤ Mfor
all n∈N.
Step 3: For n<N, we have |an| ≤ Mand for n≥N, we have |an|≤|an|2≤
M. Therefore, |an| ≤ Mfor all n∈N.
Step 4: Now, we have |an| ≤ Mfor all n∈N. This implies that the series
P∞
n=1 |an|converges by the Comparison Test.
Step 5: Thus, since P∞
n=1 |an|converges, the series P∞
n=1 anconverges ab-
solutely.
Question 32
Question
Determine the convergence of the series ∞
X
n=1
2n3+ 3n2
n4+ 5 .
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=2n3+ 3n2
n4+ 5 . We will compare this series with a known convergent
or divergent series.
Step 1: Calculate the limit
lim
n→∞
an
1
n2
= lim
n→∞
2n3+3n2
n4+5
1
n2
= lim
n→∞
2n+ 3
n2+5
n2
= lim
n→∞
2
n= 0
Step 2: Interpret the limit Since the limit is finite and positive, and
∞
X
n=1
1
n2converges (this is a p-series with p= 2 >1), by the Limit Comparison
23
Test, the given series ∞
X
n=1
2n3+ 3n2
n4+ 5 also converges.
Question 33
Question
Determine the convergence or divergence of the series P∞
n=1
(−1)n
n+ln(n).
Solution
To determine the convergence or divergence of the series, we will use the Alter-
nating Series Test (AST).
Step 1: Determine if the series satisfies the conditions of the AST
The series P∞
n=1
(−1)n
n+ln(n)is an alternating series in the form (−1)nbn, where
bn=1
n+ln(n).
We need to check the following two conditions for the AST to apply: 1. The
sequence bn=1
n+ln(n)is positive, decreasing, and approaches 0 as napproaches
infinity.
2. The terms of the series are decreasing, i.e., bn+1 ≤bnfor all n.
Step 2: Check if the sequence satisfies the conditions of the AST
Let’s evaluate the limit as napproaches infinity of bn=1
n+ln(n)to verify the
first condition:
lim
n→∞
1
n+ ln(n)= lim
n→∞
1/n
1+1/n = 0
So, the sequence bn=1
n+ln(n)is positive, decreasing, and approaches 0 as n
approaches infinity.
To check the second condition, we calculate bn+1
bn:
bn+1
bn
=1
n+ 1 + ln(n+ 1) ·n+ ln(n)
1=n+ ln(n)
n+ 1 + ln(n+ 1)
Since n+ ln(n)> n ≥0 and n+ 1 + ln(n+ 1) > n + 1 ≥1, we have bn+1
bn<1
for all n. Therefore, the terms of the series are decreasing.
Step 3: Apply the Alternating Series Test Since the series satisfies the
conditions of the AST, we conclude that the series P∞
n=1
(−1)n
n+ln(n)converges.
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n−1
n3+ 2
24
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s compare the given series with the series P∞
n=1 1
n.
Step 1: Find the limit of the ratio Let an=n2+3n−1
n3+2 and bn=1
n. We
will consider the limit
lim
n→∞
an
bn
= lim
n→∞
n2+3n−1
n3+2
1
n
Step 2: Simplify the limit Simplify the limit to get
lim
n→∞
n3+ 3n2−n
n3+ 2 = lim
n→∞
n(n2+ 3n−1)
n3+ 2 = lim
n→∞
n3+ 3n2−n
n3+ 2 = 1
Step 3: Apply the Limit Comparison Test Since the limit is a positive
finite number, by the Limit Comparison Test, we conclude that the series
∞
X
n=1
n2+ 3n−1
n3+ 2
and ∞
X
n=1
1
n
either both converge or both diverge.
Step 4: Conclusion The series P∞
n=1 1
nis a harmonic series which diverges.
Therefore, by the Limit Comparison Test, the series
∞
X
n=1
n2+ 3n−1
n3+ 2
also diverges.
