MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 1
Liberty University
Question 1
Question
Determine whether the series ∞
X
n=1
(−1)n
nln(n+ 1) converges or diverges.
Solution
To determine the convergence of the series, we can use the Alternating Series
Test. We need to show that the terms are decreasing in magnitude and that the
limit of the terms approaches 0.
Step 1: Show that the terms are decreasing. Consider the sequence
defined by an=1
nln(n+1) . To show that the terms of the series are decreasing,
we need to show that an+1 ≤anfor all n≥1.
an+1 =1
(n+ 1) ln((n+ 1) + 1) =1
(n+ 1) ln(n+ 2)
Now, we need to compare an+1 and an:
an+1
an
=nln(n+ 1)
(n+ 1) ln(n+ 2) =ln(n+ 1)
ln(n+ 2)
To show that this ratio is less than or equal to 1, it suffices to show that
ln(n+ 1) ≤ln(n+ 2) for all n≥1. This holds true as the natural logarithm
function is increasing. Therefore, an+1 ≤anfor all n≥1.
Step 2: Show that the limit of the terms is 0. We need to check if
limn→∞
1
nln(n+1) = 0. Let’s use L’Hˆopital’s Rule:
lim
n→∞
1
nln(n+ 1) = lim
n→∞
1
ln(n+ 1) + n
n+1
= lim
n→∞
1
ln(n+ 1) + 1 = 0
Since the terms are decreasing and the limit of the terms is 0, by the Alter-
nating Series Test, the series ∞
X
n=1
(−1)n
nln(n+ 1) converges.
Question 2
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: Apply the ratio test. Let an=n!
nn. We will calculate the following
limit:
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
Step 2: Simplify the expression.
= lim
n→∞
n+ 1
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 3: Analyze the limit. Since 1
e<1, by the ratio test, the series P∞
n=1
n!
nn
converges.
Therefore, the series P∞
n=1
n!
nnconverges by the ratio test.
Question 3
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2
5n.
2
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Let an=3n+2
5n.
Step 1: Calculate the ratio R.
R= lim
n→∞
an+1
an
= lim
n→∞
3n+1+2
5n+1
3n+2
5n
= lim
n→∞
3n+1 + 2
5n+1 ·5n
3n+ 2
= lim
n→∞
3
5·5n
3n+ 2
=3
5lim
n→∞
5n
3n+ 2
=3
5.
Step 2: Determine the convergence or divergence of the series. Since the
ratio R=3
5<1, by the ratio test, the series P∞
n=1 3n+2
5nconverges.
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 can use the Ratio Test.
Step 1: Calculate the ratio limn→∞
an+1
anwhere an=n!
nn.
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+1
(n+ 1)n
= lim
n→∞
n
(1 + 1
n)n
= lim
n→∞
n
e=∞
Since the limit is greater than 1, we can conclude that the series P∞
n=1
n!
nn
diverges by the Ratio Test.
Question 5
Question
Determine the convergence or divergence of the series P∞
n=1
n2
2n.
3
Solution
To determine the convergence of the series P∞
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test by calculating the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2·2n
2n+1 ·n2
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)2·2n
2n+1 ·n2
= lim
n→∞
(n+ 1)2
2·n2
Step 3: Calculate the limit:
lim
n→∞
(n+ 1)2
2·n2
= lim
n→∞
n2+ 2n+ 1
2n2=1
2
Step 4: Analyze the limit: Since the limit is 1
2which is less than 1, by the
ratio test, the series P∞
n=1
n2
2nconverges.
Therefore, the series P∞
n=1
n2
2nconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
3n3+ 2.
Solution
To determine the convergence or divergence of the given series, we will use the
Limit Comparison Test.
Step 1: Let’s find the limit of the general term:
lim
n→∞
n2+ 1
3n3+ 2.
Step 2: We simplify the limit by dividing the numerator and denominator
by n3:
lim
n→∞
n2+ 1
3n3+ 2 = lim
n→∞
1
n+1
n3
3 + 2
n3
=0+0
3+0 = 0.
Therefore, since the limit of the general term is finite and positive, we can
apply the Limit Comparison Test.
4
Step 3: We choose the series P∞
n=1 1
nas a comparison series as it is a known
divergent p-series with p= 1.
Step 4: Let’s find the limit of the ratio between the general term of the
given series and the general term of the comparison series:
lim
n→∞
n2+1
3n3+2
1
n
= lim
n→∞
n3+n
3n3+ 2n= lim
n→∞
n(1 + 1
n2)
n(3 + 2
n2)=1
3.
Step 5: Since the limit is a positive finite number, and the comparison
series diverges, by the Limit Comparison Test, the given series P∞
n=1
n2+1
3n3+2 also
diverges.
Question 7
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 calculate the limit:
L= lim
n→∞
an+1
an
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression to find the limit.
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
L= lim
n→∞
n+ 1
(n+ 1)n+1 ·n
L= lim
n→∞
n
(n+ 1)n
L= lim
n→∞
1
(1 + 1/n)n
Step 3: Recall the limit limn→∞(1 + 1/n)n=e.
L=
1
e
=1
e
5
Step 4: Analyze the limit. Since 0 <1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 8
Question
Determine the convergence of the series P∞
n=1
n!
nn.
Solution
To determine the convergence of the series, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. We calculate 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!
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Use the limit of (1 + 1
n)n. As napproaches infinity, limn→∞(1 +
1
n)n=e, where eis Euler’s number.
Step 4: Determine the limit.
lim
n→∞
1
e
=1
e
Step 5: Apply the Ratio Test. Since the limit is less than 1, by the Ratio
Test, the series P∞
n=1
n!
nnconverges.
Question 9
Question
Determine the convergence or divergence of the series P∞
n=1
n2
2n.
6
Solution
To determine the convergence or divergence of the series P∞
n=1
n2
2n, we will use
the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n2
2n. We will consider the limit of
the ratio limn→∞
an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞ 1
2+1
n+1
2n2
=1
2
Step 2: Evaluate the limit. Since the limit is 1
2<1, by the Ratio Test, the
series P∞
n=1
n2
2nconverges.
Therefore, the series P∞
n=1
n2
2nconverges.
Question 10
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: 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
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
n
(1 + 1/n)n+1
7
Step 2: Simplify the expression inside the limit: Using the fact that limn→∞ 1 + 1
nn=
e, we have:
lim
n→∞
n
(1 + 1/n)n+1
= lim
n→∞
n
(1 + 1/n)(1 + 1/n)n
= lim
n→∞
n
(1 + 1/n)e
=1
elim
n→∞ |n|=∞
Step 3: Determine the convergence or divergence: Since the limit of the
ratio test is ∞, the series P∞
n=1
n!
nndiverges.
