MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 5
Liberty University
Question 1
Question
Let P∞
n=1 anbe a convergent series. Show that the series P∞
n=1
√an
nis also
convergent.
Solution
Given that P∞
n=1 anis a convergent series, we know that limn→∞ an= 0.
Step 1: We will prove that limn→∞
√an
n= 0. Since limn→∞ an= 0, we
have limn→∞ √an= 0. This implies that limn→∞
√an
n= 0.
Step 2: We will use the Limit Comparison Test to show that P∞
n=1
√an
nis
convergent. Let bn=√an
n. Consider the series P∞
n=1 bn. We have limn→∞
bn
1/n =
limn→∞ n·√an= 0. Since P∞
n=1 1
nis a divergent harmonic series, and P∞
n=1 bn
satisfies the conditions of the Limit Comparison Test, it follows that P∞
n=1
√an
n
is convergent.
Question 2
Question
Determine whether the series P∞
n=1 n2+n
2n3+3 converges or diverges.
Solution
To determine the convergence of the given series, we can use the limit compar-
ison test. Let’s denote the general term of the series as an=n2+n
2n3+3 .
Step 1: Find a suitable series for comparison
We will compare the given series to the series P∞
n=1 1
n, which is a p-series with
p= 1.
Step 2: Compute the limit
Let’s calculate the limit of the ratio limn→∞
an
1
n
:
lim
n→∞
an
1
n
= lim
n→∞
n2+n
2n3+ 3 ·n= lim
n→∞
n3+n2
2n3+ 3n
Step 3: Simplify the expression
Simplify the ratio to evaluate the limit:
lim
n→∞
n3+n2
2n3+ 3n= lim
n→∞
1 + 1
n
2 + 3
n2
=1
2
Step 4: Make the comparison
Since the limit is a positive finite number, the series P∞
n=1 n2+n
2n3+3 and the series
P∞
n=1 1
neither both converge or both diverge by the limit comparison test.
Step 5: Determine convergence
Since P∞
n=1 1
nis a divergent p-series (p= 1), and P∞
n=1 n2+n
2n3+3 behaves the same
as this series, the given series also diverges.
Therefore, the series P∞
n=1 n2+n
2n3+3 diverges.
Question 3
Question
Determine whether the series ∞
X
n=1
n2+ 1
n3+ 2 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let an=n2+1
n3+2 be the general term of the series.
Step 2: We will find a series bnthat converges and is positive for all nsuch
that limn→∞
an
bnis a positive finite number.
Let’s consider the series bn=1
n.
Step 3: Now, we will calculate the limit of an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 2 ·n
1
2
= lim
n→∞
n3+n
n3+ 2 = lim
n→∞
1 + 1
n2
1 + 2
n3
=1+0
1+0 = 1
Since the limit is a positive finite number, by the Limit Comparison Test,
the given series ∞
X
n=1
n2+ 1
n3+ 2 converges if and only if the series ∞
X
n=1
1
nconverges.
Step 4: The series ∞
X
n=1
1
nis a p-series with p= 1, which is a divergent series.
Therefore, by the Limit Comparison Test, the given series ∞
X
n=1
n2+ 1
n3+ 2 also
diverges.
Question 4
Question
Determine whether the series ∞
X
n=1
n3
3nconverges or diverges.
Solution
Step 1: We will use the Ratio Test to determine the convergence of the series.
Let an=n3
3n.
Step 2: Compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/3n+1
n3/3n
= lim
n→∞
(n+ 1)3
3n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
3n3
= lim
n→∞
n3
3n3+3n2
3n3+3n
3n3+1
3n3
= lim
n→∞
1
3+1
n+1
n2+1
3n3
=1
3
Step 3: Since the limit is less than 1, by the Ratio Test, the series ∞
X
n=1
n3
3n
converges.
3
Question 5
Question
Determine the convergence or divergence of the series ∞
X
n=1
2n+ 3n
5n.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Calculate the limit of the ratio of successive terms:
lim
n→∞
an+1
an
= lim
n→∞
2n+1+3n+1
5n+1
2n+3n
5n
lim
n→∞
2n+1 + 3n+1
5n+1 ·5n
2n+ 3n
lim
n→∞
2(2n) + 3(3n)
5(5n)·5n
2n+ 3n
lim
n→∞
2(2n) + 3(3n)
5(2n+ 3n)
Step 2: Simplify the expression in the limit:
lim
n→∞
2(2n) + 3(3n)
5(2n+ 3n)= lim
n→∞
2(2n/5n) + 3(3n/5n)
2n/5n+ 3n/5n
lim
n→∞
2(2/5)n+ 3(3/5)n
2n/5n+ 3n/5n
lim
n→∞
0+0
0+0 = 0
Step 3: Analyze the limit: Since the limit of the ratio of successive terms
is less than 1, by the ratio test, ∞
X
n=1
2n+ 3n
5nconverges.
Question 6
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
4
Solution
To decide on the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of successive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!(nn)
(n+ 1)n+1n!
= lim
n→∞
n!(n+ 1)
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
(1 + 1
n)n
Step 3: Evaluate the limit using the properties of the exponential function:
lim
n→∞
1
(1 + 1
n)n
=1
e
Step 4: Determine the convergence based on the ratio test: Since the limit
1
eis less than 1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 7
Question
Determine the convergence or divergence of the series P∞
n=1
n2+ 1
n4+ 2.
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Find the limit. Let an=n2+ 1
n4+ 2. We will find limn→∞
an
1/n2.
lim
n→∞
an
1/n2= lim
n→∞
n2+ 1
n4+ 2 ·n2
1= lim
n→∞
n4+n2
n4+ 2 = 1
Step 2: Apply the limit comparison test. Since the limit of an
1/n2
is equal to 1, we can apply the limit comparison test and compare the series
P∞
n=1
n2+ 1
n4+ 2 with the series P∞
n=1
1
n2.
5
Step 3: Conclude about convergence. The series P∞
n=1
1
n2is a con-
vergent p-series with p= 2 >1. Since P∞
n=1
n2+ 1
n4+ 2 is positively terms and
its limit compared series converges, we can conclude that the original series
P∞
n=1
n2+ 1
n4+ 2 also converges by the limit comparison test.
Question 8
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n!
nn, we can use the ratio test.
Step 1: Apply the ratio test. 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
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1
n)n
Step 2: Simplify the expression. Since limn→∞(1 + 1
n)n=e, we have:
R= lim
n→∞
n
e=∞
Step 3: Conclude the convergence. Since R=∞>1, the series diverges
by the ratio test. Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 9
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
6
Solution
We will use the ratio test to determine the convergence of the series.
Step 1: Compute the limit of the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n
Step 3: Rewrite the limit in a form that is easier to evaluate:
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 4: Compute the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=e−1
Step 5: Analyze the ratio and conclude: Since the limit of the absolute
value of the ratio of consecutive terms is e−1<1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 10
Question
Let an=(−1)n
√n+1 . Determine whether the series P∞
n=1 anconverges or diverges.
Solution
To determine whether the series P∞
n=1 anconverges or diverges, we will use the
Alternating Series Test.
Step 1: Find the terms of the series. The terms of the series are given
by an=(−1)n
√n+1 .
Step 2: Check if the terms of the series satisfy the conditions of
the Alternating Series Test. We need to verify if the following conditions
are met: i) The terms anare positive. ii) The terms anare decreasing. iii)
limn→∞ an= 0.
Step 3: Check the conditions. i) Since 1
√n+1 >0 for all n, and (−1)n
also is positive for all n,an>0 for all n. ii) To check if the terms are decreasing,
7
we observe that an+1 −an=(−1)n+1
√n+2 −(−1)n
√n+1 . We simplify this expression and
note that it is always negative, meaning the terms are decreasing. iii) Finally,
limn→∞ an= limn→∞
(−1)n
√n+1 = 0.
Step 4: Conclusion. Since all conditions of the Alternating Series Test
are satisfied, the series P∞
n=1 anconverges.
Question 11
Question
Determine whether the series ∞
X
n=1
n2+ 1
√n4+n2+ 1 converges or diverges.
Solution
To investigate the convergence of the series ∞
X
n=1
n2+ 1
√n4+n2+ 1, we will use the
limit comparison test with a known convergent series.
Step 1: Identify a known convergent series. Consider the series
∞
X
n=1
1
n3/2. This series is a p-series with p=3
2>1, so it converges.
Step 2: Compute the limit of the ratio of the given series to the
known series. Let an=n2+1
√n4+n2+1 and bn=1
n3/2. We will calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n2+1
√n4+n2+1
1
n3/2
= lim
n→∞
n3/2(n2+ 1)
√n4+n2+ 1
Step 3: Simplify the expression. Simplify the limit expression:
lim
n→∞
n3/2(n2+ 1)
√n4+n2+ 1 = lim
n→∞
n5/2+n3/2
n2+n−1/2+n−3/2= lim
n→∞
n1/2+ 1
1 + n−3/2+n−5/2= 1
Step 4: Conclusion Since lim
n→∞
an
bn
= 1, and ∞
X
n=1
1
n3/2converges, by the
limit comparison test, the given series ∞
X
n=1
n2+ 1
√n4+n2+ 1 also converges.
Question 12
Question
Let P∞
n=1 n2+3n
4n3+n2+2 be the given series. Determine whether the series converges
or diverges.