Question 35
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test. Let an=n!
nn. Then, we consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
25
Question 2
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
Step 1: Apply the ratio test. Let’s consider the limit of the ratio of consec-
utive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
Step 3: Analyze the limit.
lim
n→∞
n
n+ 1
= 1
Step 4: Determine convergence. Since the limit is equal to 1, the ratio test
is inconclusive. Therefore, we need to try a different test.
Step 5: Apply the root test. Let’s consider the root test, which involves
calculating:
lim
n→∞
n
p|an|= lim
n→∞
n
s
n!
nn
Step 6: Simplify and calculate the limit.
lim
n→∞
n
s
n!
nn
= lim
n→∞
n1/n
n
This limit simplifies to:
lim
n→∞
1
n= 0
Step 7: Determine convergence. Since the limit is less than 1, by the root
test, the series P∞
n=1 n!
nnconverges.
2
Question 3
Question
Prove or disprove the convergence of the series P∞
n=1 n2+1
n3+n.
Solution
To determine the convergence of the series P∞
n=1 n2+1
n3+n, we can use the limit com-
parison test. Specifically, we will compare it with the harmonic series P∞
n=1 1
n.
Step 1: Determine a comparison series Consider the series P∞
n=1 1
n.
Let’s simplify the given series to find a suitable comparison.
n2+ 1
n3+n=n2/n3+ 1/n3
n+ 1/n =1/n + 1/n3
1+1/n2
Step 2: Find the limit of the ratio Now, we calculate the limit as n
approaches infinity:
lim
n→∞
(1/n + 1/n3)
(1 + 1/n2)= lim
n→∞
1/n + 1/n3
1+1/n2
Step 3: Simplify the limit
lim
n→∞
1/n + 1/n3
1+1/n2=0+0
1+0 = 0
Step 4: Apply the limit comparison test Since the limit is finite and
positive, we can conclude that the given series P∞
n=1 n2+1
n3+nconverges by the
limit comparison test with P∞
n=1 1
n.
Therefore, the series P∞
n=1 n2+1
n3+nconverges.
Question 4
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
3
Step 2: Simplify the expression.
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Further simplify.
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 4: Analyze the limit. Since the limit is equal to 1, the ratio test is
inconclusive. We will try another test.
Step 5: Apply the root test. Let’s try the root test, computing:
lim
n→∞
n
p|an|= lim
n→∞
n
rn!
nn
Step 6: Simplify and analyze the limit.
= lim
n→∞
n
√n!
n
Since n!→ ∞ faster than nn→ ∞, the limit approaches 0.
Step 7: Conclusion. The root test shows that the series converges. There-
fore, the series P∞
n=1 n!
nnconverges.
Question 5
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Calculate the ratio R. Since the ratio test involves calculating
the limit of the absolute value of the ratio of consecutive terms, we have:
R= lim
n→∞
an+1
an
where an=n!
nn. Calculating an+1 and anseparately, we get:
an+1 =(n+ 1)!
(n+ 1)n+1 and an=n!
nn
4
Therefore, the ratio Rbecomes:
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio R. Simplify the ratio further as follows:
R= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
R= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
R= lim
n→∞
nn
(n+ 1)n
R= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit of R. Taking the limit as napproaches infinity,
we get:
R= lim
n→∞ n
n+ 1n
=1
e
Step 4: Determine the convergence or divergence of the series. If R < 1, the
series converges. Otherwise, it diverges. Since R=1
e<1, the series P∞
n=1 n!
nn
converges by the ratio test.
Question 6
Question
Determine whether the series
∞
X
n=1
n2+ 3n
n4+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
bn=1
n2.
Step 1: Find the limit of the ratio of the two series.
lim
n→∞ n2+3n
n4+1
1
n2= lim
n→∞
n4+ 3n3
n4+ 1 = 1
5
Step 2: State the Limit Comparison Test. If limn→∞
an
bnexists and is
a positive finite number, then either both series P∞
n=1 anand P∞
n=1 bnconverge
or both diverge.
