MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 8
Liberty University
Question 1
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: Apply the ratio test. Let an=n!
nn. We will compute the limit of
the ratio an+1
anas napproaches infinity.
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)!nn
n!(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 2: Simplify the ratio.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Calculate the limit. Using the limit definition of e, we have:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Determine convergence. Since 1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 2
Question
Determine whether the series ∞
X
n=1
n2+ 1
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 1
3n3+ 2, we will use the Limit
Comparison Test.
Step 1: Let’s find a series that we can easily determine the convergence. We
will use the series ∞
X
n=1
1
n, which is the harmonic series and is known to diverge.
Step 2: Compute the limit of the ratio of the two series:
lim
n→∞
n2+1
3n3+2
1
n
= lim
n→∞
n3+n
3n3+ 2 = lim
n→∞
1 + 1
n2
3 + 2
n3
=1
3
Step 3: Since the limit is a finite nonzero value, the given series ∞
X
n=1
n2+ 1
3n3+ 2
converges or diverges exactly like the series ∞
X
n=1
1
n. Therefore, the series ∞
X
n=1
n2+ 1
3n3+ 2
diverges.
Question 3
Question
Determine whether the series ∞
X
n=1
n2+ 3n+ 1
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test. Let an=n2+3n+1
3n3+2 .
2
Step 1: Find the limit of an
1
n2
.
lim
n→∞
an
1
n2
= lim
n→∞
n2+ 3n+ 1
3n3+ 2 ·n2
Step 2: Simplify the expression in the limit.
lim
n→∞
n2+ 3n+ 1
3n3+ 2 ·n2= lim
n→∞
1 + 3
n+1
n2
3 + 2
n3·n2= lim
n→∞
1
3n2= 0
Since the limit is finite and positive, we conclude that the series ∞
X
n=1
n2+ 3n+ 1
3n3+ 2
converges by the Limit Comparison Test with ∞
X
n=1
1
n2.
Question 4
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: Calculate the ratio Rof consecutive terms. Let an=n!
nn. Then,
the ratio of consecutive terms is given by:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1 = lim
n→∞
n+ 1
(1 + 1
n)n
Step 3: Evaluate the limit.
lim
n→∞
n+ 1
(1 + 1
n)n= lim
n→∞
n+ 1
1 + 1
nn= lim
n→∞
n+ 1
n+1
nn= lim
n→∞
n+ 1
1=∞
Step 4: Determine the convergence of the series. Since the limit of the ratio
Ris infinity, the series P∞
n=1
n!
nndiverges by the ratio test.
Question 5
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
3
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Compute the ratio R. The ratio test states that if limn→∞
an+1
an
=
R, then the series converges if R < 1 and diverges if R > 1.
Compute the ratio R:
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
n+ 1
n
= lim
n→∞ 1−1
n+ 1n
=1
e.
Step 2: Analyze the ratio R. Since R=1
e<1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2n
5n.
Solution
To determine the convergence or divergence of the series P∞
n=1 3n+2n
5n, we will
use the Ratio Test.
4
Step 1: Compute the ratio an+1
an
.
an+1
an
=
3n+1+2n+1
5n+1
3n+2n
5n
=
3·3n+2·2n
5n+1
3n+2n
5n
=3·3n
5n+ 2 ·2n
5n
3n
5n+2n
5n
=3(3
5)n+ 2(2
5)n
(3
5)n+ (2
5)n
=3(3
5)n
(3
5)n+2(2
5)n
(3
5)n
= 3 + 2
32
3n
.
Step 2: Find the limit of the ratio as napproaches infinity.
lim
n→∞
an+1
an
= lim
n→∞ 3 + 2
32
3n
= 3 + 2
3lim
n→∞ 2
3n
= 3 + 0
= 3.
Step 3: Apply the Ratio Test. Since the limit of the ratio is 3, which is
greater than 1, the series P∞
n=1 3n+2n
5ndiverges.
Therefore, the given series P∞
n=1 3n+2n
5nis divergent.
Question 7
Question
Determine the convergence or divergence of the series:
∞
X
n=1
3n+ 4n
5n
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
5
Step 1: Apply the ratio test. Let an=3n+4n
5n, then
lim
n→∞
an+1
an
= lim
n→∞
3n+1+4n+1
5n+1
3n+4n
5n
= lim
n→∞
3·3n+4·4n
5·5n
3n+4n
5n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n+ 4n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n+ 4n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n1 + 4
3n
=4
3<1
Step 2: Analyze the limit. Since the limit of the absolute value of the ratio
is less than 1, by the ratio test, the series P∞
n=1 3n+4n
5nconverges.
Therefore, the original series converges.
Question 8
Question
Determine the convergence or divergence of the series ∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series ∞
X
n=1
n!
nn, we will use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute 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 2: Simplify the ratio.
an+1
an
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!=nn
(n+ 1)n=n
n+ 1n
6
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=e−1
Step 4: Analyze the limit. Since the ratio limn→∞ n
n+ 1n
=e−1<1,
by the ratio test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnis convergent.
Question 9
Question
Determine whether the series P∞
n=1
n3
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n3
2n, we will use the ratio test.
Step 1: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1+3/n + 3/n2+ 1/n3
2
=1
2.
Step 3: Apply the ratio test: Since the limit is 1
2, by the ratio test, if the
limit is less than 1, then the series converges. Therefore, the series P∞
n=1
n3
2n
converges.
Question 10
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
7
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Compute the limit of the ratio test. Let an=n!
nn. We consider the
limit
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
(1 + 1
n)n
= lim
n→∞
n+ 1
e
Step 2: Analyze the limit. Since the limit is ∞
e, which is greater than 1,
the series diverges by the ratio test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 11
Question
Determine whether the series P∞
n=1
(−1)n
√nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
(−1)n
√n, we can use the Alter-
nating Series Test.
Step 1: Check for the conditions of the Alternating Series Test. Let an=
1
√n. Notice that anis positive and decreasing for all n.
Step 2: Verify the limit condition of the Alternating Series Test. Calculate
limn→∞
1
√n= 0. Since the limit is 0, we have satisfied the second condition of
the Alternating Series Test.
Step 3: Apply the Alternating Series Test. Since anis positive, decreasing,
and the limit of anas napproaches infinity is 0, the series P∞
n=1
(−1)n
√nconverges
by the Alternating Series Test.
Therefore, the series P∞
n=1
(−1)n
√nconverges.
Question 12
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we will use the Ratio Test.
8
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Then, the
ratio of consecutive terms is:
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→∞
nn
(n+ 1)n
Step 3: Determine the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Apply the Ratio Test. Since the limit is less than 1, by the Ratio
Test, the series P∞
n=1
n!
nnconverges.
Therefore, the series converges.
Question 13
Question
Determine whether the series P∞
n=1 1
n2ln nconverges or diverges.
Solution
To determine the convergence of the series, we will use the Integral Test. Let
f(x) = 1
x2ln x, where x≥2.