Therefore, the series diverges.
Question 11
Question
Determine whether the series P∞
n=1 3n2+1
n4+2 converges or diverges.
Solution
To determine whether the series converges or diverges, we will use the Limit
Comparison Test. Let’s denote the n-th term of the series as an=3n2+1
n4+2 .
Step 1: Find a series Pbnthat is easier to work with. We will choose
bn=1
n2since this is a known convergent series.
Step 2: Compute the limit of the ratio anbn.
lim
n→∞
an
bn
= lim
n→∞
3n2+ 1
n4+ 2 ·n2
1= lim
n→∞
3 + 1
n2
1 + 2
n4
=3
1= 3
Step 3: Make a conclusion based on the limit. Since the limit is a finite
positive number, by the Limit Comparison Test, the convergence of P∞
n=1 anis
the same as the convergence of P∞
n=1 bn. Since P∞
n=1 1
n2is convergent (by the
p-series test with p= 2 >1), our original series P∞
n=1 3n2+1
n4+2 also converges.
Question 12
Question
Determine whether the series ∞
X
n=1
n2+ 1
n3+ 1 converges or diverges.
8
Solution
To determine the convergence of the given series, we can use the limit compar-
ison test.
Step 1: Let’s choose a series bnthat we know converges. We will use the
harmonic series ∞
X
n=1
1
n.
Step 2: Find the limit limn→∞
n2+1
n3+1
1
n
.
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1
= lim
n→∞
1 + 1
n2
1 + 1
n3
=1+0
1+0
= 1
Step 3: Since the limit is a positive finite number, we can conclude that the
given series ∞
X
n=1
n2+ 1
n3+ 1 behaves similarly to the harmonic series which diverges.
Therefore, by the limit comparison test, ∞
X
n=1
n2+ 1
n3+ 1 also diverges.
Question 13
Question
Determine the convergence or divergence of the series
∞
X
n=1
√n3+ 1
n2+ 1 .
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s choose a series to compare. We consider the series P∞
n=1 1
n,
which is a harmonic series that diverges.
Step 2: We will calculate the limit
lim
n→∞
√n3+1
n2+1
1
n
.
9
Step 3: Simplifying the expression, we have
lim
n→∞
√n3+1
n2+1
1
n
= lim
n→∞
n√n3+ 1
n(n2+ 1) = lim
n→∞
√n3+ 1
n2+ 1 .
Step 4: Let’s find the limit of the expression. Since
lim
n→∞
√n3+ 1
n2+ 1 = lim
n→∞
√n3+ 1/n2
1+1/n2,
we apply L’Hˆopital’s Rule to the numerator to get
lim
n→∞
√n3+ 1
n2+ 1 = lim
n→∞
(3n2/2)/(2√n3+ 1)
−2/n3= lim
n→∞ −3n4√n3+ 1
4(n3+ 1) =−3
4.
Step 5: Since the limit is finite and non-zero, by the Limit Comparison
Test, since the limit is a constant between 0 and ∞, the series P∞
n=1
√n3+1
n2+1
diverges.
Question 14
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: Apply the Ratio Test. Let an=n!
nn. We will consider the limit
L= lim
n→∞
an+1
an
10
L= 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+1/n
= 1
Step 2: Evaluate the limit L. Since L= 1, the Ratio Test is inconclusive.
Therefore, we cannot determine the convergence or divergence of the series from
this test alone.
Step 3: Further analysis. Notice that lim
n→∞
n!
nn= 0 by Stirling’s approx-
imation. Therefore, n!
nnapproaches 0 as napproaches infinity and the series
converges by the Test for Divergence.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 15
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: Calculate the ratio Let an=n!
nn. We compute the ratio Ras
follows:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio Simplifying the ratio, we have:
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
11
R= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Evaluate the limit Using the fact that limn→∞ 1 + 1
nn=e, we
get:
R=
1
e
=1
e
Step 4: Conclusion Since R=1
e<1, by the ratio test, the series P∞
n=1
n!
nn
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: Compute the ratio of consecutive terms. Let an=n!
nn. We compute
limn→∞
an+1
an:
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
Step 2: Simplify and analyze the limit. Using L’Hopital’s Rule, we can
simplify the limit further:
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e
Step 3: Determine the convergence of the series. Since the limit is less
than 1, by the Ratio Test, the series P∞
n=1
n!
nnconverges. Therefore, the series
converges.
Question 17
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2·2n
n!
12
Solution
To determine the convergence or divergence of the series, we will use the Ratio
Test. The Ratio Test states that if limn→∞
an+1
an<1, then the series converges;
if the limit is greater than 1, then the series diverges; and if the limit equals 1,
the test is inconclusive.
Step 1: Let’s compute the ratio
an+1
an∥.
an+1
an
=
(n+ 1)2·2n+1
(n+ 1)! ·n!
n2·2n
=
(n+ 1)2·2
(n+ 1)(n)·2
=
n+ 1
n
= 1 + 1
n
Step 2: Now, take the limit of the ratio as napproaches infinity.
lim
n→∞ 1 + 1
n= 1
Step 3: Since the limit is equal to 1, the Ratio Test is inconclusive. We will
perform further analysis.
Step 4: We observe that the terms in the series decrease in magnitude for
large n.
n2·2n
n!=n2
1·2n
n·(n−1) ···2·1≈0
as nbecomes large.
Step 5: Since the terms of the series approach zero for large n, we can
conclude that the series converges by the Ratio Test.
Question 18
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 can use the Ratio Test.
13
Step 1: Apply the Ratio Test. Let an=n!
nn. We will consider the limit:
L= 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→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
=1
e.
Step 2: Analyze the limit. Since L=1
eis a finite positive number, we can
conclude that the series ∞
X
n=1
n!
nnconverges by the Ratio Test.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 19
Question
Determine whether the series ∞
X
n=1
n2
2nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the Ratio Test.
The Ratio Test states that if limn→∞
an+1
an
=L, then: - If L < 1, the series
converges absolutely. - If L > 1 or L=∞, the series diverges. - If L= 1, the
test is inconclusive.
Step 1: Compute the ratio an+1
an
. Let an=n2
2n. Then an+1 =(n+ 1)2
2n+1 .
Hence, an+1
an
=(n+ 1)2
2n+1 ·2n
n2=(n+ 1)2
2n2.