Step 3: Conclusion Since limn→∞
an
bn= 1 (a positive finite number), by
the Limit Comparison Test, the series P∞
n=1 n2+3n
n4+1 has the same convergence
behavior as P∞
n=1 1
n2, which is a convergent p-series with p= 2. Therefore, the
given series converges.
Question 7
Question
Determine whether the series ∞
X
n=1
2n
3n+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
2n
3n+ 1, we can use the Ratio
Test.
Step 1: Find the limit. Let an=2n
3n+1 . We compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
2n+1
3n+1 + 1 ·3n+ 1
2n
= lim
n→∞
2
3 + 1
3n
=2
3.
Step 2: Apply the Ratio Test. - If limn→∞
an+1
an<1, then the se-
ries converges. - If limn→∞
an+1
an>1 or limn→∞
an+1
an= 1, then the series
diverges.
Since limn→∞
an+1
an=2
3<1, by the Ratio Test, the series ∞
X
n=1
2n
3n+ 1
converges.
Question 8
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
6
Solution
Let’s use the ratio test to determine the convergence of the series. Recall that
for a series P∞
n=1 an, if
lim
n→∞
an+1
an
=L,
then the series converges if L < 1, and diverges if L > 1.
Step 1: Find the ratio of consecutive terms We have an=n!
nn. Let’s
find an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=n+ 1
(n+ 1)n+1 ·n=n
(n+ 1)n
Step 2: Determine the limit Now, let’s compute the limit as napproaches
infinity:
lim
n→∞
n
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n= lim
n→∞
n
e=∞
Step 3: Conclusion Since the limit of an+1
anis infinity, the series diverges
by the ratio test. Therefore, the series
∞
X
n=1
n!
nn
diverges.
Question 9
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn, then we consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplifying the above expression by dividing the numerator by the denominator
gives:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!nn
(n+ 1)n+1n!
Step 2: Simplify the expression. Further simplifying the expression, we get:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
7
Step 3: Evaluate the limit. As ntends to infinity, the expression n
n+1 tends
to 1, and hence the limit simplifies to:
lim
n→∞ n
n+ 1n
= lim
n→∞ |1|= 1
Step 4: Apply the Ratio Test. Since the limit is equal to 1, we cannot make
any conclusion about convergence or divergence of the series using the Ratio
Test. We will need to use a different test.
In this case, we can use the Root Test to determine the convergence.
Therefore, the convergence of the series P∞
n=1 n!
nnis inconclusive by the Ratio
Test.
Question 10
Question
Determine whether the series ∞
X
n=1
n3
2n
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
Step 2: Simplify the expression and compute the limit:
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1+3/n + 3/n2+ 1/n3
2
=1
2
8
Step 3: Analyze the limit: Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n3
2n
converges.
Therefore, the series converges.
Question 11
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Using the
ratio test, we find:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Analyze the result. Since 1
e<1, the ratio test tells us that the
series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 12
Question
Determine whether the series
∞
X
n=1
n2+n+ 1
n2+ 2n+ 1
converges or diverges.
9
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series
bn=1
n
Step 1: Find the limit of the ratio of the given series and bnWe
will find the limit of the ratio an
bnas napproaches infinity:
lim
n→∞
n2+n+1
n2+2n+1
1
n
Simplify the expression:
lim
n→∞
n3+n2+n
n2+ 2n+ 1
lim
n→∞
n3+n2+n
n2+ 2n+ 1 = lim
n→∞
n3+n2+n
n2+ 2n+ 1 = lim
n→∞
1 + 1
n+1
n2
1 + 2
n+1
n2
= 1
Step 2: Determine the convergence based on the limit of the ratio
Since the limit is a positive finite number, we can conclude that the given series
has the same convergence behavior as the series P1
n, which is a divergent p-
series with p= 1. Therefore, by the limit comparison test, the given series
∞
X
n=1
n2+n+ 1
n2+ 2n+ 1
diverges.