Step 1: Find the integral Compute the indefinite integral of f(x):
Z1
x2ln xdx
Step 2: Perform u-substitution Let u= ln x, then du =1
xdx. The
integral becomes:
Z1
u2du =−1
u+C=−1
ln x+C
Step 3: Apply the Integral Test Since the integral R∞
2
1
x2ln xdx simplifies
to [−1
ln x]∞
2, we need to determine if this improper integral converges:
lim
b→∞ −1
ln b+1
ln 2
It can be observed that the limit does not exist and hence, the integral
diverges.
Step 4: Conclusion Since the integral R1
x2ln xdx diverges, by the Integral
Test, the series P∞
n=1 1
n2ln nalso diverges. Thus, the series diverges.
9
Question 14
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 calculate the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
L= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
=1
e
Step 3: Determine the convergence. The ratio test states that if L < 1,
then the series converges. Since L=1
e<1, the series P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n
2n3+ 5 .
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will compare the given series to a simpler series. Let’s consider
the series ∞
X
n=1
1
n.
We will calculate the limit
L= lim
n→∞
n2+3n
2n3+5
1
n
= lim
n→∞
n3+ 3n2
2n3+ 5 ·n
1= lim
n→∞
n3+ 3n2
2n3+ 5 .
10
Step 2: We can simplify the expression inside the limit as follows:
L= lim
n→∞
n3+ 3n2
2n3+ 5 = lim
n→∞
n3(1 + 3
n)
n3(2 + 5
n3)= lim
n→∞
1 + 3
n
2 + 5
n3
=1
2.
Step 3: Now, we will determine the convergence of the series using the
limit comparison test. Since L > 0, and L < ∞, the given series P∞
n=1
n2+3n
2n3+5
converges since it behaves similarly to the convergent series P∞
n=1 1
n.
Therefore, the series P∞
n=1
n2+3n
2n3+5 converges.
Question 16
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: Calculate the ratio of consecutive terms. Let an=n!
nn. We will
compute the ratio:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio and evaluate the limit.
L= 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+ 1n
Step 3: Determine the limit.
L= lim
n→∞ n
n+ 1n
=1
limn→∞ 1 + 1
nn=1
e
Step 4: Interpret the result. Since L= 1/e < 1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 17
Question
Determine whether the series ∞
X
n=1
n!
(n2)!
converges or diverges.
11
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)!/(n2+ 2n+ 1)!
n!/(n2)!
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n2+ 2n+ 1)!
n!/(n2)!
= lim
n→∞
(n+ 1)n2!
(n2+ 2n+ 1)!
Step 3: Further simplify the ratio: Let’s compute the limit of the individual
terms in the ratio:
lim
n→∞
n+ 1
n2+ 2n+ 1 = lim
n→∞
n+ 1
(n+ 1)2= lim
n→∞
1
n+ 1 = 0
Step 4: Analyze the limit: The limit of the ratio is 0. Since the limit is less
than 1, by the Ratio Test, the series
∞
X
n=1
n!
(n2)!
converges.
Question 18
Question
Let {an}be a sequence of positive numbers such that P∞
n=1 anconverges. Define
bn=nan
1+n2. Determine whether the series P∞
n=1 bnconverges.
Solution
Step 1: We will start by showing that limn→∞ bn= 0.
Since limn→∞ an= 0 (as the series P∞
n=1 anconverges), we have:
lim
n→∞ bn= lim
n→∞
nan
1 + n2= 0 ·0
1+0 = 0.
Step 2: Next, we will investigate the convergence of the series P∞
n=1 bnusing
the Limit Comparison Test.
Let cn=1
n, then:
lim
n→∞
bn
cn
= lim
n→∞
nan
(1 + n2)n= lim
n→∞
an
1 + n2= 0.
Step 3: Since P∞
n=1 cn=P∞
n=1 1
nis a harmonic series which diverges, and
limn→∞
bn
cn= 0, the series P∞
n=1 bnalso diverges by the Limit Comparison Test.
12
Question 19
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. We consider the limit:
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
Step 2: Simplify the limit.
= lim
n→∞
1
n+ 1 = 0
Step 3: 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.
Question 20
Question
Determine the convergence of the series ∞
X
n=1
n2+ 1
3n3+ 5.
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s compare the given series to a simpler series whose convergence is
known.
Step 1: Calculate the limit of the ratio of terms. Let an=n2+1
3n3+5 be
the general term of the series. We will calculate the following limit:
L= lim
n→∞
an
1
n
= lim
n→∞
n2+ 1
3n3+ 5 ·n
1
Step 2: Simplify the limit expression.
L= lim
n→∞
n3+n
3n3+ 5 = lim
n→∞
1 + 1
n2
3 + 5
n3
=1
3
13
Step 3: Analyze the limit. Since L=1
3>0, the limit exists.
Step 4: Conclusion. According to the limit comparison test, the series
∞
X
n=1
n2+ 1
3n3+ 5 will converge if the series ∞
X
n=1
1
nconverges. Since ∞
X
n=1
1
nis the
harmonic series, which is a well-known divergent series, our given series also
diverges.
Question 21
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
nan
1+nan.
Solution
To determine the convergence of the series P∞
n=1
nan
1+nan, we will use the Limit
Comparison Test.
Step 1: (Calculating the limit) Let bn=nan
1+nan. We will calculate the limit
of the ratio bn
anas napproaches infinity.
lim
n→∞
bn
an
= lim
n→∞
nan
1+nan
an
= lim
n→∞
n
1 + nan
=1
0+= +∞
Step 2: (Comparison to a divergent series) Since limn→∞
bn
anis divergent, by
the Limit Comparison Test, if P∞
n=1 anconverges, then P∞
n=1
nan
1+nandiverges.
Therefore, the series P∞
n=1
nan
1+nanis divergent.
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 given series, we will use the
ratio test.
Step 1: Compute the limit of the ratio test.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplify the expression.
= lim
n→∞
(n+ 1)! ·nn
(n+ 1)n+1 ·n!
= lim
n→∞
nn
(n+ 1)n
14
Step 2: Calculate the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
Using the limit laws, we have
=1
e
Step 3: Interpret the result. Since the limit 1
eis less than 1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
Question 23
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 can 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→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(1 + 1/n)n
Step 3: Use the fact that (1+1/n)nconverges to eas napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
n+ 1
(1 + 1/n)n
=1
e.
Step 4: Apply the ratio test: Since the limit is less than 1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
Therefore, the given series converges.
15
Question 24
Question
Consider the series P∞
n=1
n!
nn. Determine whether the series converges or di-
verges.
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. We will
compute:
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
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
Step 3: Apply limit properties.
lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
Step 4: Use the fact that limn→∞ 1 + 1
nn=e. Therefore, the limit
reduces to:
lim
n→∞
1
e=1
e
Step 5: Apply the ratio test. Since 1
e<1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 25
Question
Let P∞
n=1 anbe a convergent series with an>0 for all n. Define bn=
√a1·a2·. . . ·an. Show that the series P∞
n=1 bnconverges.
16
Solution
Step 1: Notice that since bn=√a1·a2·. . . ·an, we can rewrite bnas pQn
k=1 ak.
Step 2: By the properties of square roots and infinite products, we have
bn=√a1·√a2·. . . ·√an.
Step 3: Since P∞
n=1 anis a convergent series, we know that limn→∞ an= 0.