8
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s choose a series to compare with the given series.
Step 1: Choose a Comparison Series Let’s consider the series P∞
n=1 1
n
as our comparison series.
Step 2: Find the Limit Let’s find the limit of the ratio of the terms of
the two series:
lim
n→∞
n2+3n
4n3+n2+2
1
n
= lim
n→∞
n3+ 3n2
4n3+n2+ 2n.
Step 3: Simplify the Limit Simplify the limit:
= lim
n→∞
n3(1 + 3
n)
n3(4 + 1
n+2
n2)= lim
n→∞
1 + 3
n
4 + 1
n+2
n2
=1
4.
Step 4: Conclusion Since the limit is a finite positive number, the given
series P∞
n=1 n2+3n
4n3+n2+2 converges by the limit comparison test with the series
P∞
n=1 1
n.
Question 13
Question
Let an=n2+3n
2nfor n≥1. Determine whether the series P∞
n=1 anconverges or
diverges.
Solution
To determine the convergence of the series P∞
n=1 an, we will use the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms.
lim
n→∞
an+1
an
= lim
n→∞
(n+1)2+3(n+1)
2n+1
n2+3n
2n
Step 2: Simplify the ratio.
= lim
n→∞
(n2+ 2n+ 1 + 3n+ 3) ·2n
(n2+ 3n)·2n+1
= lim
n→∞
(n2+ 5n+ 4) ·2n
2(n2+ 3n)·2n
= lim
n→∞
n2+ 5n+ 4
2n2+ 6n
9
Step 3: Find the limit.
= lim
n→∞
n2(1 + 5
n+4
n2)
n2(2 + 6
n)
= lim
n→∞
1 + 5
n+4
n2
2 + 6
n
=1
2
Step 4: Interpret the result. Since the limit is 1
2<1, by the ratio test, the
series P∞
n=1 anconverges.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. Let an=n!
nn.
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
n−n
=1
e,
where we used the fact that limn→∞ 1 + 1
nn=e.
Step 2: Apply the ratio test.
If R < 1, then the series P∞
n=1 anconverges.
If R > 1 or R=∞, then the series P∞
n=1 andiverges.
If R= 1, the test is inconclusive.
Since R=1
e<1, the series P∞
n=1 n!
nnconverges by the ratio test.
10
Question 15
Question
Determine whether the series ∞
X
n=1
(−1)n
n2+ (−1)nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
(−1)n
n2+ (−1)n, we will use the
Alternating Series Test.
Step 1: Determine the terms of the series The terms of the series are
given by an=(−1)n
n2+(−1)n.
Step 2: Verify the conditions of the Alternating Series Test We
need to verify that the sequence {an}is decreasing and that limn→∞ an= 0.
For an=(−1)n
n2+(−1)n, we have that an+1 −an=(−1)n+1
(n+1)2+(−1)n+1 −(−1)n
n2+(−1)n.
Simplifying, we get:
an+1 −an=(−1)n+1(n2+ (−1)n)−(−1)n((n+ 1)2+ (−1)n+1)
(n2+ (−1)n)((n+ 1)2+ (−1)n+1)
As (n+ 1)2−n2= 2n+ 1 >0 for all n, the denominator of an+1 −anis
positive. The numerator simplifies to 2n+ 1 >0, hence an+1 −an>0.
Thus, the sequence {an}is decreasing.
Next, we find limn→∞ an:
lim
n→∞
(−1)n
n2+ (−1)n= 0
Step 3: Conclude using the Alternating Series Test Since the se-
quence {an}is decreasing and limn→∞ an= 0, the series ∞
X
n=1
(−1)n
n2+ (−1)ncon-
verges by the Alternating Series Test.
Question 16
Question
Determine the convergence or divergence of the series P∞
n=1 n3
2n.
11
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
2n, we will use
the Ratio Test.
Step 1: Compute the limit involved in the Ratio Test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
Step 2: Simplify the above limit:
lim
n→∞
(n+ 1)3
2(n3)
= 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: If the limit in Step 2 is less than 1, then the
series converges. Since 1
2<1, the series P∞
n=1 n3
2nconverges.
Therefore, the series P∞
n=1 n3
2nconverges.
Question 17
Question
Determine whether the series P∞
n=1 n2+3n
2n4+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series P∞
n=1 n2+3n
2n4+5 .
Step 1: Find the limit of the ratio of the given series to a known series. Let
an=n2+3n
2n4+5 and bn=1
n2. We will find the limit of an
bnas napproaches infinity.
lim
n→∞
an
bn
= lim
n→∞
n2+3n
2n4+5
1
n2
= lim
n→∞
n4+ 3n3
2n2+ 5n2= lim
n→∞
1 + 3
n
2 + 5
n2
=1
2
Step 2: Determine convergence. Since the limit of an
bnis a finite positive
number, both series either converge or diverge. Since P∞
n=1 1
n2converges (it is
ap-series with p= 2 >1), by the limit comparison test, the series P∞
n=1 n2+3n
2n4+5
also converges.
Therefore, the series P∞
n=1 n2+3n
2n4+5 converges.
Question 18
Question
Determine the convergence or divergence of the series P∞
n=1 3n+2n
2n+n2.
12
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. We will compare the given series to a known series whose
convergence is already known.
Step 1: Find a comparable series Let’s consider the series P∞
n=1 3n
2n.
Step 2: Calculate the limit Compute the limit:
lim
n→∞
3n+2n
2n+n2
3n
2n
= lim
n→∞
3n+ 2n
2n+n2·2n
3n
Step 3: Simplify the limit Simplify the expression:
= lim
n→∞
1 + 2n
3n
1 + n
2n2
Step 4: Find the appropriate subsequence limits Notice that as n
approaches infinity, both 2n
3nand n
2n2tend towards zero. Thus, the limit
becomes:
=1+0
1+0 = 1
Step 5: Apply the limit comparison test Since the limit is a finite non-
zero value, the series P∞
n=1 3n+2n
2n+n2converges if and only if the series P∞
n=1 3n
2n
converges.
Step 6: Determine the convergence of the comparable series The
series P∞
n=1 3n
2nis a geometric series with common ratio r=3
2which converges
since |r|<1, and thus our original series converges by the comparison test.
Therefore, the series P∞
n=1 3n+2n
2n+n2converges.
Question 19
Question
Let (an) be a sequence such that an=n
2n. Determine whether the series
P∞
n=1 anconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 an=P∞
n=1 n
2n, we will use
the ratio test.
Step 1: Find the limit: Compute the limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)/2n+1
n/2n
= lim
n→∞
n+ 1
2(n)
lim
n→∞
n+ 1
2n
=1
2
13
Step 2: Apply the ratio test: Since 1
2<1, by the ratio test, the series
P∞
n=1 n
2nconverges.
Therefore, the given series P∞
n=1 n
2nconverges.
Question 20
Question
Determine whether the series P∞
n=1 n2+3
n3+2n+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series an=n2+3
n3+2n+1 .
Step 1: Find a series bnthat is easier to work with.
Let’s choose the series bn=1
n, which is a known series.
Step 2: Find the limit of the ratio anbn.
lim
n→∞
an
bn
= lim
n→∞
n2+3
n3+2n+1
1
n
= lim
n→∞
n2+ 3
n3+ 2n+ 1 ·n
= lim
n→∞
n3+ 3n
n3+ 2n+ 1
= 1
Step 3: Make a conclusion based on the limit.
Since limn→∞
an
bn= 1, and P∞
n=1 1
nis a divergent harmonic series, by the
Limit Comparison Test, the given series P∞
n=1 n2+3
n3+2n+1 also diverges. Hence,
the series P∞
n=1 n2+3
n3+2n+1 diverges.
Question 21
Question
Determine whether the series ∞
X
n=1
2n+ 3n
5nconverges or diverges.
14
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the limit of the ratio.
L= lim
n→∞
an+1
an
,
where an=2n+3n
5n.
Step 2: Find an+1 and an.
an+1 =2n+1 + 3n+1
5n+1 and an=2n+ 3n
5n.
Step 3: Calculate the ratio an+1
an.
an+1
an
=
2n+1+3n+1
5n+1
2n+3n
5n
=(2n+1 + 3n+1)·5n
5n+1 ·(2n+ 3n)
=2·(2n)·5n+ 3 ·(3n)·5n
5·(2n)·5n+ 5 ·(3n)·5n
=2·2n·5n+ 3 ·3n·5n
5·2n·5n+ 5 ·3n·5n
=2·2n·5n+ 3 ·3n·5n
2·5·2n·5n+ 3 ·5·3n·5n
=2·2
5n+ 3 ·3
5n
2+3·3
5n.
Step 4: Find the limit of the ratio.
L= lim
n→∞
2·2
5n+ 3 ·3
5n
2+3·3
5n
=0+0
2+0 = 0.
Step 5: Analyze the limit L. Since L < 1, by the ratio test, the series
∞
X
n=1
2n+ 3n
5nconverges.
Therefore, the series is convergent.
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
2n3+3 .
15
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Find a comparable series Let’s consider the series P∞
n=1 1
n. This
is a well-known series whose convergence behavior is known.