14
Step 2: Take the limit as napproaches infinity.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
1 + 1
n
2
=1
2
Step 3: Determine the convergence of the series. Since 1
2<1, by the Ratio
Test, the series ∞
X
n=1
n2
2nconverges absolutely.
Question 20
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, we will use the Limit
Comparison Test. Let’s consider the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3
and compare it with the series
∞
X
n=1
1
n.
Step 1: Calculate the following limit:
lim
n→∞
(n3+ 2n+ 1)/n4
1/n = lim
n→∞
n3+ 2n+ 1
n3= 1.
Step 2: Since the limit is a positive finite value, by the Limit Comparison
Test, the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3
has the same convergence behavior as
∞
X
n=1
1
n.
15
Step 3: The series
∞
X
n=1
1
n
is known to be a harmonic series, which diverges.
Therefore, by the Limit Comparison Test, the given series
∞
X
n=1
n3+ 2n+ 1
n4+ 3
also diverges.
Question 21
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n!
nn,
we will use the ratio test. Let’s denote the general term of the series as an=n!
nn.
Step 1: Compute the ratio Rof consecutive terms:
R= lim
n→∞
an+1
an
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
R= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
R= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
Step 2: Simplify the expression by expanding and canceling terms:
R= lim
n→∞
nn+1 +nn
nn+1 +n·nn
16
R= lim
n→∞
1+1/n
1+1/n
R= lim
n→∞ 1=1
Step 3: Apply the ratio test: - If R < 1, the series converges. - If R > 1,
the series diverges. - If R= 1, the ratio test is inconclusive.
Since R= 1, the ratio test is inconclusive. Therefore, we need to try another
test. Let’s try the root test.
Step 4: Compute the root test:
R= lim
n→∞
n
s
n!
nn
R= lim
n→∞
n
√n!
n
Step 5: Apply Stirling’s approximation to simplify the root test.
R= lim
n→∞
n
e
n= lim
n→∞
1
e
Step 6: Conclude the convergence of the series: Since 1
e<1, the series
P∞
n=1
n!
nnconverges by the root test.
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 P∞
n=1
n!
nn, we will use
the ratio test.
Step 1: Calculate the ratio R.
R= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Simplify the ratio R.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
R= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
17
R= lim
n→∞
n+ 1
(n+ 1)n+1
R= lim
n→∞
1
(1 + 1/n)n
Step 3: Find the limit of R.
R= lim
n→∞
1
(1 + 1/n)n=1
e
where e≈2.71828 is Euler’s number.
Step 4: Determine convergence or divergence based on the ratio R. Since
R=1
e<1, by the ratio test, the series P∞
n=1
n!
nnconverges.
Question 23
Question
Determine the convergence of the series
∞
X
n=1
5n+ 3n
2·4n
Solution
Let’s start by simplifying the series.
Step 1: Rewrite the series in a more convenient form.
∞
X
n=1
5n+ 3n
2·4n=∞
X
n=1
1
25
4n
+1
23
4n
Step 2: Split the series into two separate series.
∞
X
n=1
1
25
4n
+∞
X
n=1
1
23
4n
Step 3: Check the convergence of each series separately using the ratio test.
For the first series,
lim
n→∞
an+1
an
= lim
n→∞
1
25
4n+1
·2·4n
1=5
4lim
n→∞ 5
4n
=5
4·0=0
Since the limit is less than 1, the series P∞
n=1 1
25
4nconverges.
For the second series,
lim
n→∞
bn+1
bn
= lim
n→∞
1
23
4n+1
·2·4n
1=3
4lim
n→∞ 3
4n
=3
4·0=0
Since the limit is less than 1, the series P∞
n=1 1
23
4nconverges.
Step 4: Since both series converge, the original series converges.
18
Question 24
Question
Determine the convergence or divergence of the series P∞
n=1
n2
3n.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Apply the ratio test. Let an=n2
3n. We will compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/3n+1
n2/3n
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)2
3(n2)
= lim
n→∞
n2+ 2n+ 1
3n2
= lim
n→∞
1 + 2
n+1
n2
3
=1
3
Step 3: Analyze the limit. Since L < 1, by the ratio test, the series P∞
n=1
n2
3n
converges.
Therefore, the given series is convergent.
Question 25
Question
Determine whether the series P∞
n=1
n2+1
2n3−n+5 converges or diverges.
Solution
Step 1: We will use the Limit Comparison Test to determine the convergence
of the series. Let’s consider the series P∞
n=1
n2+1
2n3−n+5 and the series P∞
n=1 1
n.
Step 2: We will find the limit of the ratio of the two series as napproaches
infinity:
lim
n→∞
n2+1
2n3−n+5
1
n
= lim
n→∞
n3+n
2n3−n+ 5.
Step 3: Simplifying the expression, we get:
lim
n→∞
n3+n
2n3−n+ 5 = lim
n→∞
1 + 1
n2
2−1
n2+5
n3
.
Step 4: Since the highest power of nin the denominator is n3, the limit
converges to 1
2.
Step 5: Since P∞
n=1 1
nis a harmonic series which diverges, and the limit of
the ratio is a positive finite number, by the Limit Comparison Test, the series
P∞
n=1
n2+1
2n3−n+5 also diverges.
19
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
1
n(ln n)p
where pis a positive constant.
Solution
Step 1: To determine the convergence of the series, we will use the Integral Test.
Let
f(x) = 1
x(ln x)p
be a continuous, positive, and decreasing function for x≥3 (since ln xis positive
for x > 1).
Step 2: We will consider the improper integral
Z∞
3
1
x(ln x)pdx
Step 3: Let u= ln xsuch that du =1
xdx, then the integral becomes
Z∞
3
1
x(ln x)pdx =Z∞
ln(3)
1
updu = lim
t→∞ Zt
ln(3)
u−pdu
Step 4: The antiderivative of u−pis u−p+1
−p+1 , thus the integral becomes
lim
t→∞ t−p+1
−p+ 1 −(ln(3))−p+1
−p+ 1
Step 5: The integral converges if and only if the limit above is finite. This
limit is finite if and only if p > 1 by p-series test.
Step 6: Therefore, the series
∞
X
n=1
1
n(ln n)p
converges if p > 1 and diverges otherwise.
Question 27
Question
Determine the convergence of the series
∞
X
n=1
n2
2n
20
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Recall that the ratio test states that if limn→∞
an+1
an<1, then the series Pan
converges absolutely.