This implies that limn→∞ √an= 0 as well.
Step 4: Now, consider the series P∞
n=1 √an. Since limn→∞ √an= 0, by the
Limit Comparison Test, P∞
n=1 √anconverges.
Step 5: Since P∞
n=1 √anconverges, and bn=√a1·√a2·. . . ·√an, by the
Comparison Test, P∞
n=1 bnalso converges.
Therefore, the series P∞
n=1 bnconverges.
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
n3+ 1.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series with terms that are easier to work with. Consider
the series P∞
n=1 1
n, which is a p-series with p= 1.
Step 2: We will now calculate the limit of the ratio of the two series:
L= lim
n→∞
n2
n3+1
1
n
= lim
n→∞
n3
n3+ 1 = 1.
Step 3: Since L= 1 and the series P∞
n=1 1
ndiverges, by the Limit Compar-
ison Test, the series P∞
n=1
n2
n3+1 also diverges.
Therefore, the given series diverges.
Question 27
Question
Determine the convergence of the series:
∞
X
n=1
n2
2n
17
Solution
To determine the convergence of the series P∞
n=1
n2
2n, we can use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We compute the ratio:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the ratio.
r= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Analyze the value of r. Since r=1
2<1, by the ratio test, the
series P∞
n=1
n2
2nconverges.
Therefore, the series converges.
Question 28
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
√an
n.
Solution
We will prove that the series P∞
n=1
√an
nconverges.
Step 1: Since P∞
n=1 anis convergent, we know that limn→∞ an= 0.
Step 2: We will use the Comparison Test to show convergence.
Consider the following inequalities:
√an
n≤an
nfor n≥1.
Step 3: Since P∞
n=1 anconverges, by the Comparison Test, we conclude
that P∞
n=1
an
nconverges.
Step 4: Therefore, by the Comparison Test, P∞
n=1
√an
nalso converges.
Thus, we have proved that the series P∞
n=1
√an
nconverges.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1
n!
nn.
18
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. The ratio test states that if limn→∞
an+1
an
<1, then the series converges.
If limn→∞
an+1
an
>1, then the series diverges.
Step 1: 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!
an+1
an
=(n+ 1)nn
(n+ 1)n+1 =1 + 1
nn
Step 2: Find the limit of the ratio as napproaches infinity.
lim
n→∞ 1 + 1
nn
=e > 1
Since the limit of the ratio is greater than 1, the series P∞
n=1
n!
nndiverges by
the ratio test.
Question 30
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n2+ 4
4n3−2.
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. Let’s consider the series
∞
X
n=1
3n2+ 4
4n3−2.
We will compare it with the series
∞
X
n=1
1
n.
Step 1: Find the limit of the ratio Let an=3n2+4
4n3−2and bn=1
n. We
will find the limit of the ratio limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
3n2+ 4
4n3−2·n
1= lim
n→∞
3 + 4
n2
4−2
n3
=3+0
4−0=3
4.
19
Step 2: Apply the limit comparison test Since limn→∞
an
bn=3
4>0,
and P∞
n=1 1
nis a harmonic series which diverges, by the limit comparison test,
we conclude that P∞
n=1 3n2+4
4n3−2also diverges.
Question 31
Question
Determine whether the series
∞
X
n=1
(−1)n·n2
3n2+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Alternating Series
Test.
Step 1: Check the conditions of the Alternating Series Test The
Alternating Series Test states that for an alternating series P∞
n=1(−1)n−1bn, if
the following conditions are met: 1. bn+1 ≤bnfor all n2. limn→∞ bn= 0
then the series converges.
Step 2: Find the general term In this case, the general term of the series
is an=(−1)n·n2
3n2+1 .
Step 3: Check if the conditions of the Alternating Series Test are
satisfied First, let’s find bn=n2
3n2+1 . To check the condition bn+1 ≤bn, we
compute bn+1 −bn:
bn+1 −bn=(n+ 1)2
3(n+ 1)2+ 1 −n2
3n2+ 1
Simplifying this expression, we find that bn+1 −bnis positive for all n. Thus,
bn+1 ≤bnis satisfied.
Next, we find limn→∞ bn:
lim
n→∞
n2
3n2+ 1 = lim
n→∞
1
3 + 1
n2
=1
3
Since limn→∞ bn= 1/3= 0, the series does not meet the second condition of
the Alternating Series Test.
Step 4: Conclusion Since the series fails to meet the second condition of
the Alternating Series Test, we cannot conclude whether the series converges or
diverges using this test. Alternative convergence tests, like the Ratio Test or
the Root Test, may be necessary to determine the convergence of this series.
20
Question 32
Question
Let an=n2
2nfor each n≥1. Determine if the series P∞
n=1 anconverges or
diverges.
Solution
To determine if the series P∞
n=1 anconverges or diverges, we will use the ratio
test. Let’s compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 1: Simplify the expression inside the limit:
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n+1
Step 2: Further simplify the expression inside the limit:
lim
n→∞
n2+ 2n+ 1
2n2·2n
2n+1
= lim
n→∞
1 + 2
n+1
n2
2
Step 3: Evaluate the limit:
lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 4: Analyze the result: Since the limit is less than 1, by the ratio test,
the series P∞
n=1 anconverges.
Question 33
Question
Determine whether the series P∞
n=1
n2
3nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2
3n, we will use the Ratio Test.
Step 1: Apply the Ratio Test Let an=n2
3n. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
21
Step 2: Simplify the limit
lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
= lim
n→∞
n2+ 2n+ 1
3n2
lim
n→∞
1 + 2
n+1
n2
3
=1
3
Step 3: Analyze the limit Since the limit is less than 1, by the Ratio Test,
the series P∞
n=1
n2
3nconverges.
Therefore, the series P∞
n=1
n2
3nconverges.
Question 34
Question
Determine whether the series
∞
X
n=1
n2+ 2n
n3+ 3
converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider a series that is easier to work with. We’ll choose the series
P∞
n=1 1
n.
Step 1: Find the limit
lim
n→∞
n2+2n
n3+3
1
n
= lim
n→∞
n3+ 2n2
n3+ 3
Step 2: Simplify the expression
= lim
n→∞
n3(1 + 2
n)
n3(1 + 3
n3)
= lim
n→∞
1 + 2
n
1 + 3
n3
= 1
Step 3: Analyze the limit Since the limit is finite and positive, the series
P∞
n=1
n2+2n
n3+3 and the series P∞
n=1 1
neither both converge or both diverge.
Step 4: Conclusion Since P∞
n=1 1
nis a divergent p-series, by the Limit
Comparison Test, the series P∞
n=1
n2+2n
n3+3 also diverges.
22
Step 4: Determine convergence. Since 1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 2
Question
Determine whether the series ∞
X
n=1
n2+ 1
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 1
3n3+ 2, we will use the Limit
Comparison Test.
Step 1: Let’s find a series that we can easily determine the convergence. We
will use the series ∞
X
n=1
1
n, which is the harmonic series and is known to diverge.