Question 13
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Let’s evaluate the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression inside the limit:
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
10
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Taking the limit as napproaches infinity:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e
Step 4: Apply the ratio test: If the limit is less than 1, the series converges;
if the limit is greater than 1, the series diverges. Since the limit is 1
e<1, the
series P∞
n=1 n!
nnconverges by the ratio test.
Question 14
Question
Consider the series ∞
X
n=1
1
n3ln2n. Determine if the series converges or diverges.
Solution
To determine the convergence of the series, we will use the Integral Test. Let
f(x) = 1
x3ln2x.
Step 1: Find the integral We will calculate the integral of f(x) over the
interval [2,∞).
Z∞
2
1
x3ln2xdx
Step 2: Evaluate the integral Let u= ln x, then du =1
xdx. The integral
becomes Z1
u2du =−1
u=−1
ln x
∞
2
Applying the Fundamental Theorem of Calculus:
−lim
b→∞
1
ln b+1
ln 2
Step 3: Check for convergence Since the improper integral converges,
by the Integral Test, the series ∞
X
n=1
1
n3ln2nalso converges.
11
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3 .
Solution
To determine the convergence or divergence of the series P∞
n=1 n3+2n+1
n4+3 , we will
use the Limit Comparison Test.
Step 1: Let’s find the limit of the ratio of the given series term to the term
of a known convergent or divergent series.
Consider the series P∞
n=1 1
n, which is a known divergent p-series when p= 1.
Let an=n3+2n+1
n4+3 and bn=1
n. We want to find the limit of an
bnas napproaches
infinity.
lim
n→∞
an
bn
= lim
n→∞
n3+ 2n+ 1
n4+ 3 ·n
1= lim
n→∞
n4+ 2n2+n
n4+ 3 .
Dividing by the highest power of nin the denominator, we get:
= lim
n→∞
1+2/n2+ 1/n3
1+3/n4=1
1= 1.
Step 2: Now, we compare the limit of an
bnto a known series.
Since the limit is finite and non-zero, we apply the Limit Comparison Test.
Since the harmonic series diverges, and P∞
n=1 1
ndiverges, we conclude that
P∞
n=1 n3+2n+1
n4+3 also diverges.
Therefore, the series P∞
n=1 n3+2n+1
n4+3 diverges.
Question 16
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
12
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!nn
n!(n+ 1)n+1 =n+ 1
(1 + n)n=n+ 1
(1 + 1/n)n.
Step 2: Examine the limit of the ratio. Taking the limit as napproaches
infinity,
lim
n→∞
n+ 1
(1 + 1/n)n= lim
n→∞
1+1/n
(1 + 1/n)n= lim
n→∞
1
e=1
e,
where we have used the limit definition of the number e.
Step 3: Apply the ratio test. Since the limit of the ratio is 1
e<1, by the
ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 17
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine whether the series P∞
n=1 n!
nnconverges or diverges, we will use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit of
the ratio an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Evaluate the limit. Let L= limn→∞
nn
(n+1)n. Rewrite this limit in
a suitable form for further evaluation.
L= lim
n→∞
1
(1 + 1
n)n
13
Step 4: Find the limit using the fact that limn→∞(1 + 1
n)n=e(a standard
result from calculus).
L=1
e
Step 5: Analyze the ratio. Since L=1
e<1, the ratio test implies that the
series P∞
n=1 n!
nnis absolutely convergent.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 18
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2
4n.
Solution
To determine the convergence or divergence of the given series, we will first
rewrite it in a more convenient form.
Step 1: Rewrite the series
∞
X
n=1
3n+ 2
4n=∞
X
n=1 3n
4n+2
4n=∞
X
n=1 3
4n
+∞
X
n=1
2
4n.
Now, we have split the series into two separate series, both of which we need
to analyze for convergence.
Step 2: Analyze the first series For the first series,
∞
X
n=1 3
4n
,
we have a geometric series with a= 3/4 and r= 3/4. This series converges if
|r|<1.
Since |3/4|<1, the series P∞
n=1 3
4nconverges.