Step 2: Let an=n2
2n. We compute the ratio an+1
an:
an+1
an
=(n+ 1)2/2n+1
n2/2n=(n+ 1)2
2n2
Step 3: Simplifying the ratio further:
(n+ 1)2
2n2=n2+ 2n+ 1
2n2=1
2+1
n+1
2n2
Step 4: As napproaches infinity, the terms 1
nand 1
2n2shrink to zero leaving
the ratio as 1
2.
Step 5: Since 1
2<1, by the ratio test, the series Pn2
2nconverges absolutely.
Question 28
Question
Consider the series P∞
n=1
n2+1
n3+2 . Determine whether the series converges or
diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n3+2 , we can use the limit
comparison test.
Step 1: Let’s find a series bnsuch that limn→∞
an
bn=L > 0, where an=
n2+1
n3+2 .
Step 2: Simplify the expression an
bn. Let bn=1
n. Then,
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 2 ·n= lim
n→∞
n3+n
n3+ 2n= 1
Step 3: Since 0 < L < ∞, both series P∞
n=1
n2+1
n3+2 and P∞
n=1 1
neither both
converge or both diverge. Since P∞
n=1 1
ndiverges (it’s actually the harmonic
series which is a divergent p-series with p= 1), the given series P∞
n=1
n2+1
n3+2 also
diverges by the limit comparison test.
Therefore, the series P∞
n=1
n2+1
n3+2 diverges.
21
Question 29
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 R:
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→∞ 1 + 1
nn
=e,
where we have used the limit definition of e.
Step 2: Evaluate the limit R: Since R=e > 1, the series P∞
n=1
n!
nndiverges
by the Ratio Test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 30
Question
Let (an) be a sequence such that limn→∞
an+1
an=L. Determine the convergence
or divergence of the series P∞
n=1 anbased on the value of L.
Solution
We will use the Ratio Test to determine the convergence or divergence of the
series P∞
n=1 an.
Step 1: Ratio Test Consider the series P∞
n=1 an. Let L= limn→∞
an+1
an.
Step 2: Convergence for L < 1 If L < 1, then the series P∞
n=1 an
converges absolutely.
22
Step 3: Divergence for L > 1or L=∞If L > 1 or L=∞, then the
series P∞
n=1 andiverges.
Step 4: Inconclusive for L= 1 If L= 1, the Ratio Test is inconclusive.
Therefore, the series P∞
n=1 anconverges if L < 1, diverges if L > 1 or L=∞,
and the test is inconclusive if L= 1.
Question 31
Question
Determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+n+ 1.
Solution
To determine the convergence or divergence of the given series, we will use the
limit comparison test with the series ∞
X
n=1
1
n.
Step 1: Compute the limit of the ratio of the two series. Let an=
√n3+ 1
n2+n+ 1 and bn=1
n. We will consider the limit
lim
n→∞
an
bn
= lim
n→∞
√n3+1
n2+n+1
1
n
.
Step 2: Simplify the expression.
lim
n→∞
√n3+ 1
n2+n+ 1 ·n
1= lim
n→∞
n√n3+ 1
n2+n+ 1.
Step 3: Use the limit comparison test. As lim
n→∞
n√n3+ 1
n2+n+ 1 = lim
n→∞
√n3+ 1
n+ 1 =
1 (by dividing the numerator and denominator by n2and taking the limit), we
can conclude that ∞
X
n=1
√n3+ 1
n2+n+ 1 converges by the limit comparison test with
the convergent series ∞
X
n=1
1
n.
Therefore, the original series ∞
X
n=1
√n3+ 1
n2+n+ 1 converges.
23
Question 32
Question
For the series P∞
n=1(−1)nn
n2+1 , determine whether it converges absolutely, con-
verges conditionally, or diverges.
Solution
To determine the convergence of the series P∞
n=1(−1)nn
n2+1 , we will first exam-
ine the absolute convergence by considering the series P∞
n=1 (−1)nn
n2+1 .
Step 1: Find the absolute value of the terms in the series.
(−1)nn
n2+ 1
=n
n2+ 1
Step 2: Determine the convergence of the series P∞
n=1
n
n2+1 using the limit
comparison test with P∞
n=1 1
n.
We have:
lim
n→∞
n
n2+1
1
n
= lim
n→∞
n2
n2+ 1 = 1
Therefore, by the limit comparison test, since P∞
n=1 1
ndiverges (harmonic
series), the series P∞
n=1
n
n2+1 also diverges.
Step 3: Since the absolute series diverges, we need to investigate whether
the original series P∞
n=1(−1)nn
n2+1 converges conditionally.
Step 4: Check if the series satisfies the conditions of the alternating series
test.
The terms are alternately positive and negative, and the absolute values of
the terms decrease monotonically to zero as nincreases.
Step 5: Therefore, by the alternating series test, the series P∞
n=1(−1)nn
n2+1
converges conditionally.
Question 33
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 can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. Then, the Ratio Test states
that if limn→∞
an+1
an<1, then the series P∞
n=1 anconverges; if limn→∞
an+1
an>
1, then the series diverges; and if limn→∞
an+1
an= 1, the test is inconclusive.
24
Step 2: Compute limn→∞
an+1
an.
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
Step 3: Simplify the expression and evaluate the limit.
= lim
n→∞ 1 + 1
n n
n+ 1n
= lim
n→∞ 1 + 1
n 1
1+1/nn
= 1 ·1
e=1
e
Step 4: Analyze the limit. Since 1
e<1, by the Ratio Test, the series
P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 34
Question
Determine whether the series P∞
n=1
n2+1
n3+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n3+1 , we can use the limit
comparison test. Let’s choose the series P∞
n=1 1
nas our comparison series.
Step 1: Find the limit of the ratio of the given series and the
comparison series. Consider the limit:
lim
n→∞
n2+1
n3+1
1
n
Simplify the expression:
lim
n→∞
n3+n
n3+ 1 = lim
n→∞
1 + 1
n2
1 + 1
n3
= 1
25
Step 2: Conclusion from the limit comparison test. Since the limit
is a finite positive value, by the limit comparison test, we can conclude that the
series P∞
n=1
n2+1
n3+1 behaves the same as P∞
n=1 1
n.
Step 3: Determine the convergence of the comparison series. The
series P∞
n=1 1
nis the harmonic series which is known to diverge.
Step 4: Final conclusion. Since the given series behaves the same as
the divergent harmonic series, the series P∞
n=1
n2+1
n3+1 also diverges by the limit
comparison test.