Step 2: Compute the limit of the ratio of the two series:
lim
n→∞
n2+1
3n3+2
1
n
= lim
n→∞
n3+n
3n3+ 2 = lim
n→∞
1 + 1
n2
3 + 2
n3
=1
3
Step 3: Since the limit is a finite nonzero value, the given series ∞
X
n=1
n2+ 1
3n3+ 2
converges or diverges exactly like the series ∞
X
n=1
1
n. Therefore, the series ∞
X
n=1
n2+ 1
3n3+ 2
diverges.
Question 3
Question
Determine whether the series ∞
X
n=1
n2+ 3n+ 1
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test. Let an=n2+3n+1
3n3+2 .
2
Step 1: Find the limit of an
1
n2
.
lim
n→∞
an
1
n2
= lim
n→∞
n2+ 3n+ 1
3n3+ 2 ·n2
Step 2: Simplify the expression in the limit.
lim
n→∞
n2+ 3n+ 1
3n3+ 2 ·n2= lim
n→∞
1 + 3
n+1
n2
3 + 2
n3·n2= lim
n→∞
1
3n2= 0
Since the limit is finite and positive, we conclude that the series ∞
X
n=1
n2+ 3n+ 1
3n3+ 2
converges by the Limit Comparison Test with ∞
X
n=1
1
n2.
Question 4
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: Calculate the ratio Rof consecutive terms. Let an=n!
nn. Then,
the ratio of consecutive terms is given by:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1 = lim
n→∞
n+ 1
(1 + 1
n)n
Step 3: Evaluate the limit.
lim
n→∞
n+ 1
(1 + 1
n)n= lim
n→∞
n+ 1
1 + 1
nn= lim
n→∞
n+ 1
n+1
nn= lim
n→∞
n+ 1
1=∞
Step 4: Determine the convergence of the series. Since the limit of the ratio
Ris infinity, the series P∞
n=1
n!
nndiverges by the ratio test.
Question 5
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
3
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Compute the ratio R. The ratio test states that if limn→∞
an+1
an
=
R, then the series converges if R < 1 and diverges if R > 1.
Compute the ratio R:
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
n+ 1
n
= lim
n→∞ 1−1
n+ 1n
=1
e.
Step 2: Analyze the ratio R. Since R=1
e<1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2n
5n.
Solution
To determine the convergence or divergence of the series P∞
n=1 3n+2n
5n, we will
use the Ratio Test.
4
Step 1: Compute the ratio an+1
an
.
an+1
an
=
3n+1+2n+1
5n+1
3n+2n
5n
=
3·3n+2·2n
5n+1
3n+2n
5n
=3·3n
5n+ 2 ·2n
5n
3n
5n+2n
5n
=3(3
5)n+ 2(2
5)n
(3
5)n+ (2
5)n
=3(3
5)n
(3
5)n+2(2
5)n
(3
5)n
= 3 + 2
32
3n
.
Step 2: Find the limit of the ratio as napproaches infinity.
lim
n→∞
an+1
an
= lim
n→∞ 3 + 2
32
3n
= 3 + 2
3lim
n→∞ 2
3n
= 3 + 0
= 3.
Step 3: Apply the Ratio Test. Since the limit of the ratio is 3, which is
greater than 1, the series P∞
n=1 3n+2n
5ndiverges.
Therefore, the given series P∞
n=1 3n+2n
5nis divergent.
Question 7
Question
Determine the convergence or divergence of the series:
∞
X
n=1
3n+ 4n
5n
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
5
Step 1: Apply the ratio test. Let an=3n+4n
5n, then
lim
n→∞
an+1
an
= lim
n→∞
3n+1+4n+1
5n+1
3n+4n
5n
= lim
n→∞
3·3n+4·4n
5·5n
3n+4n
5n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n+ 4n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n+ 4n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n1 + 4
3n
=4
3<1
Step 2: Analyze the limit. Since the limit of the absolute value of the ratio
is less than 1, by the ratio test, the series P∞
n=1 3n+4n
5nconverges.
Therefore, the original series converges.
Question 8
Question
Determine the convergence or divergence of the series ∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series ∞
X
n=1
n!
nn, we will use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute 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 2: Simplify the ratio.
an+1
an
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!=nn
(n+ 1)n=n
n+ 1n
6
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=e−1
Step 4: Analyze the limit. Since the ratio limn→∞ n
n+ 1n
=e−1<1,
by the ratio test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnis convergent.
Question 9
Question
Determine whether the series P∞
n=1
n3
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n3
2n, we will use the ratio test.
Step 1: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1+3/n + 3/n2+ 1/n3
2
=1
2.
Step 3: Apply the ratio test: Since the limit is 1
2, by the ratio test, if the
limit is less than 1, then the series converges. Therefore, the series P∞
n=1
n3
2n
converges.
Question 10
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
7
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Compute the limit of the ratio test. Let an=n!
nn. We consider the
limit
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
(1 + 1
n)n
= lim
n→∞
n+ 1
e
Step 2: Analyze the limit. Since the limit is ∞
e, which is greater than 1,
the series diverges by the ratio test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 11
Question
Determine whether the series P∞
n=1
(−1)n
√nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
(−1)n
√n, we can use the Alter-
nating Series Test.
Step 1: Check for the conditions of the Alternating Series Test. Let an=
1
√n. Notice that anis positive and decreasing for all n.
Step 2: Verify the limit condition of the Alternating Series Test. Calculate
limn→∞
1
√n= 0. Since the limit is 0, we have satisfied the second condition of
the Alternating Series Test.
Step 3: Apply the Alternating Series Test. Since anis positive, decreasing,
and the limit of anas napproaches infinity is 0, the series P∞
n=1
(−1)n
√nconverges
by the Alternating Series Test.
Therefore, the series P∞
n=1
(−1)n
√nconverges.
Question 12
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we will use the Ratio Test.
8
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Then, the
ratio of consecutive terms is:
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→∞
nn
(n+ 1)n
Step 3: Determine the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Apply the Ratio Test. Since the limit is less than 1, by the Ratio
Test, the series P∞
n=1
n!
nnconverges.
Therefore, the series converges.
Question 13
Question
Determine whether the series P∞
n=1 1
n2ln nconverges or diverges.
Solution
To determine the convergence of the series, we will use the Integral Test. Let
f(x) = 1
x2ln x, where x≥2.
Step 1: Find the integral Compute the indefinite integral of f(x):
Z1
x2ln xdx
Step 2: Perform u-substitution Let u= ln x, then du =1
xdx. The
integral becomes:
Z1
u2du =−1
u+C=−1
ln x+C
Step 3: Apply the Integral Test Since the integral R∞
2
1
x2ln xdx simplifies
to [−1
ln x]∞
2, we need to determine if this improper integral converges:
lim
b→∞ −1
ln b+1
ln 2
It can be observed that the limit does not exist and hence, the integral
diverges.
Step 4: Conclusion Since the integral R1
x2ln xdx diverges, by the Integral
Test, the series P∞
n=1 1
n2ln nalso diverges. Thus, the series diverges.
9
Question 14
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 calculate the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
L= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
=1
e
Step 3: Determine the convergence. The ratio test states that if L < 1,
then the series converges. Since L=1
e<1, the series P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n
2n3+ 5 .