Step 2: Calculate the limit We will calculate the limit:
L= lim
n→∞
an
bn
= lim
n→∞
n2+1
2n3+3
1
n
= lim
n→∞
n3+n
2n3+ 3 = lim
n→∞
1 + 1
n2
2 + 3
n3
=1
2
Step 3: Verify convergence Since 0 < L < ∞, by the limit comparison
test, P∞
n=1 n2+1
2n3+3 has the same convergence behavior as P∞
n=1 1
n. Since P∞
n=1 1
n
is a harmonic series which diverges, P∞
n=1 n2+1
2n3+3 also diverges.
Question 23
Question
Determine whether the series ∞
X
n=0
(−1)n
√n+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=0
(−1)n
√n+ 1, we will use the Alter-
nating Series Test.
Step 1: Check the conditions of the Alternating Series Test. The
Alternating Series Test states that if the terms {an}in the series ∞
X
n=0
(−1)nan
satisfy the following two conditions: 1. an+1 ≤anfor all n. 2. limn→∞ an= 0.
Then the series converges.
Step 2: Verify the conditions of the Alternating Series Test. Let
an=1
√n+1 . We need to show that an+1 ≤anfor all nand limn→∞ an= 0.
For the first condition: an=1
√n+1 ,an+1 =1
√n+2 .
To show an+1 ≤an:1
√n+2 ≤1
√n+1 ,√n+ 1 ≤√n+ 2, n+ 1 ≤n+ 2. This
is always true, so the first condition is satisfied.
For the second condition: limn→∞
1
√n+1 = 0.
Step 3: Conclude convergence. Since both conditions of the Alternating
Series Test are satisfied, the series ∞
X
n=0
(−1)n
√n+ 1 converges by the Alternating
Series Test.
16
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3+n
3n4+ 4.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a suitable series to compare with the given series. We
can consider ∞
X
n=1
1
n.
Step 2: Compute the limit of the ratio of the terms of the two series:
lim
n→∞
n3+n
3n4+4
1
n
= lim
n→∞
n4+n2
3n4+ 4 =1
3.
Step 3: Since the limit is a positive finite value, by the Limit Comparison
Test, the given series P∞
n=1 n3+n
3n4+4 converges if and only if the series P∞
n=1 1
n
converges.
Step 4: The harmonic series P∞
n=1 1
nis a p-series with p= 1, which is
divergent.
Step 5: Therefore, by the Limit Comparison Test, the given series P∞
n=1 n3+n
3n4+4
also diverges.
Question 25
Question
Prove whether the series ∞
X
n=1
n2
n3+ 2n+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
17
Step 1: Consider the series
∞
X
n=1
n2
n3+ 2n+ 1
and let an=n2
n3+2n+1 . We will find the limit of an
1
n2
as napproaches infinity.
Step 2: Compute the limit:
lim
n→∞
an
1
n2
= lim
n→∞
n2
n3+2n+1
1
n2
= lim
n→∞
n4
n3+ 2n+ 1
Step 3: Simplify the expression:
= lim
n→∞
n
1 + 2
n2+1
n4
= lim
n→∞
n
1=∞
Step 4: Since the limit is infinite, and 1
n2is a convergent p-series where
p= 2 >1, by the limit comparison test, the original series
∞
X
n=1
n2
n3+ 2n+ 1
diverges.
Question 26
Question
Determine whether the series ∞
X
n=1
n2
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
3n3+ 2, we will use the limit
comparison test.
Step 1: Find the limit. Let an=n2
3n3+2 . We will find limn→∞
an
1/n .
lim
n→∞
an
1/n = lim
n→∞
n2
3n3+ 2 ·n
1= lim
n→∞
n3
3n3+ 2
Divide both the numerator and denominator by n3:
lim
n→∞
n3
3n3+ 2 = lim
n→∞
1
3+2/n3=1
3
Step 2: Conclusion based on the limit. Since lim
n→∞
an
1/n =1
3= 0, the
series ∞
X
n=1
n2
3n3+ 2 diverges by the limit comparison test.
18
Question 27
Question
Determine whether the series
∞
X
n=1
n2+ 5n+ 1
n3+ 2
converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Let’s consider the series P∞
n=1 n2+5n+1
n3+2 and the series P∞
n=1 1
n. We
will find the limit limn→∞
an
bn, where an=n2+5n+1
n3+2 and bn=1
n.
lim
n→∞
an
bn
= lim
n→∞ n·n2+ 5n+ 1
n3+ 2 = lim
n→∞
n3+ 5n2+n
n3+ 2 = lim
n→∞
1 + 5
n+1
n2
1 + 2
n3
= 1
Since the limit is a nonzero finite number, we can conclude that the given series
and the harmonic series either both converge or both diverge.
Step 2: Since the harmonic series diverges, by the limit comparison test,
the original series P∞
n=1 n2+5n+1
n3+2 also diverges. Thus, the given series diverges.
Question 28
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 calculate
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n
Step 2: Simplify the limit.
lim
n→∞
(n+ 1) ·nn
(n+ 1)n= lim
n→∞
nn
(n+ 1)n−1= lim
n→∞ n
n+ 1n−1
=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 given series converges.
19
Question 29
Question
Let an=n2
2n. Determine whether the series P∞
n=1 anconverges or diverges.
Solution
To determine whether the series P∞
n=1 anconverges or diverges, we will use the
Ratio Test.
Step 1: Calculate the limit of the absolute ratio.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the limit.
= lim
n→∞
(n+ 1)2
2n2·2n
2n+1
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n·2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Analyze the limit. Since the limit is 1
2<1, by the Ratio Test, the
series P∞
n=1 anconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 30
Question
Determine whether the series P∞
n=1 n2
n3+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
n3+1 , we will use the Limit
Comparison Test.
Step 1: Let an=n2
n3+1 . We will find a series bnthat we know the conver-
gence of, and then calculate the following limit:
lim
n→∞
an
bn
20
Step 2: Let’s choose bn=1
n. Now, calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n2
n3+ 1 ·n
1= lim
n→∞
n3
n3+ 1
Step 3: To evaluate this limit, we divide the numerator and denominator
by n3:
lim
n→∞
n3
n3+ 1 = lim
n→∞
1
1 + 1
n3
= 1
Step 4: Since the limit is a finite positive number, we can conclude that
the series P∞
n=1 n2
n3+1 converges by the Limit Comparison Test, as it behaves
similarly to the convergent series P∞
n=1 1
n.
Question 31
Question
Determine whether the series P∞
n=1 n3+2n
n4+3 converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. We will compare the given series to a known series whose convergence is
known.
Step 1: Find a known series to compare Let’s consider the series
P∞
n=1 1
n. This series is known to diverge (Harmonic series).
Step 2: Calculate the limit We will calculate the limit of the ratio of the
terms of the two series.
lim
n→∞
n3+2n
n4+3
1
n
Step 3: Simplify the limit expression Simplify the expression and find
the limit.
lim
n→∞
n4+ 2n2
n4+ 3 = lim
n→∞
1 + 2
n2
1 + 3
n4
= 1
Step 4: Conclusion Since the limit is a finite positive number, by the
limit comparison test, the given series P∞
n=1 n3+2n
n4+3 has the same convergence
behavior as the divergent harmonic series P∞
n=1 1
n. Therefore, the given series
diverges.
Question 32
Question
Determine whether the series P∞
n=1 n2+n+1
2n3+3 converges or diverges.
21
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s choose a series to compare with. Consider the series P∞
n=1 1
n
which is known to be a p-series with p= 1. This will be our benchmark series.
Step 2: We will calculate the following limit:
lim
n→∞
n2+n+1
2n3+3
1
n
= lim
n→∞
n3+n2+n
2n2+ 3/n = lim
n→∞
1+1/n + 1/n2
2+3/n2=1+0+0
2+0 =1
2
Step 3: Since the limit is a positive finite value, by the Limit Compari-
son Test, the series P∞
n=1 n2+n+1
2n3+3 converges if and only if the series P∞
n=1 1
n
converges.
Step 4: Since the series P∞
n=1 1
nis a harmonic series and diverges, we
conclude that the given series P∞
n=1 n2+n+1
2n3+3 also diverges.
Question 33
Question
Determine whether the series P∞
n=1 n2+n
n4+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+n
n4+1 , we can use the compar-
ison test.
Step 1: Find a suitable series for comparison Notice that for n≥1,
we have n2+n
n4+ 1 ≤n2+n
n4=1
n2.
Step 2: Determine the convergence of the series P∞
n=1 1
n2The series
P∞
n=1 1
n2is a known convergent series (it is a p-series with p= 2 >1).
Step 3: Apply the comparison test Since n2+n
n4+1 ≤1
n2for all n≥1
and P∞
n=1 1
n2converges, by the comparison test, the series P∞
n=1 n2+n
n4+1 also
converges.
Therefore, the series P∞
n=1 n2+n
n4+1 converges.
Question 34
Question
Determine the convergence or divergence of the series P∞
n=1
(−1)n
√n+(−1)n.
22
Solution
To determine the convergence or divergence of the series, we will use the Alter-
nating Series Test.
Step 1: Identify the terms of the series.
The terms of the series are an=(−1)n
√n+(−1)n.
Step 2: Check the conditions of the Alternating Series Test.
We need to verify if the sequence {an}is (i) decreasing and (ii) approaching
zero as napproaches infinity.
Step 3: Show that the sequence {an}is decreasing.