Step 3: Analyze the second series For the second series,
∞
X
n=1
2
4n,
we observe that this is also a geometric series with a= 2/4 and r= 1/4. This
series converges if |r|<1.
Since |1/4|<1, the series P∞
n=1 2
4nconverges.
Step 4: Conclusion Since both series P∞
n=1 3
4nand P∞
n=1 2
4nconverge,
the original series P∞
n=1 3n+2
4nalso converges.
14
Question 19
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the Ratio Test.
Step 1: Compute the ratio r: Let an=n!
nn. Then we have:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
Step 2: Simplify the expression: We know that limn→∞(1 + 1/n)n=e, so:
r= lim
n→∞
n
e=∞
Step 3: Analyze the limit: Since the limit of the ratio test is ∞, the series
∞
X
n=1
n!
nndiverges.
Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 20
Question
Let {an}be a sequence such that an>0 for all nand limn→∞ an= 0. Determine
the convergence or divergence of the series
∞
X
n=1
an
1 + a1+a2+··· +an
15
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Find a suitable series for comparison. Let’s consider the series
P∞
n=1
an
nfor comparison. Notice that an
1+a1+a2+···+an>an
nsince 1 + a1+a2+
··· +an> n for all n.
Step 2: Compute the limit of the ratio. Let bn=an
n. We will calculate
lim
n→∞
an/n
bn
= lim
n→∞
an
n·n
an
= lim
n→∞ 1=1.
Step 3: Apply the Limit Comparison Test. Since P∞
n=1 bnconverges if
and only if P∞
n=1 an/(1+a1+a2+···+an) converges, and we found the limit of
the ratio to be 1, by the Limit Comparison Test, the series P∞
n=1
an
1+a1+a2+···+an
converges if and only if P∞
n=1
an
nconverges.
Therefore, since P∞
n=1
an
nis a p-series with p= 1, it diverges. Thus, the
given series P∞
n=1
an
1+a1+a2+···+analso diverges.
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the given series, we will use the
ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nnbe the
16
general term of the series. Then, we compute:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e= 0
Step 2: Apply the ratio test. Since the limit of the ratio of consecutive
terms is 1
e= 0, the series P∞
n=1 n!
nndiverges by the ratio test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. We compute
the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)n!/(n+ 1)n(n+ 1)
n!
= lim
n→∞
1
(1 + 1/n)n
Step 3: Evaluate the limit: We recognize that the limit above is the defini-
tion of the constant e. Therefore,
lim
n→∞
an+1
an
=1
e
Step 4: Apply the ratio test: The series converges if the limit is less than
1 and diverges if the limit is greater than 1. Since 1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
17
Question 23
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify further:
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Apply the ratio test: Since 1
e<1, the series P∞
n=1 n!
nnconverges by
the ratio test.
Question 24
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Let’s apply the ratio test to the series. The ratio test states: If
limn→∞
an+1
an=L, then the series converges if L < 1 and diverges if L > 1.
18
Step 2: Calculate the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify the expression:
an+1
an
=(n+ 1)nn
(n+ 1)n+1 =n
n+ 1n
Step 4: Compute the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 5: Since the limit is 1
ewhich is less than 1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 25
Question
Determine whether the series P∞
n=1 2n2+3
n3+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test with the series P∞
n=1 1
n, which is a known series.
Step 1: Let’s find the limit of the ratio of the two series:
lim
n→∞
2n2+3
n3+1
1
n
= lim
n→∞
2n3+ 3n
n3+ 1 ·n
1= lim
n→∞
2 + 3
n2
1 + 1
n3
= 2
Step 2: Since the limit is a finite positive number (= 2), by the Limit Com-
parison Test, we conclude that P∞
n=1 2n2+3
n3+1 and P∞
n=1 1
neither both converge
or both diverge.
Step 3: Since the harmonic series P∞
n=1 1
ndiverges (as a p-series with
p= 1), we can conclude that the given series P∞
n=1 2n2+3
n3+1 also diverges by the
Limit Comparison Test.
Question 26
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
2n3+5 .