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will compare the given series to a simpler series. Let’s consider
the series ∞
X
n=1
1
n.
We will calculate the limit
L= lim
n→∞
n2+3n
2n3+5
1
n
= lim
n→∞
n3+ 3n2
2n3+ 5 ·n
1= lim
n→∞
n3+ 3n2
2n3+ 5 .
10
Step 2: We can simplify the expression inside the limit as follows:
L= lim
n→∞
n3+ 3n2
2n3+ 5 = lim
n→∞
n3(1 + 3
n)
n3(2 + 5
n3)= lim
n→∞
1 + 3
n
2 + 5
n3
=1
2.
Step 3: Now, we will determine the convergence of the series using the
limit comparison test. Since L > 0, and L < ∞, the given series P∞
n=1
n2+3n
2n3+5
converges since it behaves similarly to the convergent series P∞
n=1 1
n.
Therefore, the series P∞
n=1
n2+3n
2n3+5 converges.
Question 16
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: Calculate the ratio of consecutive terms. Let an=n!
nn. We will
compute the ratio:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio and evaluate the limit.
L= 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+ 1n
Step 3: Determine the limit.
L= lim
n→∞ n
n+ 1n
=1
limn→∞ 1 + 1
nn=1
e
Step 4: Interpret the result. Since L= 1/e < 1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 17
Question
Determine whether the series ∞
X
n=1
n!
(n2)!
converges or diverges.
11
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)!/(n2+ 2n+ 1)!
n!/(n2)!
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n2+ 2n+ 1)!
n!/(n2)!
= lim
n→∞
(n+ 1)n2!
(n2+ 2n+ 1)!
Step 3: Further simplify the ratio: Let’s compute the limit of the individual
terms in the ratio:
lim
n→∞
n+ 1
n2+ 2n+ 1 = lim
n→∞
n+ 1
(n+ 1)2= lim
n→∞
1
n+ 1 = 0
Step 4: Analyze the limit: The limit of the ratio is 0. Since the limit is less
than 1, by the Ratio Test, the series
∞
X
n=1
n!
(n2)!
converges.
Question 18
Question
Let {an}be a sequence of positive numbers such that P∞
n=1 anconverges. Define
bn=nan
1+n2. Determine whether the series P∞
n=1 bnconverges.
Solution
Step 1: We will start by showing that limn→∞ bn= 0.
Since limn→∞ an= 0 (as the series P∞
n=1 anconverges), we have:
lim
n→∞ bn= lim
n→∞
nan
1 + n2= 0 ·0
1+0 = 0.
Step 2: Next, we will investigate the convergence of the series P∞
n=1 bnusing
the Limit Comparison Test.
Let cn=1
n, then:
lim
n→∞
bn
cn
= lim
n→∞
nan
(1 + n2)n= lim
n→∞
an
1 + n2= 0.
Step 3: Since P∞
n=1 cn=P∞
n=1 1
nis a harmonic series which diverges, and
limn→∞
bn
cn= 0, the series P∞
n=1 bnalso diverges by the Limit Comparison Test.
12
Question 19
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. We consider the limit:
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
Step 2: Simplify the limit.
= lim
n→∞
1
n+ 1 = 0
Step 3: 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.
Question 20
Question
Determine the convergence of the series ∞
X
n=1
n2+ 1
3n3+ 5.
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s compare the given series to a simpler series whose convergence is
known.
Step 1: Calculate the limit of the ratio of terms. Let an=n2+1
3n3+5 be
the general term of the series. We will calculate the following limit:
L= lim
n→∞
an
1
n
= lim
n→∞
n2+ 1
3n3+ 5 ·n
1
Step 2: Simplify the limit expression.
L= lim
n→∞
n3+n
3n3+ 5 = lim
n→∞
1 + 1
n2
3 + 5
n3
=1
3
13
Step 3: Analyze the limit. Since L=1
3>0, the limit exists.
Step 4: Conclusion. According to the limit comparison test, the series
∞
X
n=1
n2+ 1
3n3+ 5 will converge if the series ∞
X
n=1
1
nconverges. Since ∞
X
n=1
1
nis the
harmonic series, which is a well-known divergent series, our given series also
diverges.
Question 21
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
nan
1+nan.
Solution
To determine the convergence of the series P∞
n=1
nan
1+nan, we will use the Limit
Comparison Test.
Step 1: (Calculating the limit) Let bn=nan
1+nan. We will calculate the limit
of the ratio bn
anas napproaches infinity.
lim
n→∞
bn
an
= lim
n→∞
nan
1+nan
an
= lim
n→∞
n
1 + nan
=1
0+= +∞
Step 2: (Comparison to a divergent series) Since limn→∞
bn
anis divergent, by
the Limit Comparison Test, if P∞
n=1 anconverges, then P∞
n=1
nan
1+nandiverges.
Therefore, the series P∞
n=1
nan
1+nanis divergent.
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 given series, we will use the
ratio test.
Step 1: Compute the limit of the ratio test.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplify the expression.
= lim
n→∞
(n+ 1)! ·nn
(n+ 1)n+1 ·n!
= lim
n→∞
nn
(n+ 1)n
14
Step 2: Calculate the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
Using the limit laws, we have
=1
e
Step 3: Interpret the result. Since the limit 1
eis less than 1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
Question 23
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 can 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→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(1 + 1/n)n
Step 3: Use the fact that (1+1/n)nconverges to eas napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
n+ 1
(1 + 1/n)n
=1
e.
Step 4: Apply the ratio test: Since the limit is less than 1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
Therefore, the given series converges.
15
Question 24
Question
Consider the series P∞
n=1
n!
nn. Determine whether the series converges or di-
verges.
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. We will
compute:
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
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
Step 3: Apply limit properties.
lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
Step 4: Use the fact that limn→∞ 1 + 1
nn=e. Therefore, the limit
reduces to:
lim
n→∞
1
e=1
e
Step 5: Apply the ratio test. Since 1
e<1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 25
Question
Let P∞
n=1 anbe a convergent series with an>0 for all n. Define bn=
√a1·a2·. . . ·an. Show that the series P∞
n=1 bnconverges.
16
Solution
Step 1: Notice that since bn=√a1·a2·. . . ·an, we can rewrite bnas pQn
k=1 ak.
Step 2: By the properties of square roots and infinite products, we have
bn=√a1·√a2·. . . ·√an.
Step 3: Since P∞
n=1 anis a convergent series, we know that limn→∞ an= 0.
This implies that limn→∞ √an= 0 as well.
Step 4: Now, consider the series P∞
n=1 √an. Since limn→∞ √an= 0, by the
Limit Comparison Test, P∞
n=1 √anconverges.
Step 5: Since P∞
n=1 √anconverges, and bn=√a1·√a2·. . . ·√an, by the
Comparison Test, P∞
n=1 bnalso converges.
Therefore, the series P∞
n=1 bnconverges.
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
n3+ 1.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series with terms that are easier to work with. Consider
the series P∞
n=1 1
n, which is a p-series with p= 1.
Step 2: We will now calculate the limit of the ratio of the two series:
L= lim
n→∞
n2
n3+1
1
n
= lim
n→∞
n3
n3+ 1 = 1.