Consider the difference between consecutive terms:
an+1 −an=(−1)n+1
√n+ 1 + (−1)n+1 −(−1)n
√n+ (−1)n
=(−1)n+1√n+ (−1)2n−(−1)n(√n+ 1 + (−1)n+1)
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
=−√n−1 + √n+ 1 + 1
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
=√n+ 1 −√n
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
The numerator is positive, so an+1 > an, which means the sequence {an}is
decreasing.
Step 4: Show that the sequence {an}approaches zero.
We need to take the limit as napproaches infinity of anto show that it ap-
proaches zero.
lim
n→∞
(−1)n
√n+ (−1)n= 0
Therefore, the sequence {an}approaches zero.
Step 5: Conclude by the Alternating Series Test.
Since the sequence {an}is decreasing and approaches zero, and all conditions
of the Alternating Series Test are satisfied, the series P∞
n=1 anis convergent.
Question 35
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+ 3n
2n3+n+ 1.
23
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Consider the series ∞
X
n=1
n2+ 3n
2n3+n+ 1 and a known series ∞
X
n=1
1
np
which diverges for p≤1.
Step 2: Let’s find the limit of the ratio of the terms of the given series and
the known series:
L= lim
n→∞
n2+3n
2n3+n+1
1
n
Step 3: Simplifying the expression by dividing each term by the leading
term in the denominator:
L= lim
n→∞
n2+3n
n(2n2+1/n+1/n)
1/n = lim
n→∞
n2+ 3n
n(2n2+ 2) = lim
n→∞
n+ 3
2n2+ 2
Step 4: The limit Lcan be simplified by dividing each term by the highest
power in the denominator:
L= lim
n→∞
1 + 3
n
2 + 2
n2
=1+0
2+0 =1
2
Step 5: Since 0 <1
2<∞, by the limit comparison test, as p= 1 and L > 0,
the series ∞
X
n=1
n2+ 3n
2n3+n+ 1 diverges.
24
Question 5
Question
Determine the convergence or divergence of the series ∞
X
n=1
2n+ 3n
5n.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Calculate the limit of the ratio of successive terms:
lim
n→∞
an+1
an
= lim
n→∞
2n+1+3n+1
5n+1
2n+3n
5n
lim
n→∞
2n+1 + 3n+1
5n+1 ·5n
2n+ 3n
lim
n→∞
2(2n) + 3(3n)
5(5n)·5n
2n+ 3n
lim
n→∞
2(2n) + 3(3n)
5(2n+ 3n)
Step 2: Simplify the expression in the limit:
lim
n→∞
2(2n) + 3(3n)
5(2n+ 3n)= lim
n→∞
2(2n/5n) + 3(3n/5n)
2n/5n+ 3n/5n
lim
n→∞
2(2/5)n+ 3(3/5)n
2n/5n+ 3n/5n
lim
n→∞
0+0
0+0 = 0
Step 3: Analyze the limit: Since the limit of the ratio of successive terms
is less than 1, by the ratio test, ∞
X
n=1
2n+ 3n
5nconverges.
Question 6
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
4
Solution
To decide on the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of successive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!(nn)
(n+ 1)n+1n!
= lim
n→∞
n!(n+ 1)
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
(1 + 1
n)n
Step 3: Evaluate the limit using the properties of the exponential function:
lim
n→∞
1
(1 + 1
n)n
=1
e
Step 4: Determine the convergence based on the ratio test: Since the limit
1
eis less than 1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 7
Question
Determine the convergence or divergence of the series P∞
n=1
n2+ 1
n4+ 2.
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Find the limit. Let an=n2+ 1
n4+ 2. We will find limn→∞
an
1/n2.
lim
n→∞
an
1/n2= lim
n→∞
n2+ 1
n4+ 2 ·n2
1= lim
n→∞
n4+n2
n4+ 2 = 1
Step 2: Apply the limit comparison test. Since the limit of an
1/n2
is equal to 1, we can apply the limit comparison test and compare the series
P∞
n=1
n2+ 1
n4+ 2 with the series P∞
n=1
1
n2.
5
Step 3: Conclude about convergence. The series P∞
n=1
1
n2is a con-
vergent p-series with p= 2 >1. Since P∞
n=1
n2+ 1
n4+ 2 is positively terms and
its limit compared series converges, we can conclude that the original series
P∞
n=1
n2+ 1
n4+ 2 also converges by the limit comparison test.
Question 8
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n!
nn, we can use the ratio test.
Step 1: Apply the ratio test. 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
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1
n)n
Step 2: Simplify the expression. Since limn→∞(1 + 1
n)n=e, we have:
R= lim
n→∞
n
e=∞
Step 3: Conclude the convergence. Since R=∞>1, the series diverges
by the ratio test. Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 9
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
6
Solution
We will use the ratio test to determine the convergence of the series.
Step 1: Compute the limit of the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n
Step 3: Rewrite the limit in a form that is easier to evaluate:
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 4: Compute the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=e−1
Step 5: Analyze the ratio and conclude: Since the limit of the absolute
value of the ratio of consecutive terms is e−1<1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 10
Question
Let an=(−1)n
√n+1 . Determine whether the series P∞
n=1 anconverges or diverges.
Solution
To determine whether the series P∞
n=1 anconverges or diverges, we will use the
Alternating Series Test.
Step 1: Find the terms of the series. The terms of the series are given
by an=(−1)n
√n+1 .
Step 2: Check if the terms of the series satisfy the conditions of
the Alternating Series Test. We need to verify if the following conditions
are met: i) The terms anare positive. ii) The terms anare decreasing. iii)
limn→∞ an= 0.
Step 3: Check the conditions. i) Since 1
√n+1 >0 for all n, and (−1)n
also is positive for all n,an>0 for all n. ii) To check if the terms are decreasing,
7
we observe that an+1 −an=(−1)n+1
√n+2 −(−1)n
√n+1 . We simplify this expression and
note that it is always negative, meaning the terms are decreasing. iii) Finally,
limn→∞ an= limn→∞
(−1)n
√n+1 = 0.
Step 4: Conclusion. Since all conditions of the Alternating Series Test
are satisfied, the series P∞
n=1 anconverges.
Question 11
Question
Determine whether the series ∞
X
n=1
n2+ 1
√n4+n2+ 1 converges or diverges.
Solution
To investigate the convergence of the series ∞
X
n=1
n2+ 1
√n4+n2+ 1, we will use the
limit comparison test with a known convergent series.
Step 1: Identify a known convergent series. Consider the series
∞
X
n=1
1
n3/2. This series is a p-series with p=3
2>1, so it converges.
Step 2: Compute the limit of the ratio of the given series to the
known series. Let an=n2+1
√n4+n2+1 and bn=1
n3/2. We will calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n2+1
√n4+n2+1
1
n3/2
= lim
n→∞
n3/2(n2+ 1)
√n4+n2+ 1
Step 3: Simplify the expression. Simplify the limit expression:
lim
n→∞
n3/2(n2+ 1)
√n4+n2+ 1 = lim
n→∞
n5/2+n3/2
n2+n−1/2+n−3/2= lim
n→∞
n1/2+ 1
1 + n−3/2+n−5/2= 1
Step 4: Conclusion Since lim
n→∞
an
bn
= 1, and ∞
X
n=1
1
n3/2converges, by the
limit comparison test, the given series ∞
X
n=1
n2+ 1
√n4+n2+ 1 also converges.
Question 12
Question
Let P∞
n=1 n2+3n
4n3+n2+2 be the given series. Determine whether the series converges
or diverges.
8
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s choose a series to compare with the given series.
Step 1: Choose a Comparison Series Let’s consider the series P∞
n=1 1
n
as our comparison series.
Step 2: Find the Limit Let’s find the limit of the ratio of the terms of
the two series:
lim
n→∞
n2+3n
4n3+n2+2
1
n
= lim
n→∞
n3+ 3n2
4n3+n2+ 2n.
Step 3: Simplify the Limit Simplify the limit:
= lim
n→∞
n3(1 + 3
n)
n3(4 + 1
n+2
n2)= lim
n→∞
1 + 3
n
4 + 1
n+2
n2
=1
4.
Step 4: Conclusion Since the limit is a finite positive number, the given
series P∞
n=1 n2+3n
4n3+n2+2 converges by the limit comparison test with the series
P∞
n=1 1
n.
Question 13
Question
Let an=n2+3n
2nfor n≥1. Determine whether the series P∞
n=1 anconverges or
diverges.
Solution
To determine the convergence of the series P∞
n=1 an, we will use the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms.
lim
n→∞
an+1
an
= lim
n→∞
(n+1)2+3(n+1)
2n+1
n2+3n
2n
Step 2: Simplify the ratio.
= lim
n→∞
(n2+ 2n+ 1 + 3n+ 3) ·2n
(n2+ 3n)·2n+1
= lim
n→∞
(n2+ 5n+ 4) ·2n
2(n2+ 3n)·2n
= lim
n→∞
n2+ 5n+ 4
2n2+ 6n
9
Step 3: Find the limit.
= lim
n→∞
n2(1 + 5
n+4
n2)
n2(2 + 6
n)
= lim
n→∞
1 + 5
n+4
n2
2 + 6
n
=1
2
Step 4: Interpret the result. Since the limit is 1
2<1, by the ratio test, the
series P∞
n=1 anconverges.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. Let an=n!
nn.
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
n−n
=1
e,
where we used the fact that limn→∞ 1 + 1
nn=e.
Step 2: Apply the ratio test.