Question 35
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the limit of the ratio test. Let an=n!
nn. Then, the ratio
test gives us:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
= lim
n→∞
n+ 1
(1 + 1
n)n·(n+ 1)
Step 2: Simplify the expression inside the limit.
lim
n→∞
n+ 1
(1 + 1
n)n·(n+ 1)
= lim
n→∞
1
(1 + 1
n)n
Step 3: Evaluate the limit.
lim
n→∞
1
(1 + 1
n)n
=1
e
where we used the fact that the limit of 1 + 1
nnis eas napproaches infinity.
Step 4: Determine the convergence of the series. Since 1
e<1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
26
Since the terms are decreasing and the limit of the terms is 0, by the Alter-
nating Series Test, the series ∞
X
n=1
(−1)n
nln(n+ 1) converges.
Question 2
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: Apply the ratio test. Let an=n!
nn. We will calculate the following
limit:
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
Step 2: Simplify the expression.
= lim
n→∞
n+ 1
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 3: Analyze the limit. Since 1
e<1, by the ratio test, the series P∞
n=1
n!
nn
converges.
Therefore, the series P∞
n=1
n!
nnconverges by the ratio test.
Question 3
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n+ 2
5n.
2
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Let an=3n+2
5n.
Step 1: Calculate the ratio R.
R= lim
n→∞
an+1
an
= lim
n→∞
3n+1+2
5n+1
3n+2
5n
= lim
n→∞
3n+1 + 2
5n+1 ·5n
3n+ 2
= lim
n→∞
3
5·5n
3n+ 2
=3
5lim
n→∞
5n
3n+ 2
=3
5.
Step 2: Determine the convergence or divergence of the series. Since the
ratio R=3
5<1, by the ratio test, the series P∞
n=1 3n+2
5nconverges.
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 can use the Ratio Test.
Step 1: Calculate the ratio limn→∞
an+1
anwhere an=n!
nn.
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+1
(n+ 1)n
= lim
n→∞
n
(1 + 1
n)n
= lim
n→∞
n
e=∞
Since the limit is greater than 1, we can conclude that the series P∞
n=1
n!
nn
diverges by the Ratio Test.
Question 5
Question
Determine the convergence or divergence of the series P∞
n=1
n2
2n.
3
Solution
To determine the convergence of the series P∞
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test by calculating the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2·2n
2n+1 ·n2
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)2·2n
2n+1 ·n2
= lim
n→∞
(n+ 1)2
2·n2
Step 3: Calculate the limit:
lim
n→∞
(n+ 1)2
2·n2
= lim
n→∞
n2+ 2n+ 1
2n2=1
2
Step 4: Analyze the limit: Since the limit is 1
2which is less than 1, by the
ratio test, the series P∞
n=1
n2
2nconverges.
Therefore, the series P∞
n=1
n2
2nconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
3n3+ 2.
Solution
To determine the convergence or divergence of the given series, we will use the
Limit Comparison Test.
Step 1: Let’s find the limit of the general term:
lim
n→∞
n2+ 1
3n3+ 2.
Step 2: We simplify the limit by dividing the numerator and denominator
by n3:
lim
n→∞
n2+ 1
3n3+ 2 = lim
n→∞
1
n+1
n3
3 + 2
n3
=0+0
3+0 = 0.
Therefore, since the limit of the general term is finite and positive, we can
apply the Limit Comparison Test.
4
Step 3: We choose the series P∞
n=1 1
nas a comparison series as it is a known
divergent p-series with p= 1.
Step 4: Let’s find the limit of the ratio between the general term of the
given series and the general term of the comparison series:
lim
n→∞
n2+1
3n3+2
1
n
= lim
n→∞
n3+n
3n3+ 2n= lim
n→∞
n(1 + 1
n2)
n(3 + 2
n2)=1
3.
Step 5: Since the limit is a positive finite number, and the comparison
series diverges, by the Limit Comparison Test, the given series P∞
n=1
n2+1
3n3+2 also
diverges.
Question 7
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 calculate the limit:
L= lim
n→∞
an+1
an
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression to find the limit.
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
L= lim
n→∞
n+ 1
(n+ 1)n+1 ·n
L= lim
n→∞
n
(n+ 1)n
L= lim
n→∞
1
(1 + 1/n)n
Step 3: Recall the limit limn→∞(1 + 1/n)n=e.
L=
1
e
=1
e
5
Step 4: Analyze the limit. Since 0 <1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 8
Question
Determine the convergence of the series P∞
n=1
n!
nn.
Solution
To determine the convergence of the series, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. We calculate 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!
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Use the limit of (1 + 1
n)n. As napproaches infinity, limn→∞(1 +
1
n)n=e, where eis Euler’s number.
Step 4: Determine the limit.
lim
n→∞
1
e
=1
e
Step 5: Apply the Ratio Test. Since the limit is less than 1, by the Ratio
Test, the series P∞
n=1
n!
nnconverges.
Question 9
Question
Determine the convergence or divergence of the series P∞
n=1
n2
2n.
6
Solution
To determine the convergence or divergence of the series P∞
n=1
n2
2n, we will use
the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n2
2n. We will consider the limit of
the ratio limn→∞
an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞ 1
2+1
n+1
2n2
=1
2
Step 2: Evaluate the limit. Since the limit is 1
2<1, by the Ratio Test, the
series P∞
n=1
n2
2nconverges.
Therefore, the series P∞
n=1
n2
2nconverges.
Question 10
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: 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
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
n
(1 + 1/n)n+1
7
Step 2: Simplify the expression inside the limit: Using the fact that limn→∞ 1 + 1
nn=
e, we have:
lim
n→∞
n
(1 + 1/n)n+1
= lim
n→∞
n
(1 + 1/n)(1 + 1/n)n
= lim
n→∞
n
(1 + 1/n)e
=1
elim
n→∞ |n|=∞
Step 3: Determine the convergence or divergence: Since the limit of the
ratio test is ∞, the series P∞
n=1
n!
nndiverges.
Therefore, the series diverges.
Question 11
Question
Determine whether the series P∞
n=1 3n2+1
n4+2 converges or diverges.
Solution
To determine whether the series converges or diverges, we will use the Limit
Comparison Test. Let’s denote the n-th term of the series as an=3n2+1
n4+2 .
Step 1: Find a series Pbnthat is easier to work with. We will choose
bn=1
n2since this is a known convergent series.