Step 3: Since L= 1 and the series P∞
n=1 1
ndiverges, by the Limit Compar-
ison Test, the series P∞
n=1
n2
n3+1 also diverges.
Therefore, the given series diverges.
Question 27
Question
Determine the convergence of the series:
∞
X
n=1
n2
2n
17
Solution
To determine the convergence of the series P∞
n=1
n2
2n, we can use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We compute the ratio:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the ratio.
r= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Analyze the value of r. Since r=1
2<1, by the ratio test, the
series P∞
n=1
n2
2nconverges.
Therefore, the series converges.
Question 28
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
√an
n.
Solution
We will prove that the series P∞
n=1
√an
nconverges.
Step 1: Since P∞
n=1 anis convergent, we know that limn→∞ an= 0.
Step 2: We will use the Comparison Test to show convergence.
Consider the following inequalities:
√an
n≤an
nfor n≥1.
Step 3: Since P∞
n=1 anconverges, by the Comparison Test, we conclude
that P∞
n=1
an
nconverges.
Step 4: Therefore, by the Comparison Test, P∞
n=1
√an
nalso converges.
Thus, we have proved that the series P∞
n=1
√an
nconverges.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1
n!
nn.
18
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. The ratio test states that if limn→∞
an+1
an
<1, then the series converges.
If limn→∞
an+1
an
>1, then the series diverges.
Step 1: 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!
an+1
an
=(n+ 1)nn
(n+ 1)n+1 =1 + 1
nn
Step 2: Find the limit of the ratio as napproaches infinity.
lim
n→∞ 1 + 1
nn
=e > 1
Since the limit of the ratio is greater than 1, the series P∞
n=1
n!
nndiverges by
the ratio test.
Question 30
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n2+ 4
4n3−2.
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. Let’s consider the series
∞
X
n=1
3n2+ 4
4n3−2.
We will compare it with the series
∞
X
n=1
1
n.
Step 1: Find the limit of the ratio Let an=3n2+4
4n3−2and bn=1
n. We
will find the limit of the ratio limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
3n2+ 4
4n3−2·n
1= lim
n→∞
3 + 4
n2
4−2
n3
=3+0
4−0=3
4.
19
Step 2: Apply the limit comparison test Since limn→∞
an
bn=3
4>0,
and P∞
n=1 1
nis a harmonic series which diverges, by the limit comparison test,
we conclude that P∞
n=1 3n2+4
4n3−2also diverges.
Question 31
Question
Determine whether the series
∞
X
n=1
(−1)n·n2
3n2+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the Alternating Series
Test.
Step 1: Check the conditions of the Alternating Series Test The
Alternating Series Test states that for an alternating series P∞
n=1(−1)n−1bn, if
the following conditions are met: 1. bn+1 ≤bnfor all n2. limn→∞ bn= 0
then the series converges.
Step 2: Find the general term In this case, the general term of the series
is an=(−1)n·n2
3n2+1 .
Step 3: Check if the conditions of the Alternating Series Test are
satisfied First, let’s find bn=n2
3n2+1 . To check the condition bn+1 ≤bn, we
compute bn+1 −bn:
bn+1 −bn=(n+ 1)2
3(n+ 1)2+ 1 −n2
3n2+ 1
Simplifying this expression, we find that bn+1 −bnis positive for all n. Thus,
bn+1 ≤bnis satisfied.
Next, we find limn→∞ bn:
lim
n→∞
n2
3n2+ 1 = lim
n→∞
1
3 + 1
n2
=1
3
Since limn→∞ bn= 1/3= 0, the series does not meet the second condition of
the Alternating Series Test.
Step 4: Conclusion Since the series fails to meet the second condition of
the Alternating Series Test, we cannot conclude whether the series converges or
diverges using this test. Alternative convergence tests, like the Ratio Test or
the Root Test, may be necessary to determine the convergence of this series.
20
Question 32
Question
Let an=n2
2nfor each n≥1. Determine if the series P∞
n=1 anconverges or
diverges.
Solution
To determine if the series P∞
n=1 anconverges or diverges, we will use the ratio
test. Let’s compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 1: Simplify the expression inside the limit:
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n+1
Step 2: Further simplify the expression inside the limit:
lim
n→∞
n2+ 2n+ 1
2n2·2n
2n+1
= lim
n→∞
1 + 2
n+1
n2
2
Step 3: Evaluate the limit:
lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 4: Analyze the result: Since the limit is less than 1, by the ratio test,
the series P∞
n=1 anconverges.
Question 33
Question
Determine whether the series P∞
n=1
n2
3nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2
3n, we will use the Ratio Test.
Step 1: Apply the Ratio Test Let an=n2
3n. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
21
Step 2: Simplify the limit
lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
= lim
n→∞
n2+ 2n+ 1
3n2
lim
n→∞
1 + 2
n+1
n2
3
=1
3
Step 3: Analyze the limit Since the limit is less than 1, by the Ratio Test,
the series P∞
n=1
n2
3nconverges.
Therefore, the series P∞
n=1
n2
3nconverges.
Question 34
Question
Determine whether the series
∞
X
n=1
n2+ 2n
n3+ 3
converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider a series that is easier to work with. We’ll choose the series
P∞
n=1 1
n.
Step 1: Find the limit
lim
n→∞
n2+2n
n3+3
1
n
= lim
n→∞
n3+ 2n2
n3+ 3
Step 2: Simplify the expression
= lim
n→∞
n3(1 + 2
n)
n3(1 + 3
n3)
= lim
n→∞
1 + 2
n
1 + 3
n3
= 1
Step 3: Analyze the limit Since the limit is finite and positive, the series
P∞
n=1
n2+2n
n3+3 and the series P∞
n=1 1
neither both converge or both diverge.
Step 4: Conclusion Since P∞
n=1 1
nis a divergent p-series, by the Limit
Comparison Test, the series P∞
n=1
n2+2n
n3+3 also diverges.
22
Step 4: Determine convergence. Since 1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 2
Question
Determine whether the series ∞
X
n=1
n2+ 1
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 1
3n3+ 2, we will use the Limit
Comparison Test.
Step 1: Let’s find a series that we can easily determine the convergence. We
will use the series ∞
X
n=1
1
n, which is the harmonic series and is known to diverge.
Step 2: Compute the limit of the ratio of the two series:
lim
n→∞
n2+1
3n3+2
1
n
= lim
n→∞
n3+n
3n3+ 2 = lim
n→∞
1 + 1
n2
3 + 2
n3
=1
3
Step 3: Since the limit is a finite nonzero value, the given series ∞
X
n=1
n2+ 1
3n3+ 2
converges or diverges exactly like the series ∞
X
n=1
1
n. Therefore, the series ∞
X
n=1
n2+ 1
3n3+ 2
diverges.
Question 3
Question
Determine whether the series ∞
X
n=1
n2+ 3n+ 1
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test. Let an=n2+3n+1
3n3+2 .