If R < 1, then the series P∞
n=1 anconverges.
If R > 1 or R=∞, then the series P∞
n=1 andiverges.
If R= 1, the test is inconclusive.
Since R=1
e<1, the series P∞
n=1 n!
nnconverges by the ratio test.
10
Question 15
Question
Determine whether the series ∞
X
n=1
(−1)n
n2+ (−1)nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
(−1)n
n2+ (−1)n, we will use the
Alternating Series Test.
Step 1: Determine the terms of the series The terms of the series are
given by an=(−1)n
n2+(−1)n.
Step 2: Verify the conditions of the Alternating Series Test We
need to verify that the sequence {an}is decreasing and that limn→∞ an= 0.
For an=(−1)n
n2+(−1)n, we have that an+1 −an=(−1)n+1
(n+1)2+(−1)n+1 −(−1)n
n2+(−1)n.
Simplifying, we get:
an+1 −an=(−1)n+1(n2+ (−1)n)−(−1)n((n+ 1)2+ (−1)n+1)
(n2+ (−1)n)((n+ 1)2+ (−1)n+1)
As (n+ 1)2−n2= 2n+ 1 >0 for all n, the denominator of an+1 −anis
positive. The numerator simplifies to 2n+ 1 >0, hence an+1 −an>0.
Thus, the sequence {an}is decreasing.
Next, we find limn→∞ an:
lim
n→∞
(−1)n
n2+ (−1)n= 0
Step 3: Conclude using the Alternating Series Test Since the se-
quence {an}is decreasing and limn→∞ an= 0, the series ∞
X
n=1
(−1)n
n2+ (−1)ncon-
verges by the Alternating Series Test.
Question 16
Question
Determine the convergence or divergence of the series P∞
n=1 n3
2n.
11
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
2n, we will use
the Ratio Test.
Step 1: Compute the limit involved in the Ratio Test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
Step 2: Simplify the above limit:
lim
n→∞
(n+ 1)3
2(n3)
= 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: If the limit in Step 2 is less than 1, then the
series converges. Since 1
2<1, the series P∞
n=1 n3
2nconverges.
Therefore, the series P∞
n=1 n3
2nconverges.
Question 17
Question
Determine whether the series P∞
n=1 n2+3n
2n4+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series P∞
n=1 n2+3n
2n4+5 .
Step 1: Find the limit of the ratio of the given series to a known series. Let
an=n2+3n
2n4+5 and bn=1
n2. We will find the limit of an
bnas napproaches infinity.
lim
n→∞
an
bn
= lim
n→∞
n2+3n
2n4+5
1
n2
= lim
n→∞
n4+ 3n3
2n2+ 5n2= lim
n→∞
1 + 3
n
2 + 5
n2
=1
2
Step 2: Determine convergence. Since the limit of an
bnis a finite positive
number, both series either converge or diverge. Since P∞
n=1 1
n2converges (it is
ap-series with p= 2 >1), by the limit comparison test, the series P∞
n=1 n2+3n
2n4+5
also converges.
Therefore, the series P∞
n=1 n2+3n
2n4+5 converges.
Question 18
Question
Determine the convergence or divergence of the series P∞
n=1 3n+2n
2n+n2.
12
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. We will compare the given series to a known series whose
convergence is already known.
Step 1: Find a comparable series Let’s consider the series P∞
n=1 3n
2n.
Step 2: Calculate the limit Compute the limit:
lim
n→∞
3n+2n
2n+n2
3n
2n
= lim
n→∞
3n+ 2n
2n+n2·2n
3n
Step 3: Simplify the limit Simplify the expression:
= lim
n→∞
1 + 2n
3n
1 + n
2n2
Step 4: Find the appropriate subsequence limits Notice that as n
approaches infinity, both 2n
3nand n
2n2tend towards zero. Thus, the limit
becomes:
=1+0
1+0 = 1
Step 5: Apply the limit comparison test Since the limit is a finite non-
zero value, the series P∞
n=1 3n+2n
2n+n2converges if and only if the series P∞
n=1 3n
2n
converges.
Step 6: Determine the convergence of the comparable series The
series P∞
n=1 3n
2nis a geometric series with common ratio r=3
2which converges
since |r|<1, and thus our original series converges by the comparison test.
Therefore, the series P∞
n=1 3n+2n
2n+n2converges.
Question 19
Question
Let (an) be a sequence such that an=n
2n. Determine whether the series
P∞
n=1 anconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 an=P∞
n=1 n
2n, we will use
the ratio test.
Step 1: Find the limit: Compute the limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)/2n+1
n/2n
= lim
n→∞
n+ 1
2(n)
lim
n→∞
n+ 1
2n
=1
2
13
Step 2: Apply the ratio test: Since 1
2<1, by the ratio test, the series
P∞
n=1 n
2nconverges.
Therefore, the given series P∞
n=1 n
2nconverges.
Question 20
Question
Determine whether the series P∞
n=1 n2+3
n3+2n+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series an=n2+3
n3+2n+1 .
Step 1: Find a series bnthat is easier to work with.
Let’s choose the series bn=1
n, which is a known series.
Step 2: Find the limit of the ratio anbn.
lim
n→∞
an
bn
= lim
n→∞
n2+3
n3+2n+1
1
n
= lim
n→∞
n2+ 3
n3+ 2n+ 1 ·n
= lim
n→∞
n3+ 3n
n3+ 2n+ 1
= 1
Step 3: Make a conclusion based on the limit.
Since limn→∞
an
bn= 1, and P∞
n=1 1
nis a divergent harmonic series, by the
Limit Comparison Test, the given series P∞
n=1 n2+3
n3+2n+1 also diverges. Hence,
the series P∞
n=1 n2+3
n3+2n+1 diverges.
Question 21
Question
Determine whether the series ∞
X
n=1
2n+ 3n
5nconverges or diverges.
14
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the limit of the ratio.
L= lim
n→∞
an+1
an
,
where an=2n+3n
5n.
Step 2: Find an+1 and an.
an+1 =2n+1 + 3n+1
5n+1 and an=2n+ 3n
5n.
Step 3: Calculate the ratio an+1
an.
an+1
an
=
2n+1+3n+1
5n+1
2n+3n
5n
=(2n+1 + 3n+1)·5n
5n+1 ·(2n+ 3n)
=2·(2n)·5n+ 3 ·(3n)·5n
5·(2n)·5n+ 5 ·(3n)·5n
=2·2n·5n+ 3 ·3n·5n
5·2n·5n+ 5 ·3n·5n
=2·2n·5n+ 3 ·3n·5n
2·5·2n·5n+ 3 ·5·3n·5n
=2·2
5n+ 3 ·3
5n
2+3·3
5n.
Step 4: Find the limit of the ratio.
L= lim
n→∞
2·2
5n+ 3 ·3
5n
2+3·3
5n
=0+0
2+0 = 0.
Step 5: Analyze the limit L. Since L < 1, by the ratio test, the series
∞
X
n=1
2n+ 3n
5nconverges.
Therefore, the series is convergent.
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
2n3+3 .
15
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Find a comparable series Let’s consider the series P∞
n=1 1
n. This
is a well-known series whose convergence behavior is known.
Step 2: Calculate the limit We will calculate the limit:
L= lim
n→∞
an
bn
= lim
n→∞
n2+1
2n3+3
1
n
= lim
n→∞
n3+n
2n3+ 3 = lim
n→∞
1 + 1
n2
2 + 3
n3
=1
2
Step 3: Verify convergence Since 0 < L < ∞, by the limit comparison
test, P∞
n=1 n2+1
2n3+3 has the same convergence behavior as P∞
n=1 1
n. Since P∞
n=1 1
n
is a harmonic series which diverges, P∞
n=1 n2+1
2n3+3 also diverges.
Question 23
Question
Determine whether the series ∞
X
n=0
(−1)n
√n+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=0
(−1)n
√n+ 1, we will use the Alter-
nating Series Test.
Step 1: Check the conditions of the Alternating Series Test. The
Alternating Series Test states that if the terms {an}in the series ∞
X
n=0
(−1)nan
satisfy the following two conditions: 1. an+1 ≤anfor all n. 2. limn→∞ an= 0.
Then the series converges.
Step 2: Verify the conditions of the Alternating Series Test. Let
an=1
√n+1 . We need to show that an+1 ≤anfor all nand limn→∞ an= 0.
For the first condition: an=1
√n+1 ,an+1 =1
√n+2 .
To show an+1 ≤an:1
√n+2 ≤1
√n+1 ,√n+ 1 ≤√n+ 2, n+ 1 ≤n+ 2. This
is always true, so the first condition is satisfied.
For the second condition: limn→∞
1
√n+1 = 0.
Step 3: Conclude convergence. Since both conditions of the Alternating
Series Test are satisfied, the series ∞
X
n=0
(−1)n
√n+ 1 converges by the Alternating
Series Test.
16
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3+n
3n4+ 4.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a suitable series to compare with the given series. We
can consider ∞
X
n=1
1
n.
Step 2: Compute the limit of the ratio of the terms of the two series:
lim
n→∞
n3+n
3n4+4
1
n
= lim
n→∞
n4+n2
3n4+ 4 =1
3.
Step 3: Since the limit is a positive finite value, by the Limit Comparison
Test, the given series P∞
n=1 n3+n
3n4+4 converges if and only if the series P∞
n=1 1
n
converges.