Step 2: Compute the limit of the ratio anbn.
lim
n→∞
an
bn
= lim
n→∞
3n2+ 1
n4+ 2 ·n2
1= lim
n→∞
3 + 1
n2
1 + 2
n4
=3
1= 3
Step 3: Make a conclusion based on the limit. Since the limit is a finite
positive number, by the Limit Comparison Test, the convergence of P∞
n=1 anis
the same as the convergence of P∞
n=1 bn. Since P∞
n=1 1
n2is convergent (by the
p-series test with p= 2 >1), our original series P∞
n=1 3n2+1
n4+2 also converges.
Question 12
Question
Determine whether the series ∞
X
n=1
n2+ 1
n3+ 1 converges or diverges.
8
Solution
To determine the convergence of the given series, we can use the limit compar-
ison test.
Step 1: Let’s choose a series bnthat we know converges. We will use the
harmonic series ∞
X
n=1
1
n.
Step 2: Find the limit limn→∞
n2+1
n3+1
1
n
.
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1
= lim
n→∞
1 + 1
n2
1 + 1
n3
=1+0
1+0
= 1
Step 3: Since the limit is a positive finite number, we can conclude that the
given series ∞
X
n=1
n2+ 1
n3+ 1 behaves similarly to the harmonic series which diverges.
Therefore, by the limit comparison test, ∞
X
n=1
n2+ 1
n3+ 1 also diverges.
Question 13
Question
Determine the convergence or divergence of the series
∞
X
n=1
√n3+ 1
n2+ 1 .
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s choose a series to compare. We consider the series P∞
n=1 1
n,
which is a harmonic series that diverges.
Step 2: We will calculate the limit
lim
n→∞
√n3+1
n2+1
1
n
.
9
Step 3: Simplifying the expression, we have
lim
n→∞
√n3+1
n2+1
1
n
= lim
n→∞
n√n3+ 1
n(n2+ 1) = lim
n→∞
√n3+ 1
n2+ 1 .
Step 4: Let’s find the limit of the expression. Since
lim
n→∞
√n3+ 1
n2+ 1 = lim
n→∞
√n3+ 1/n2
1+1/n2,
we apply L’Hˆopital’s Rule to the numerator to get
lim
n→∞
√n3+ 1
n2+ 1 = lim
n→∞
(3n2/2)/(2√n3+ 1)
−2/n3= lim
n→∞ −3n4√n3+ 1
4(n3+ 1) =−3
4.
Step 5: Since the limit is finite and non-zero, by the Limit Comparison
Test, since the limit is a constant between 0 and ∞, the series P∞
n=1
√n3+1
n2+1
diverges.
Question 14
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: Apply the Ratio Test. Let an=n!
nn. We will consider the limit
L= lim
n→∞
an+1
an
10
L= 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+1/n
= 1
Step 2: Evaluate the limit L. Since L= 1, the Ratio Test is inconclusive.
Therefore, we cannot determine the convergence or divergence of the series from
this test alone.
Step 3: Further analysis. Notice that lim
n→∞
n!
nn= 0 by Stirling’s approx-
imation. Therefore, n!
nnapproaches 0 as napproaches infinity and the series
converges by the Test for Divergence.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 15
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: Calculate the ratio Let an=n!
nn. We compute the ratio Ras
follows:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio Simplifying the ratio, we have:
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
11
R= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Evaluate the limit Using the fact that limn→∞ 1 + 1
nn=e, we
get:
R=
1
e
=1
e
Step 4: Conclusion Since R=1
e<1, by the ratio test, the series P∞
n=1
n!
nn
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: Compute the ratio of consecutive terms. Let an=n!
nn. We compute
limn→∞
an+1
an:
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
Step 2: Simplify and analyze the limit. Using L’Hopital’s Rule, we can
simplify the limit further:
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e
Step 3: Determine the convergence of the series. Since the limit is less
than 1, by the Ratio Test, the series P∞
n=1
n!
nnconverges. Therefore, the series
converges.
Question 17
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2·2n
n!
12
Solution
To determine the convergence or divergence of the series, we will use the Ratio
Test. The Ratio Test states that if limn→∞
an+1
an<1, then the series converges;
if the limit is greater than 1, then the series diverges; and if the limit equals 1,
the test is inconclusive.
Step 1: Let’s compute the ratio
an+1
an∥.
an+1
an
=
(n+ 1)2·2n+1
(n+ 1)! ·n!
n2·2n
=
(n+ 1)2·2
(n+ 1)(n)·2
=
n+ 1
n
= 1 + 1
n
Step 2: Now, take the limit of the ratio as napproaches infinity.
lim
n→∞ 1 + 1
n= 1
Step 3: Since the limit is equal to 1, the Ratio Test is inconclusive. We will
perform further analysis.
Step 4: We observe that the terms in the series decrease in magnitude for
large n.
n2·2n
n!=n2
1·2n
n·(n−1) ···2·1≈0
as nbecomes large.
Step 5: Since the terms of the series approach zero for large n, we can
conclude that the series converges by the Ratio Test.
Question 18
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 can use the Ratio Test.
13
Step 1: Apply the Ratio Test. Let an=n!
nn. We will consider the limit:
L= 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→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
=1
e.
Step 2: Analyze the limit. Since L=1
eis a finite positive number, we can
conclude that the series ∞
X
n=1
n!
nnconverges by the Ratio Test.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 19
Question
Determine whether the series ∞
X
n=1
n2
2nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the Ratio Test.
The Ratio Test states that if limn→∞
an+1
an
=L, then: - If L < 1, the series
converges absolutely. - If L > 1 or L=∞, the series diverges. - If L= 1, the
test is inconclusive.
Step 1: Compute the ratio an+1
an
. Let an=n2
2n. Then an+1 =(n+ 1)2
2n+1 .
Hence, an+1
an
=(n+ 1)2
2n+1 ·2n
n2=(n+ 1)2
2n2.
14
Step 2: Take the limit as napproaches infinity.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
1 + 1
n
2
=1
2
Step 3: Determine the convergence of the series. Since 1
2<1, by the Ratio
Test, the series ∞
X
n=1
n2
2nconverges absolutely.
Question 20
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, we will use the Limit
Comparison Test. Let’s consider the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3
and compare it with the series
∞
X
n=1
1
n.
Step 1: Calculate the following limit:
lim
n→∞
(n3+ 2n+ 1)/n4
1/n = lim
n→∞
n3+ 2n+ 1
n3= 1.
Step 2: Since the limit is a positive finite value, by the Limit Comparison
Test, the series
∞
X
n=1
n3+ 2n+ 1
n4+ 3
has the same convergence behavior as
∞
X
n=1
1
n.