2
Step 1: Find the limit of an
1
n2
.
lim
n→∞
an
1
n2
= lim
n→∞
n2+ 3n+ 1
3n3+ 2 ·n2
Step 2: Simplify the expression in the limit.
lim
n→∞
n2+ 3n+ 1
3n3+ 2 ·n2= lim
n→∞
1 + 3
n+1
n2
3 + 2
n3·n2= lim
n→∞
1
3n2= 0
Since the limit is finite and positive, we conclude that the series ∞
X
n=1
n2+ 3n+ 1
3n3+ 2
converges by the Limit Comparison Test with ∞
X
n=1
1
n2.
Question 4
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: Calculate the ratio Rof consecutive terms. Let an=n!
nn. Then,
the ratio of consecutive terms is given by:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1 = lim
n→∞
n+ 1
(1 + 1
n)n
Step 3: Evaluate the limit.
lim
n→∞
n+ 1
(1 + 1
n)n= lim
n→∞
n+ 1
1 + 1
nn= lim
n→∞
n+ 1
n+1
nn= lim
n→∞
n+ 1
1=∞
Step 4: Determine the convergence of the series. Since the limit of the ratio
Ris infinity, the series P∞
n=1
n!
nndiverges by the ratio test.
Question 5
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
3
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Compute the ratio R. The ratio test states that if limn→∞
an+1
an
=
R, then the series converges if R < 1 and diverges if R > 1.
Compute the ratio R:
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
n+ 1
n
= lim
n→∞ 1−1
n+ 1n
=1
e.
Step 2: Analyze the ratio R. Since R=1
e<1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2n
5n.
Solution
To determine the convergence or divergence of the series P∞
n=1 3n+2n
5n, we will
use the Ratio Test.
4
Step 1: Compute the ratio an+1
an
.
an+1
an
=
3n+1+2n+1
5n+1
3n+2n
5n
=
3·3n+2·2n
5n+1
3n+2n
5n
=3·3n
5n+ 2 ·2n
5n
3n
5n+2n
5n
=3(3
5)n+ 2(2
5)n
(3
5)n+ (2
5)n
=3(3
5)n
(3
5)n+2(2
5)n
(3
5)n
= 3 + 2
32
3n
.
Step 2: Find the limit of the ratio as napproaches infinity.
lim
n→∞
an+1
an
= lim
n→∞ 3 + 2
32
3n
= 3 + 2
3lim
n→∞ 2
3n
= 3 + 0
= 3.
Step 3: Apply the Ratio Test. Since the limit of the ratio is 3, which is
greater than 1, the series P∞
n=1 3n+2n
5ndiverges.
Therefore, the given series P∞
n=1 3n+2n
5nis divergent.
Question 7
Question
Determine the convergence or divergence of the series:
∞
X
n=1
3n+ 4n
5n
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
5
Step 1: Apply the ratio test. Let an=3n+4n
5n, then
lim
n→∞
an+1
an
= lim
n→∞
3n+1+4n+1
5n+1
3n+4n
5n
= lim
n→∞
3·3n+4·4n
5·5n
3n+4n
5n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n+ 4n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n+ 4n
= lim
n→∞
3·3
5n+ 4 ·4
5n
3n1 + 4
3n
=4
3<1
Step 2: Analyze the limit. Since the limit of the absolute value of the ratio
is less than 1, by the ratio test, the series P∞
n=1 3n+4n
5nconverges.
Therefore, the original series converges.
Question 8
Question
Determine the convergence or divergence of the series ∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series ∞
X
n=1
n!
nn, we will use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute 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 2: Simplify the ratio.
an+1
an
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!=nn
(n+ 1)n=n
n+ 1n
6
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=e−1
Step 4: Analyze the limit. Since the ratio limn→∞ n
n+ 1n
=e−1<1,
by the ratio test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnis convergent.
Question 9
Question
Determine whether the series P∞
n=1
n3
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n3
2n, we will use the ratio test.
Step 1: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1+3/n + 3/n2+ 1/n3
2
=1
2.
Step 3: Apply the ratio test: Since the limit is 1
2, by the ratio test, if the
limit is less than 1, then the series converges. Therefore, the series P∞
n=1
n3
2n
converges.
Question 10
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
7
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Compute the limit of the ratio test. Let an=n!
nn. We consider the
limit
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
(1 + 1
n)n
= lim
n→∞
n+ 1
e
Step 2: Analyze the limit. Since the limit is ∞
e, which is greater than 1,
the series diverges by the ratio test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 11
Question
Determine whether the series P∞
n=1
(−1)n
√nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1
(−1)n
√n, we can use the Alter-
nating Series Test.
Step 1: Check for the conditions of the Alternating Series Test. Let an=
1
√n. Notice that anis positive and decreasing for all n.
Step 2: Verify the limit condition of the Alternating Series Test. Calculate
limn→∞
1
√n= 0. Since the limit is 0, we have satisfied the second condition of
the Alternating Series Test.
Step 3: Apply the Alternating Series Test. Since anis positive, decreasing,
and the limit of anas napproaches infinity is 0, the series P∞
n=1
(−1)n
√nconverges
by the Alternating Series Test.
Therefore, the series P∞
n=1
(−1)n
√nconverges.
Question 12
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we will use the Ratio Test.
8
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Then, the
ratio of consecutive terms is:
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→∞
nn
(n+ 1)n
Step 3: Determine the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Apply the Ratio Test. Since the limit is less than 1, by the Ratio
Test, the series P∞
n=1
n!
nnconverges.
Therefore, the series converges.
Question 13
Question
Determine whether the series P∞
n=1 1
n2ln nconverges or diverges.
Solution
To determine the convergence of the series, we will use the Integral Test. Let
f(x) = 1
x2ln x, where x≥2.
Step 1: Find the integral Compute the indefinite integral of f(x):
Z1
x2ln xdx
Step 2: Perform u-substitution Let u= ln x, then du =1
xdx. The
integral becomes:
Z1
u2du =−1
u+C=−1
ln x+C
Step 3: Apply the Integral Test Since the integral R∞
2
1
x2ln xdx simplifies
to [−1
ln x]∞
2, we need to determine if this improper integral converges:
lim
b→∞ −1
ln b+1
ln 2
It can be observed that the limit does not exist and hence, the integral
diverges.
Step 4: Conclusion Since the integral R1
x2ln xdx diverges, by the Integral
Test, the series P∞
n=1 1
n2ln nalso diverges. Thus, the series diverges.
9
Question 14
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 calculate the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
L= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
=1
e
Step 3: Determine the convergence. The ratio test states that if L < 1,
then the series converges. Since L=1
e<1, the series P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n
2n3+ 5 .
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will compare the given series to a simpler series. Let’s consider
the series ∞
X
n=1
1
n.
We will calculate the limit
L= lim
n→∞
n2+3n
2n3+5
1
n
= lim
n→∞
n3+ 3n2
2n3+ 5 ·n
1= lim
n→∞
n3+ 3n2
2n3+ 5 .
10
Step 2: We can simplify the expression inside the limit as follows:
L= lim
n→∞
n3+ 3n2
2n3+ 5 = lim
n→∞
n3(1 + 3
n)
n3(2 + 5
n3)= lim
n→∞
1 + 3
n
2 + 5
n3
=1
2.