19
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s consider the series P∞
n=1 n2+1
2n3+5 and compare it with
the convergent series P∞
n=1 1
n.
Step 1: Find the limit Let an=n2+1
2n3+5 and bn=1
n. We want to find
limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
2n3+ 5 ·n
1= lim
n→∞
n3+n
2n3+ 5
Step 2: Simplify the expression
= lim
n→∞
n3+n
2n3+ 5 = lim
n→∞
1 + 1
n2
2 + 5
n3
=1
2
Since the limit is a positive finite value, we can apply the Limit Comparison
Test.
Step 3: Apply the Limit Comparison Test Since limn→∞
an
bnexists and
is a positive finite value, we can conclude that either both series Panand Pbn
converge or both diverge. Since Pbn=P1
nis a p-series with p= 1 >0, we
know that Pbnconverges. Therefore, by the Limit Comparison Test, the series
P∞
n=1 n2+1
2n3+5 also converges.
Question 27
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test: Let an=n!
nn. Compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
20
Step 3: Evaluate the limit:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Check the ratio: Since the limit of the ratio is 1
e<1, by the ratio
test, the series P∞
n=1 n!
nnconverges.
Question 28
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
following statement: If limn→∞ an= 0, then the series P∞
n=1 1
nanalso converges.
Solution
To prove or disprove the statement, we will provide a counterexample.
Step 1: Counterexample Consider the series defined by an=1
n2. It’s
clear that limn→∞ an= 0 since limn→∞
1
n2= 0. However, the series P∞
n=1 an=
P∞
n=1 1
n2converges (by the p-series test with p= 2).
Step 2: Series P∞
n=1 1
nanNow let’s consider the series P∞
n=1 1
nan=P∞
n=1 1
n·1
n2
=
P∞
n=1 n. This harmonic series P∞
n=1 ndiverges (since it’s a well-known result).
Step 3: Conclusion As seen in Step 1 and Step 2, we have found a
counterexample that disproves the statement. Therefore, the statement ”If
limn→∞ an= 0, then the series P∞
n=1 1
nanalso converges” is false.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To analyze the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
21
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We evaluate:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1
= lim
n→∞
(n+ 1) n
n+ 1n
= lim
n→∞
(n+ 1) 1
1 + 1
nn
= lim
n→∞(n+ 1) ·lim
n→∞ 1
1 + 1
nn
=∞ · 1
e
=∞.
Step 2: Analyze the limit. Since the limit is greater than 1, the series
P∞
n=1 n!
nndiverges by the Ratio Test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 30
Question
Determine whether the series
∞
X
n=1
n
n3+√n+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Comparison Test.
Step 1: We will find a series that converges and is larger than the given
series. Notice that for all n≥1, we have
n
n3+√n+ 1 ≤n
n3=1
n2
Therefore, we will compare our series to P∞
n=1 1
n2.
Step 2: We will show that the series P∞
n=1 1
n2converges. This is a p-series
with p= 2 >1, so by the p-series test, P∞
n=1 1
n2converges.
Step 3: Using the Comparison Test, since n
n3+√n+1 ≤1
n2and P∞
n=1 1
n2
converges, then by the Comparison Test, P∞
n=1 n
n3+√n+1 also converges.
Therefore, the given series converges by the Comparison Test.
22
Question 31
Question
Let {an}be a sequence such that limn→∞ an= 0 and P∞
n=1 anconverges, but
P∞
n=1 |an|2diverges. Prove that P∞
n=1 anconverges absolutely.
Solution
Step 1: Since limn→∞ an= 0, there exists some N∈Nsuch that |an|<1 for
all n≥N. Therefore, for n≥N, we have |an|≤|an|2.
Step 2: Consider the series P∞
n=1 an. Since P∞
n=1 anconverges, the sequence
{an}must be bounded. Thus, there exists some M > 0 such that |an| ≤ Mfor
all n∈N.