15
Step 3: The series
∞
X
n=1
1
n
is known to be a harmonic series, which diverges.
Therefore, by the Limit Comparison Test, the given series
∞
X
n=1
n3+ 2n+ 1
n4+ 3
also diverges.
Question 21
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n!
nn,
we will use the ratio test. Let’s denote the general term of the series as an=n!
nn.
Step 1: Compute the ratio Rof consecutive terms:
R= lim
n→∞
an+1
an
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
R= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
R= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
Step 2: Simplify the expression by expanding and canceling terms:
R= lim
n→∞
nn+1 +nn
nn+1 +n·nn
16
R= lim
n→∞
1+1/n
1+1/n
R= lim
n→∞ 1=1
Step 3: Apply the ratio test: - If R < 1, the series converges. - If R > 1,
the series diverges. - If R= 1, the ratio test is inconclusive.
Since R= 1, the ratio test is inconclusive. Therefore, we need to try another
test. Let’s try the root test.
Step 4: Compute the root test:
R= lim
n→∞
n
s
n!
nn
R= lim
n→∞
n
√n!
n
Step 5: Apply Stirling’s approximation to simplify the root test.
R= lim
n→∞
n
e
n= lim
n→∞
1
e
Step 6: Conclude the convergence of the series: Since 1
e<1, the series
P∞
n=1
n!
nnconverges by the root test.
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 P∞
n=1
n!
nn, we will use
the ratio test.
Step 1: Calculate the ratio R.
R= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Simplify the ratio R.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
R= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
17
R= lim
n→∞
n+ 1
(n+ 1)n+1
R= lim
n→∞
1
(1 + 1/n)n
Step 3: Find the limit of R.
R= lim
n→∞
1
(1 + 1/n)n=1
e
where e≈2.71828 is Euler’s number.
Step 4: Determine convergence or divergence based on the ratio R. Since
R=1
e<1, by the ratio test, the series P∞
n=1
n!
nnconverges.
Question 23
Question
Determine the convergence of the series
∞
X
n=1
5n+ 3n
2·4n
Solution
Let’s start by simplifying the series.
Step 1: Rewrite the series in a more convenient form.
∞
X
n=1
5n+ 3n
2·4n=∞
X
n=1
1
25
4n
+1
23
4n
Step 2: Split the series into two separate series.
∞
X
n=1
1
25
4n
+∞
X
n=1
1
23
4n
Step 3: Check the convergence of each series separately using the ratio test.
For the first series,
lim
n→∞
an+1
an
= lim
n→∞
1
25
4n+1
·2·4n
1=5
4lim
n→∞ 5
4n
=5
4·0=0
Since the limit is less than 1, the series P∞
n=1 1
25
4nconverges.
For the second series,
lim
n→∞
bn+1
bn
= lim
n→∞
1
23
4n+1
·2·4n
1=3
4lim
n→∞ 3
4n
=3
4·0=0
Since the limit is less than 1, the series P∞
n=1 1
23
4nconverges.
Step 4: Since both series converge, the original series converges.
18
Question 24
Question
Determine the convergence or divergence of the series P∞
n=1
n2
3n.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Apply the ratio test. Let an=n2
3n. We will compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/3n+1
n2/3n
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)2
3(n2)
= lim
n→∞
n2+ 2n+ 1
3n2
= lim
n→∞
1 + 2
n+1
n2
3
=1
3
Step 3: Analyze the limit. Since L < 1, by the ratio test, the series P∞
n=1
n2
3n
converges.
Therefore, the given series is convergent.
Question 25
Question
Determine whether the series P∞
n=1
n2+1
2n3−n+5 converges or diverges.
Solution
Step 1: We will use the Limit Comparison Test to determine the convergence
of the series. Let’s consider the series P∞
n=1
n2+1
2n3−n+5 and the series P∞
n=1 1
n.
Step 2: We will find the limit of the ratio of the two series as napproaches
infinity:
lim
n→∞
n2+1
2n3−n+5
1
n
= lim
n→∞
n3+n
2n3−n+ 5.
Step 3: Simplifying the expression, we get:
lim
n→∞
n3+n
2n3−n+ 5 = lim
n→∞
1 + 1
n2
2−1
n2+5
n3
.
Step 4: Since the highest power of nin the denominator is n3, the limit
converges to 1
2.
Step 5: Since P∞
n=1 1
nis a harmonic series which diverges, and the limit of
the ratio is a positive finite number, by the Limit Comparison Test, the series
P∞
n=1
n2+1
2n3−n+5 also diverges.
19
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
1
n(ln n)p
where pis a positive constant.
Solution
Step 1: To determine the convergence of the series, we will use the Integral Test.
Let
f(x) = 1
x(ln x)p
be a continuous, positive, and decreasing function for x≥3 (since ln xis positive
for x > 1).
Step 2: We will consider the improper integral
Z∞
3
1
x(ln x)pdx
Step 3: Let u= ln xsuch that du =1
xdx, then the integral becomes
Z∞
3
1
x(ln x)pdx =Z∞
ln(3)
1
updu = lim
t→∞ Zt
ln(3)
u−pdu
Step 4: The antiderivative of u−pis u−p+1
−p+1 , thus the integral becomes
lim
t→∞ t−p+1
−p+ 1 −(ln(3))−p+1
−p+ 1
Step 5: The integral converges if and only if the limit above is finite. This
limit is finite if and only if p > 1 by p-series test.
Step 6: Therefore, the series
∞
X
n=1
1
n(ln n)p
converges if p > 1 and diverges otherwise.
Question 27
Question
Determine the convergence of the series
∞
X
n=1
n2
2n
20
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Recall that the ratio test states that if limn→∞
an+1
an<1, then the series Pan
converges absolutely.
Step 2: Let an=n2
2n. We compute the ratio an+1
an:
an+1
an
=(n+ 1)2/2n+1
n2/2n=(n+ 1)2
2n2
Step 3: Simplifying the ratio further:
(n+ 1)2
2n2=n2+ 2n+ 1
2n2=1
2+1
n+1
2n2
Step 4: As napproaches infinity, the terms 1
nand 1
2n2shrink to zero leaving
the ratio as 1
2.
Step 5: Since 1
2<1, by the ratio test, the series Pn2
2nconverges absolutely.
Question 28
Question
Consider the series P∞
n=1
n2+1
n3+2 . Determine whether the series converges or
diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n3+2 , we can use the limit
comparison test.