Step 3: Now, we will determine the convergence of the series using the
limit comparison test. Since L > 0, and L < ∞, the given series P∞
n=1
n2+3n
2n3+5
converges since it behaves similarly to the convergent series P∞
n=1 1
n.
Therefore, the series P∞
n=1
n2+3n
2n3+5 converges.
Question 16
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: Calculate the ratio of consecutive terms. Let an=n!
nn. We will
compute the ratio:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio and evaluate the limit.
L= 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+ 1n
Step 3: Determine the limit.
L= lim
n→∞ n
n+ 1n
=1
limn→∞ 1 + 1
nn=1
e
Step 4: Interpret the result. Since L= 1/e < 1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 17
Question
Determine whether the series ∞
X
n=1
n!
(n2)!
converges or diverges.
11
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)!/(n2+ 2n+ 1)!
n!/(n2)!
Step 2: Simplify the ratio:
lim
n→∞
(n+ 1)!/(n2+ 2n+ 1)!
n!/(n2)!
= lim
n→∞
(n+ 1)n2!
(n2+ 2n+ 1)!
Step 3: Further simplify the ratio: Let’s compute the limit of the individual
terms in the ratio:
lim
n→∞
n+ 1
n2+ 2n+ 1 = lim
n→∞
n+ 1
(n+ 1)2= lim
n→∞
1
n+ 1 = 0
Step 4: Analyze the limit: The limit of the ratio is 0. Since the limit is less
than 1, by the Ratio Test, the series
∞
X
n=1
n!
(n2)!
converges.
Question 18
Question
Let {an}be a sequence of positive numbers such that P∞
n=1 anconverges. Define
bn=nan
1+n2. Determine whether the series P∞
n=1 bnconverges.
Solution
Step 1: We will start by showing that limn→∞ bn= 0.
Since limn→∞ an= 0 (as the series P∞
n=1 anconverges), we have:
lim
n→∞ bn= lim
n→∞
nan
1 + n2= 0 ·0
1+0 = 0.
Step 2: Next, we will investigate the convergence of the series P∞
n=1 bnusing
the Limit Comparison Test.
Let cn=1
n, then:
lim
n→∞
bn
cn
= lim
n→∞
nan
(1 + n2)n= lim
n→∞
an
1 + n2= 0.
Step 3: Since P∞
n=1 cn=P∞
n=1 1
nis a harmonic series which diverges, and
limn→∞
bn
cn= 0, the series P∞
n=1 bnalso diverges by the Limit Comparison Test.
12
Question 19
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. We consider the limit:
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
Step 2: Simplify the limit.
= lim
n→∞
1
n+ 1 = 0
Step 3: 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.
Question 20
Question
Determine the convergence of the series ∞
X
n=1
n2+ 1
3n3+ 5.
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s compare the given series to a simpler series whose convergence is
known.
Step 1: Calculate the limit of the ratio of terms. Let an=n2+1
3n3+5 be
the general term of the series. We will calculate the following limit:
L= lim
n→∞
an
1
n
= lim
n→∞
n2+ 1
3n3+ 5 ·n
1
Step 2: Simplify the limit expression.
L= lim
n→∞
n3+n
3n3+ 5 = lim
n→∞
1 + 1
n2
3 + 5
n3
=1
3
13
Step 3: Analyze the limit. Since L=1
3>0, the limit exists.
Step 4: Conclusion. According to the limit comparison test, the series
∞
X
n=1
n2+ 1
3n3+ 5 will converge if the series ∞
X
n=1
1
nconverges. Since ∞
X
n=1
1
nis the
harmonic series, which is a well-known divergent series, our given series also
diverges.
Question 21
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
nan
1+nan.
Solution
To determine the convergence of the series P∞
n=1
nan
1+nan, we will use the Limit
Comparison Test.
Step 1: (Calculating the limit) Let bn=nan
1+nan. We will calculate the limit
of the ratio bn
anas napproaches infinity.
lim
n→∞
bn
an
= lim
n→∞
nan
1+nan
an
= lim
n→∞
n
1 + nan
=1
0+= +∞
Step 2: (Comparison to a divergent series) Since limn→∞
bn
anis divergent, by
the Limit Comparison Test, if P∞
n=1 anconverges, then P∞
n=1
nan
1+nandiverges.
Therefore, the series P∞
n=1
nan
1+nanis divergent.
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 given series, we will use the
ratio test.
Step 1: Compute the limit of the ratio test.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplify the expression.
= lim
n→∞
(n+ 1)! ·nn
(n+ 1)n+1 ·n!
= lim
n→∞
nn
(n+ 1)n
14
Step 2: Calculate the limit.
lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
Using the limit laws, we have
=1
e
Step 3: Interpret the result. Since the limit 1
eis less than 1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
Question 23
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 can 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→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(1 + 1/n)n
Step 3: Use the fact that (1+1/n)nconverges to eas napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
n+ 1
(1 + 1/n)n
=1
e.
Step 4: Apply the ratio test: Since the limit is less than 1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
Therefore, the given series converges.
15
Question 24
Question
Consider the series P∞
n=1
n!
nn. Determine whether the series converges or di-
verges.
Solution
To determine the convergence of the series P∞
n=1
n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. We will
compute:
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
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
Step 3: Apply limit properties.
lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
Step 4: Use the fact that limn→∞ 1 + 1
nn=e. Therefore, the limit
reduces to:
lim
n→∞
1
e=1
e
Step 5: Apply the ratio test. Since 1
e<1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Question 25
Question
Let P∞
n=1 anbe a convergent series with an>0 for all n. Define bn=
√a1·a2·. . . ·an. Show that the series P∞
n=1 bnconverges.
16
Solution
Step 1: Notice that since bn=√a1·a2·. . . ·an, we can rewrite bnas pQn
k=1 ak.
Step 2: By the properties of square roots and infinite products, we have
bn=√a1·√a2·. . . ·√an.
Step 3: Since P∞
n=1 anis a convergent series, we know that limn→∞ an= 0.
This implies that limn→∞ √an= 0 as well.
Step 4: Now, consider the series P∞
n=1 √an. Since limn→∞ √an= 0, by the
Limit Comparison Test, P∞
n=1 √anconverges.
Step 5: Since P∞
n=1 √anconverges, and bn=√a1·√a2·. . . ·√an, by the
Comparison Test, P∞
n=1 bnalso converges.
Therefore, the series P∞
n=1 bnconverges.
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
n3+ 1.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series with terms that are easier to work with. Consider
the series P∞
n=1 1
n, which is a p-series with p= 1.
Step 2: We will now calculate the limit of the ratio of the two series:
L= lim
n→∞
n2
n3+1
1
n
= lim
n→∞
n3
n3+ 1 = 1.
Step 3: Since L= 1 and the series P∞
n=1 1
ndiverges, by the Limit Compar-
ison Test, the series P∞
n=1
n2
n3+1 also diverges.
Therefore, the given series diverges.
Question 27
Question
Determine the convergence of the series:
∞
X
n=1
n2
2n
17
Solution
To determine the convergence of the series P∞
n=1
n2
2n, we can use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We compute the ratio:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the ratio.
r= lim
n→∞
(n+ 1)2
2n2