Step 3: For n<N, we have |an| ≤ Mand for n≥N, we have |an|≤|an|2≤
M. Therefore, |an| ≤ Mfor all n∈N.
Step 4: Now, we have |an| ≤ Mfor all n∈N. This implies that the series
P∞
n=1 |an|converges by the Comparison Test.
Step 5: Thus, since P∞
n=1 |an|converges, the series P∞
n=1 anconverges ab-
solutely.
Question 32
Question
Determine the convergence of the series ∞
X
n=1
2n3+ 3n2
n4+ 5 .
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=2n3+ 3n2
n4+ 5 . We will compare this series with a known convergent
or divergent series.
Step 1: Calculate the limit
lim
n→∞
an
1
n2
= lim
n→∞
2n3+3n2
n4+5
1
n2
= lim
n→∞
2n+ 3
n2+5
n2
= lim
n→∞
2
n= 0
Step 2: Interpret the limit Since the limit is finite and positive, and
∞
X
n=1
1
n2converges (this is a p-series with p= 2 >1), by the Limit Comparison
23
Test, the given series ∞
X
n=1
2n3+ 3n2
n4+ 5 also converges.
Question 33
Question
Determine the convergence or divergence of the series P∞
n=1
(−1)n
n+ln(n).
Solution
To determine the convergence or divergence of the series, we will use the Alter-
nating Series Test (AST).
Step 1: Determine if the series satisfies the conditions of the AST
The series P∞
n=1
(−1)n
n+ln(n)is an alternating series in the form (−1)nbn, where
bn=1
n+ln(n).
We need to check the following two conditions for the AST to apply: 1. The
sequence bn=1
n+ln(n)is positive, decreasing, and approaches 0 as napproaches
infinity.
2. The terms of the series are decreasing, i.e., bn+1 ≤bnfor all n.
Step 2: Check if the sequence satisfies the conditions of the AST
Let’s evaluate the limit as napproaches infinity of bn=1
n+ln(n)to verify the
first condition:
lim
n→∞
1
n+ ln(n)= lim
n→∞
1/n
1+1/n = 0
So, the sequence bn=1
n+ln(n)is positive, decreasing, and approaches 0 as n
approaches infinity.
To check the second condition, we calculate bn+1
bn:
bn+1
bn
=1
n+ 1 + ln(n+ 1) ·n+ ln(n)
1=n+ ln(n)
n+ 1 + ln(n+ 1)
Since n+ ln(n)> n ≥0 and n+ 1 + ln(n+ 1) > n + 1 ≥1, we have bn+1
bn<1
for all n. Therefore, the terms of the series are decreasing.
Step 3: Apply the Alternating Series Test Since the series satisfies the
conditions of the AST, we conclude that the series P∞
n=1
(−1)n
n+ln(n)converges.
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n−1
n3+ 2
24
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s compare the given series with the series P∞
n=1 1
n.
Step 1: Find the limit of the ratio Let an=n2+3n−1
n3+2 and bn=1
n. We
will consider the limit
lim
n→∞
an
bn
= lim
n→∞
n2+3n−1
n3+2
1
n
Step 2: Simplify the limit Simplify the limit to get
lim
n→∞
n3+ 3n2−n
n3+ 2 = lim
n→∞
n(n2+ 3n−1)
n3+ 2 = lim
n→∞
n3+ 3n2−n
n3+ 2 = 1
Step 3: Apply the Limit Comparison Test Since the limit is a positive
finite number, by the Limit Comparison Test, we conclude that the series
∞
X
n=1
n2+ 3n−1
n3+ 2
and ∞
X
n=1
1
n
either both converge or both diverge.
Step 4: Conclusion The series P∞
n=1 1
nis a harmonic series which diverges.
Therefore, by the Limit Comparison Test, the series
∞
X
n=1
n2+ 3n−1
n3+ 2
also diverges.
Question 35
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test. Let an=n!
nn. Then, we consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
25
Step 2: Simplify the ratio.
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
nn
(n+ 1)n
Step 3: Find the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1
n)n
=1
e
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
26