Step 1: Let’s find a series bnsuch that limn→∞
an
bn=L > 0, where an=
n2+1
n3+2 .
Step 2: Simplify the expression an
bn. Let bn=1
n. Then,
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 2 ·n= lim
n→∞
n3+n
n3+ 2n= 1
Step 3: Since 0 < L < ∞, both series P∞
n=1
n2+1
n3+2 and P∞
n=1 1
neither both
converge or both diverge. Since P∞
n=1 1
ndiverges (it’s actually the harmonic
series which is a divergent p-series with p= 1), the given series P∞
n=1
n2+1
n3+2 also
diverges by the limit comparison test.
Therefore, the series P∞
n=1
n2+1
n3+2 diverges.
21
Question 29
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 R:
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→∞ 1 + 1
nn
=e,
where we have used the limit definition of e.
Step 2: Evaluate the limit R: Since R=e > 1, the series P∞
n=1
n!
nndiverges
by the Ratio Test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 30
Question
Let (an) be a sequence such that limn→∞
an+1
an=L. Determine the convergence
or divergence of the series P∞
n=1 anbased on the value of L.
Solution
We will use the Ratio Test to determine the convergence or divergence of the
series P∞
n=1 an.
Step 1: Ratio Test Consider the series P∞
n=1 an. Let L= limn→∞
an+1
an.
Step 2: Convergence for L < 1 If L < 1, then the series P∞
n=1 an
converges absolutely.
22
Step 3: Divergence for L > 1or L=∞If L > 1 or L=∞, then the
series P∞
n=1 andiverges.
Step 4: Inconclusive for L= 1 If L= 1, the Ratio Test is inconclusive.
Therefore, the series P∞
n=1 anconverges if L < 1, diverges if L > 1 or L=∞,
and the test is inconclusive if L= 1.
Question 31
Question
Determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+n+ 1.
Solution
To determine the convergence or divergence of the given series, we will use the
limit comparison test with the series ∞
X
n=1
1
n.
Step 1: Compute the limit of the ratio of the two series. Let an=
√n3+ 1
n2+n+ 1 and bn=1
n. We will consider the limit
lim
n→∞
an
bn
= lim
n→∞
√n3+1
n2+n+1
1
n
.
Step 2: Simplify the expression.
lim
n→∞
√n3+ 1
n2+n+ 1 ·n
1= lim
n→∞
n√n3+ 1
n2+n+ 1.
Step 3: Use the limit comparison test. As lim
n→∞
n√n3+ 1
n2+n+ 1 = lim
n→∞
√n3+ 1
n+ 1 =
1 (by dividing the numerator and denominator by n2and taking the limit), we
can conclude that ∞
X
n=1
√n3+ 1
n2+n+ 1 converges by the limit comparison test with
the convergent series ∞
X
n=1
1
n.
Therefore, the original series ∞
X
n=1
√n3+ 1
n2+n+ 1 converges.
23
Question 32
Question
For the series P∞
n=1(−1)nn
n2+1 , determine whether it converges absolutely, con-
verges conditionally, or diverges.
Solution
To determine the convergence of the series P∞
n=1(−1)nn
n2+1 , we will first exam-
ine the absolute convergence by considering the series P∞
n=1 (−1)nn
n2+1 .
Step 1: Find the absolute value of the terms in the series.
(−1)nn
n2+ 1
=n
n2+ 1
Step 2: Determine the convergence of the series P∞
n=1
n
n2+1 using the limit
comparison test with P∞
n=1 1
n.
We have:
lim
n→∞
n
n2+1
1
n
= lim
n→∞
n2
n2+ 1 = 1
Therefore, by the limit comparison test, since P∞
n=1 1
ndiverges (harmonic
series), the series P∞
n=1
n
n2+1 also diverges.
Step 3: Since the absolute series diverges, we need to investigate whether
the original series P∞
n=1(−1)nn
n2+1 converges conditionally.
Step 4: Check if the series satisfies the conditions of the alternating series
test.
The terms are alternately positive and negative, and the absolute values of
the terms decrease monotonically to zero as nincreases.
Step 5: Therefore, by the alternating series test, the series P∞
n=1(−1)nn
n2+1
converges conditionally.
Question 33
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 can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. Then, the Ratio Test states
that if limn→∞
an+1
an<1, then the series P∞
n=1 anconverges; if limn→∞
an+1
an>
1, then the series diverges; and if limn→∞
an+1
an= 1, the test is inconclusive.
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Step 2: Compute limn→∞
an+1
an.
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
Step 3: Simplify the expression and evaluate the limit.
= lim
n→∞ 1 + 1
n n
n+ 1n
= lim
n→∞ 1 + 1
n 1
1+1/nn
= 1 ·1
e=1
e
Step 4: Analyze the limit. Since 1
e<1, by the Ratio Test, the series
P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 34
Question
Determine whether the series P∞
n=1
n2+1
n3+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n3+1 , we can use the limit
comparison test. Let’s choose the series P∞
n=1 1
nas our comparison series.
Step 1: Find the limit of the ratio of the given series and the
comparison series. Consider the limit:
lim
n→∞
n2+1
n3+1
1
n
Simplify the expression:
lim
n→∞
n3+n
n3+ 1 = lim
n→∞
1 + 1
n2
1 + 1
n3
= 1
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Step 2: Conclusion from the limit comparison test. Since the limit
is a finite positive value, by the limit comparison test, we can conclude that the
series P∞
n=1
n2+1
n3+1 behaves the same as P∞
n=1 1
n.
Step 3: Determine the convergence of the comparison series. The
series P∞
n=1 1
nis the harmonic series which is known to diverge.
Step 4: Final conclusion. Since the given series behaves the same as
the divergent harmonic series, the series P∞
n=1
n2+1
n3+1 also diverges by the limit
comparison test.
Question 35
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the limit of the ratio test. Let an=n!
nn. Then, the ratio
test gives us:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
= lim
n→∞
n+ 1
(1 + 1
n)n·(n+ 1)
Step 2: Simplify the expression inside the limit.
lim
n→∞
n+ 1
(1 + 1
n)n·(n+ 1)
= lim
n→∞
1
(1 + 1
n)n
Step 3: Evaluate the limit.
lim
n→∞
1
(1 + 1
n)n
=1
e
where we used the fact that the limit of 1 + 1
nnis eas napproaches infinity.
Step 4: Determine the convergence of the series. Since 1
e<1, by the ratio
test, the series P∞
n=1
n!
nnconverges.
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