Step 4: The harmonic series P∞
n=1 1
nis a p-series with p= 1, which is
divergent.
Step 5: Therefore, by the Limit Comparison Test, the given series P∞
n=1 n3+n
3n4+4
also diverges.
Question 25
Question
Prove whether the series ∞
X
n=1
n2
n3+ 2n+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
17
Step 1: Consider the series
∞
X
n=1
n2
n3+ 2n+ 1
and let an=n2
n3+2n+1 . We will find the limit of an
1
n2
as napproaches infinity.
Step 2: Compute the limit:
lim
n→∞
an
1
n2
= lim
n→∞
n2
n3+2n+1
1
n2
= lim
n→∞
n4
n3+ 2n+ 1
Step 3: Simplify the expression:
= lim
n→∞
n
1 + 2
n2+1
n4
= lim
n→∞
n
1=∞
Step 4: Since the limit is infinite, and 1
n2is a convergent p-series where
p= 2 >1, by the limit comparison test, the original series
∞
X
n=1
n2
n3+ 2n+ 1
diverges.
Question 26
Question
Determine whether the series ∞
X
n=1
n2
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
3n3+ 2, we will use the limit
comparison test.
Step 1: Find the limit. Let an=n2
3n3+2 . We will find limn→∞
an
1/n .
lim
n→∞
an
1/n = lim
n→∞
n2
3n3+ 2 ·n
1= lim
n→∞
n3
3n3+ 2
Divide both the numerator and denominator by n3:
lim
n→∞
n3
3n3+ 2 = lim
n→∞
1
3+2/n3=1
3
Step 2: Conclusion based on the limit. Since lim
n→∞
an
1/n =1
3= 0, the
series ∞
X
n=1
n2
3n3+ 2 diverges by the limit comparison test.
18
Question 27
Question
Determine whether the series
∞
X
n=1
n2+ 5n+ 1
n3+ 2
converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Let’s consider the series P∞
n=1 n2+5n+1
n3+2 and the series P∞
n=1 1
n. We
will find the limit limn→∞
an
bn, where an=n2+5n+1
n3+2 and bn=1
n.
lim
n→∞
an
bn
= lim
n→∞ n·n2+ 5n+ 1
n3+ 2 = lim
n→∞
n3+ 5n2+n
n3+ 2 = lim
n→∞
1 + 5
n+1
n2
1 + 2
n3
= 1
Since the limit is a nonzero finite number, we can conclude that the given series
and the harmonic series either both converge or both diverge.
Step 2: Since the harmonic series diverges, by the limit comparison test,
the original series P∞
n=1 n2+5n+1
n3+2 also diverges. Thus, the given series diverges.
Question 28
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 calculate
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n
Step 2: Simplify the limit.
lim
n→∞
(n+ 1) ·nn
(n+ 1)n= lim
n→∞
nn
(n+ 1)n−1= lim
n→∞ n
n+ 1n−1
=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 given series converges.
19
Question 29
Question
Let an=n2
2n. Determine whether the series P∞
n=1 anconverges or diverges.
Solution
To determine whether the series P∞
n=1 anconverges or diverges, we will use the
Ratio Test.
Step 1: Calculate the limit of the absolute ratio.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the limit.
= lim
n→∞
(n+ 1)2
2n2·2n
2n+1
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n·2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Analyze the limit. Since the limit is 1
2<1, by the Ratio Test, the
series P∞
n=1 anconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 30
Question
Determine whether the series P∞
n=1 n2
n3+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
n3+1 , we will use the Limit
Comparison Test.
Step 1: Let an=n2
n3+1 . We will find a series bnthat we know the conver-
gence of, and then calculate the following limit:
lim
n→∞
an
bn
20
Step 2: Let’s choose bn=1
n. Now, calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n2
n3+ 1 ·n
1= lim
n→∞
n3
n3+ 1
Step 3: To evaluate this limit, we divide the numerator and denominator
by n3:
lim
n→∞
n3
n3+ 1 = lim
n→∞
1
1 + 1
n3
= 1
Step 4: Since the limit is a finite positive number, we can conclude that
the series P∞
n=1 n2
n3+1 converges by the Limit Comparison Test, as it behaves
similarly to the convergent series P∞
n=1 1
n.
Question 31
Question
Determine whether the series P∞
n=1 n3+2n
n4+3 converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. We will compare the given series to a known series whose convergence is
known.
Step 1: Find a known series to compare Let’s consider the series
P∞
n=1 1
n. This series is known to diverge (Harmonic series).
Step 2: Calculate the limit We will calculate the limit of the ratio of the
terms of the two series.
lim
n→∞
n3+2n
n4+3
1
n
Step 3: Simplify the limit expression Simplify the expression and find
the limit.
lim
n→∞
n4+ 2n2
n4+ 3 = lim
n→∞
1 + 2
n2
1 + 3
n4
= 1
Step 4: Conclusion Since the limit is a finite positive number, by the
limit comparison test, the given series P∞
n=1 n3+2n
n4+3 has the same convergence
behavior as the divergent harmonic series P∞
n=1 1
n. Therefore, the given series
diverges.
Question 32
Question
Determine whether the series P∞
n=1 n2+n+1
2n3+3 converges or diverges.
21
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s choose a series to compare with. Consider the series P∞
n=1 1
n
which is known to be a p-series with p= 1. This will be our benchmark series.
Step 2: We will calculate the following limit:
lim
n→∞
n2+n+1
2n3+3
1
n
= lim
n→∞
n3+n2+n
2n2+ 3/n = lim
n→∞
1+1/n + 1/n2
2+3/n2=1+0+0
2+0 =1
2
Step 3: Since the limit is a positive finite value, by the Limit Compari-
son Test, the series P∞
n=1 n2+n+1
2n3+3 converges if and only if the series P∞
n=1 1
n
converges.
Step 4: Since the series P∞
n=1 1
nis a harmonic series and diverges, we
conclude that the given series P∞
n=1 n2+n+1
2n3+3 also diverges.
Question 33
Question
Determine whether the series P∞
n=1 n2+n
n4+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+n
n4+1 , we can use the compar-
ison test.
Step 1: Find a suitable series for comparison Notice that for n≥1,
we have n2+n
n4+ 1 ≤n2+n
n4=1
n2.
Step 2: Determine the convergence of the series P∞
n=1 1
n2The series
P∞
n=1 1
n2is a known convergent series (it is a p-series with p= 2 >1).
Step 3: Apply the comparison test Since n2+n
n4+1 ≤1
n2for all n≥1
and P∞
n=1 1
n2converges, by the comparison test, the series P∞
n=1 n2+n
n4+1 also
converges.
Therefore, the series P∞
n=1 n2+n
n4+1 converges.
Question 34
Question
Determine the convergence or divergence of the series P∞
n=1
(−1)n
√n+(−1)n.
22
Solution
To determine the convergence or divergence of the series, we will use the Alter-
nating Series Test.
Step 1: Identify the terms of the series.
The terms of the series are an=(−1)n
√n+(−1)n.
Step 2: Check the conditions of the Alternating Series Test.
We need to verify if the sequence {an}is (i) decreasing and (ii) approaching
zero as napproaches infinity.
Step 3: Show that the sequence {an}is decreasing.
Consider the difference between consecutive terms:
an+1 −an=(−1)n+1
√n+ 1 + (−1)n+1 −(−1)n
√n+ (−1)n
=(−1)n+1√n+ (−1)2n−(−1)n(√n+ 1 + (−1)n+1)
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
=−√n−1 + √n+ 1 + 1
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
=√n+ 1 −√n
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
The numerator is positive, so an+1 > an, which means the sequence {an}is
decreasing.
Step 4: Show that the sequence {an}approaches zero.
We need to take the limit as napproaches infinity of anto show that it ap-
proaches zero.
lim
n→∞
(−1)n
√n+ (−1)n= 0
Therefore, the sequence {an}approaches zero.
Step 5: Conclude by the Alternating Series Test.
Since the sequence {an}is decreasing and approaches zero, and all conditions
of the Alternating Series Test are satisfied, the series P∞
n=1 anis convergent.
Question 35
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+ 3n
2n3+n+ 1.
23
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Consider the series ∞
X
n=1
n2+ 3n
2n3+n+ 1 and a known series ∞
X
n=1
1
np
which diverges for p≤1.
Step 2: Let’s find the limit of the ratio of the terms of the given series and
the known series:
L= lim
n→∞
n2+3n
2n3+n+1
1
n
Step 3: Simplifying the expression by dividing each term by the leading
term in the denominator:
L= lim
n→∞
n2+3n
n(2n2+1/n+1/n)
1/n = lim
n→∞
n2+ 3n
n(2n2+ 2) = lim
n→∞
n+ 3
2n2+ 2
Step 4: The limit Lcan be simplified by dividing each term by the highest
power in the denominator:
L= lim
n→∞
1 + 3
n
2 + 2
n2
=1+0
2+0 =1
2
Step 5: Since 0 <1
2<∞, by the limit comparison test, as p= 1 and L > 0,
the series ∞
X
n=1
n2+ 3n
2n3+n+ 1 diverges.
24
Question 5
Question
Determine the convergence or divergence of the series ∞
X
n=1
2n+ 3n
5n.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Calculate the limit of the ratio of successive terms:
lim
n→∞
an+1
an
= lim
n→∞
2n+1+3n+1
5n+1
2n+3n
5n
lim
n→∞
2n+1 + 3n+1
5n+1 ·5n
2n+ 3n
lim
n→∞
2(2n) + 3(3n)
5(5n)·5n
2n+ 3n
lim
n→∞
2(2n) + 3(3n)
5(2n+ 3n)
Step 2: Simplify the expression in the limit:
lim
n→∞
2(2n) + 3(3n)
5(2n+ 3n)= lim
n→∞
2(2n/5n) + 3(3n/5n)
2n/5n+ 3n/5n
lim
n→∞
2(2/5)n+ 3(3/5)n
2n/5n+ 3n/5n
lim
n→∞
0+0
0+0 = 0
Step 3: Analyze the limit: Since the limit of the ratio of successive terms
is less than 1, by the ratio test, ∞
X
n=1
2n+ 3n
5nconverges.
Question 6
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n!
nn
4
Solution
To decide on the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of successive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!(nn)
(n+ 1)n+1n!
= lim
n→∞
n!(n+ 1)
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
(1 + 1
n)n
Step 3: Evaluate the limit using the properties of the exponential function:
lim
n→∞
1
(1 + 1
n)n
=1
e
Step 4: Determine the convergence based on the ratio test: Since the limit
1
eis less than 1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 7
Question
Determine the convergence or divergence of the series P∞
n=1
n2+ 1
n4+ 2.
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Find the limit. Let an=n2+ 1
n4+ 2. We will find limn→∞
an
1/n2.
lim
n→∞
an
1/n2= lim
n→∞
n2+ 1
n4+ 2 ·n2
1= lim
n→∞
n4+n2
n4+ 2 = 1
Step 2: Apply the limit comparison test. Since the limit of an
1/n2
is equal to 1, we can apply the limit comparison test and compare the series
P∞
n=1
n2+ 1
n4+ 2 with the series P∞
n=1
1
n2.
5
Step 3: Conclude about convergence. The series P∞
n=1
1
n2is a con-
vergent p-series with p= 2 >1. Since P∞
n=1
n2+ 1
n4+ 2 is positively terms and
its limit compared series converges, we can conclude that the original series
P∞
n=1
n2+ 1
n4+ 2 also converges by the limit comparison test.
Question 8
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n!
nn, we can use the ratio test.
Step 1: Apply the ratio test. 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
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1
n)n
Step 2: Simplify the expression. Since limn→∞(1 + 1
n)n=e, we have:
R= lim
n→∞
n
e=∞
Step 3: Conclude the convergence. Since R=∞>1, the series diverges
by the ratio test. Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 9
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
6
Solution
We will use the ratio test to determine the convergence of the series.
Step 1: Compute the limit of the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n
Step 3: Rewrite the limit in a form that is easier to evaluate:
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 4: Compute the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=e−1
Step 5: Analyze the ratio and conclude: Since the limit of the absolute
value of the ratio of consecutive terms is e−1<1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 10
Question
Let an=(−1)n
√n+1 . Determine whether the series P∞
n=1 anconverges or diverges.
Solution
To determine whether the series P∞
n=1 anconverges or diverges, we will use the
Alternating Series Test.
Step 1: Find the terms of the series. The terms of the series are given
by an=(−1)n
√n+1 .
Step 2: Check if the terms of the series satisfy the conditions of
the Alternating Series Test. We need to verify if the following conditions
are met: i) The terms anare positive. ii) The terms anare decreasing. iii)
limn→∞ an= 0.
Step 3: Check the conditions. i) Since 1
√n+1 >0 for all n, and (−1)n
also is positive for all n,an>0 for all n. ii) To check if the terms are decreasing,
7
we observe that an+1 −an=(−1)n+1
√n+2 −(−1)n
√n+1 . We simplify this expression and
note that it is always negative, meaning the terms are decreasing. iii) Finally,
limn→∞ an= limn→∞
(−1)n
√n+1 = 0.
Step 4: Conclusion. Since all conditions of the Alternating Series Test
are satisfied, the series P∞
n=1 anconverges.
Question 11
Question
Determine whether the series ∞
X
n=1
n2+ 1
√n4+n2+ 1 converges or diverges.
Solution
To investigate the convergence of the series ∞
X
n=1
n2+ 1
√n4+n2+ 1, we will use the
limit comparison test with a known convergent series.
Step 1: Identify a known convergent series. Consider the series
∞
X
n=1
1
n3/2. This series is a p-series with p=3
2>1, so it converges.
Step 2: Compute the limit of the ratio of the given series to the
known series. Let an=n2+1
√n4+n2+1 and bn=1
n3/2. We will calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n2+1
√n4+n2+1
1
n3/2
= lim
n→∞
n3/2(n2+ 1)
√n4+n2+ 1
Step 3: Simplify the expression. Simplify the limit expression:
lim
n→∞
n3/2(n2+ 1)
√n4+n2+ 1 = lim
n→∞
n5/2+n3/2
n2+n−1/2+n−3/2= lim
n→∞
n1/2+ 1
1 + n−3/2+n−5/2= 1
Step 4: Conclusion Since lim
n→∞
an
bn
= 1, and ∞
X
n=1
1
n3/2converges, by the
limit comparison test, the given series ∞
X
n=1
n2+ 1
√n4+n2+ 1 also converges.
Question 12
Question
Let P∞
n=1 n2+3n
4n3+n2+2 be the given series. Determine whether the series converges
or diverges.
8
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s choose a series to compare with the given series.
Step 1: Choose a Comparison Series Let’s consider the series P∞
n=1 1
n
as our comparison series.
Step 2: Find the Limit Let’s find the limit of the ratio of the terms of
the two series:
lim
n→∞
n2+3n
4n3+n2+2
1
n
= lim
n→∞
n3+ 3n2
4n3+n2+ 2n.
Step 3: Simplify the Limit Simplify the limit:
= lim
n→∞
n3(1 + 3
n)
n3(4 + 1
n+2
n2)= lim
n→∞
1 + 3
n
4 + 1
n+2
n2
=1
4.
Step 4: Conclusion Since the limit is a finite positive number, the given
series P∞
n=1 n2+3n
4n3+n2+2 converges by the limit comparison test with the series
P∞
n=1 1
n.
Question 13
Question
Let an=n2+3n
2nfor n≥1. Determine whether the series P∞
n=1 anconverges or
diverges.
Solution
To determine the convergence of the series P∞
n=1 an, we will use the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms.
lim
n→∞
an+1
an
= lim
n→∞
(n+1)2+3(n+1)
2n+1
n2+3n
2n
Step 2: Simplify the ratio.
= lim
n→∞
(n2+ 2n+ 1 + 3n+ 3) ·2n
(n2+ 3n)·2n+1
= lim
n→∞
(n2+ 5n+ 4) ·2n
2(n2+ 3n)·2n
= lim
n→∞
n2+ 5n+ 4
2n2+ 6n
9
Step 3: Find the limit.
= lim
n→∞
n2(1 + 5
n+4
n2)
n2(2 + 6
n)
= lim
n→∞
1 + 5
n+4
n2
2 + 6
n
=1
2
Step 4: Interpret the result. Since the limit is 1
2<1, by the ratio test, the
series P∞
n=1 anconverges.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. Let an=n!
nn.
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
n−n
=1
e,
where we used the fact that limn→∞ 1 + 1
nn=e.
Step 2: Apply the ratio test.
If R < 1, then the series P∞
n=1 anconverges.
If R > 1 or R=∞, then the series P∞
n=1 andiverges.
If R= 1, the test is inconclusive.
Since R=1
e<1, the series P∞
n=1 n!
nnconverges by the ratio test.
10
Question 15
Question
Determine whether the series ∞
X
n=1
(−1)n
n2+ (−1)nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
(−1)n
n2+ (−1)n, we will use the
Alternating Series Test.
Step 1: Determine the terms of the series The terms of the series are
given by an=(−1)n
n2+(−1)n.
Step 2: Verify the conditions of the Alternating Series Test We
need to verify that the sequence {an}is decreasing and that limn→∞ an= 0.
For an=(−1)n
n2+(−1)n, we have that an+1 −an=(−1)n+1
(n+1)2+(−1)n+1 −(−1)n
n2+(−1)n.
Simplifying, we get:
an+1 −an=(−1)n+1(n2+ (−1)n)−(−1)n((n+ 1)2+ (−1)n+1)
(n2+ (−1)n)((n+ 1)2+ (−1)n+1)
As (n+ 1)2−n2= 2n+ 1 >0 for all n, the denominator of an+1 −anis
positive. The numerator simplifies to 2n+ 1 >0, hence an+1 −an>0.
Thus, the sequence {an}is decreasing.
Next, we find limn→∞ an:
lim
n→∞
(−1)n
n2+ (−1)n= 0
Step 3: Conclude using the Alternating Series Test Since the se-
quence {an}is decreasing and limn→∞ an= 0, the series ∞
X
n=1
(−1)n
n2+ (−1)ncon-
verges by the Alternating Series Test.
Question 16
Question
Determine the convergence or divergence of the series P∞
n=1 n3
2n.
11
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
2n, we will use
the Ratio Test.
Step 1: Compute the limit involved in the Ratio Test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
Step 2: Simplify the above limit:
lim
n→∞
(n+ 1)3
2(n3)
= 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: If the limit in Step 2 is less than 1, then the
series converges. Since 1
2<1, the series P∞
n=1 n3
2nconverges.
Therefore, the series P∞
n=1 n3
2nconverges.
Question 17
Question
Determine whether the series P∞
n=1 n2+3n
2n4+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series P∞
n=1 n2+3n
2n4+5 .
Step 1: Find the limit of the ratio of the given series to a known series. Let
an=n2+3n
2n4+5 and bn=1
n2. We will find the limit of an
bnas napproaches infinity.
lim
n→∞
an
bn
= lim
n→∞
n2+3n
2n4+5
1
n2
= lim
n→∞
n4+ 3n3
2n2+ 5n2= lim
n→∞
1 + 3
n
2 + 5
n2
=1
2
Step 2: Determine convergence. Since the limit of an
bnis a finite positive
number, both series either converge or diverge. Since P∞
n=1 1
n2converges (it is
ap-series with p= 2 >1), by the limit comparison test, the series P∞
n=1 n2+3n
2n4+5
also converges.
Therefore, the series P∞
n=1 n2+3n
2n4+5 converges.
Question 18
Question
Determine the convergence or divergence of the series P∞
n=1 3n+2n
2n+n2.
12
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. We will compare the given series to a known series whose
convergence is already known.
Step 1: Find a comparable series Let’s consider the series P∞
n=1 3n
2n.
Step 2: Calculate the limit Compute the limit:
lim
n→∞
3n+2n
2n+n2
3n
2n
= lim
n→∞
3n+ 2n
2n+n2·2n
3n
Step 3: Simplify the limit Simplify the expression:
= lim
n→∞
1 + 2n
3n
1 + n
2n2
Step 4: Find the appropriate subsequence limits Notice that as n
approaches infinity, both 2n
3nand n
2n2tend towards zero. Thus, the limit
becomes:
=1+0
1+0 = 1
Step 5: Apply the limit comparison test Since the limit is a finite non-
zero value, the series P∞
n=1 3n+2n
2n+n2converges if and only if the series P∞
n=1 3n
2n
converges.
Step 6: Determine the convergence of the comparable series The
series P∞
n=1 3n
2nis a geometric series with common ratio r=3
2which converges
since |r|<1, and thus our original series converges by the comparison test.
Therefore, the series P∞
n=1 3n+2n
2n+n2converges.
Question 19
Question
Let (an) be a sequence such that an=n
2n. Determine whether the series
P∞
n=1 anconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 an=P∞
n=1 n
2n, we will use
the ratio test.
Step 1: Find the limit: Compute the limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)/2n+1
n/2n
= lim
n→∞
n+ 1
2(n)
lim
n→∞
n+ 1
2n
=1
2
13
Step 2: Apply the ratio test: Since 1
2<1, by the ratio test, the series
P∞
n=1 n
2nconverges.
Therefore, the given series P∞
n=1 n
2nconverges.
Question 20
Question
Determine whether the series P∞
n=1 n2+3
n3+2n+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series an=n2+3
n3+2n+1 .
Step 1: Find a series bnthat is easier to work with.
Let’s choose the series bn=1
n, which is a known series.
Step 2: Find the limit of the ratio anbn.
lim
n→∞
an
bn
= lim
n→∞
n2+3
n3+2n+1
1
n
= lim
n→∞
n2+ 3
n3+ 2n+ 1 ·n
= lim
n→∞
n3+ 3n
n3+ 2n+ 1
= 1
Step 3: Make a conclusion based on the limit.
Since limn→∞
an
bn= 1, and P∞
n=1 1
nis a divergent harmonic series, by the
Limit Comparison Test, the given series P∞
n=1 n2+3
n3+2n+1 also diverges. Hence,
the series P∞
n=1 n2+3
n3+2n+1 diverges.
Question 21
Question
Determine whether the series ∞
X
n=1
2n+ 3n
5nconverges or diverges.
14
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the limit of the ratio.
L= lim
n→∞
an+1
an
,
where an=2n+3n
5n.
Step 2: Find an+1 and an.
an+1 =2n+1 + 3n+1
5n+1 and an=2n+ 3n
5n.
Step 3: Calculate the ratio an+1
an.
an+1
an
=
2n+1+3n+1
5n+1
2n+3n
5n
=(2n+1 + 3n+1)·5n
5n+1 ·(2n+ 3n)
=2·(2n)·5n+ 3 ·(3n)·5n
5·(2n)·5n+ 5 ·(3n)·5n
=2·2n·5n+ 3 ·3n·5n
5·2n·5n+ 5 ·3n·5n
=2·2n·5n+ 3 ·3n·5n
2·5·2n·5n+ 3 ·5·3n·5n
=2·2
5n+ 3 ·3
5n
2+3·3
5n.
Step 4: Find the limit of the ratio.
L= lim
n→∞
2·2
5n+ 3 ·3
5n
2+3·3
5n
=0+0
2+0 = 0.
Step 5: Analyze the limit L. Since L < 1, by the ratio test, the series
∞
X
n=1
2n+ 3n
5nconverges.
Therefore, the series is convergent.
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
2n3+3 .
15
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Find a comparable series Let’s consider the series P∞
n=1 1
n. This
is a well-known series whose convergence behavior is known.
Step 2: Calculate the limit We will calculate the limit:
L= lim
n→∞
an
bn
= lim
n→∞
n2+1
2n3+3
1
n
= lim
n→∞
n3+n
2n3+ 3 = lim
n→∞
1 + 1
n2
2 + 3
n3
=1
2
Step 3: Verify convergence Since 0 < L < ∞, by the limit comparison
test, P∞
n=1 n2+1
2n3+3 has the same convergence behavior as P∞
n=1 1
n. Since P∞
n=1 1
n
is a harmonic series which diverges, P∞
n=1 n2+1
2n3+3 also diverges.
Question 23
Question
Determine whether the series ∞
X
n=0
(−1)n
√n+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=0
(−1)n
√n+ 1, we will use the Alter-
nating Series Test.
Step 1: Check the conditions of the Alternating Series Test. The
Alternating Series Test states that if the terms {an}in the series ∞
X
n=0
(−1)nan
satisfy the following two conditions: 1. an+1 ≤anfor all n. 2. limn→∞ an= 0.
Then the series converges.
Step 2: Verify the conditions of the Alternating Series Test. Let
an=1
√n+1 . We need to show that an+1 ≤anfor all nand limn→∞ an= 0.
For the first condition: an=1
√n+1 ,an+1 =1
√n+2 .
To show an+1 ≤an:1
√n+2 ≤1
√n+1 ,√n+ 1 ≤√n+ 2, n+ 1 ≤n+ 2. This
is always true, so the first condition is satisfied.
For the second condition: limn→∞
1
√n+1 = 0.
Step 3: Conclude convergence. Since both conditions of the Alternating
Series Test are satisfied, the series ∞
X
n=0
(−1)n
√n+ 1 converges by the Alternating
Series Test.
16
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3+n
3n4+ 4.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a suitable series to compare with the given series. We
can consider ∞
X
n=1
1
n.
Step 2: Compute the limit of the ratio of the terms of the two series:
lim
n→∞
n3+n
3n4+4
1
n
= lim
n→∞
n4+n2
3n4+ 4 =1
3.
Step 3: Since the limit is a positive finite value, by the Limit Comparison
Test, the given series P∞
n=1 n3+n
3n4+4 converges if and only if the series P∞
n=1 1
n
converges.
Step 4: The harmonic series P∞
n=1 1
nis a p-series with p= 1, which is
divergent.
Step 5: Therefore, by the Limit Comparison Test, the given series P∞
n=1 n3+n
3n4+4
also diverges.
Question 25
Question
Prove whether the series ∞
X
n=1
n2
n3+ 2n+ 1
converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
17
Step 1: Consider the series
∞
X
n=1
n2
n3+ 2n+ 1
and let an=n2
n3+2n+1 . We will find the limit of an
1
n2
as napproaches infinity.
Step 2: Compute the limit:
lim
n→∞
an
1
n2
= lim
n→∞
n2
n3+2n+1
1
n2
= lim
n→∞
n4
n3+ 2n+ 1
Step 3: Simplify the expression:
= lim
n→∞
n
1 + 2
n2+1
n4
= lim
n→∞
n
1=∞
Step 4: Since the limit is infinite, and 1
n2is a convergent p-series where
p= 2 >1, by the limit comparison test, the original series
∞
X
n=1
n2
n3+ 2n+ 1
diverges.
Question 26
Question
Determine whether the series ∞
X
n=1
n2
3n3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
3n3+ 2, we will use the limit
comparison test.
Step 1: Find the limit. Let an=n2
3n3+2 . We will find limn→∞
an
1/n .
lim
n→∞
an
1/n = lim
n→∞
n2
3n3+ 2 ·n
1= lim
n→∞
n3
3n3+ 2
Divide both the numerator and denominator by n3:
lim
n→∞
n3
3n3+ 2 = lim
n→∞
1
3+2/n3=1
3
Step 2: Conclusion based on the limit. Since lim
n→∞
an
1/n =1
3= 0, the
series ∞
X
n=1
n2
3n3+ 2 diverges by the limit comparison test.
18
Question 27
Question
Determine whether the series
∞
X
n=1
n2+ 5n+ 1
n3+ 2
converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Let’s consider the series P∞
n=1 n2+5n+1
n3+2 and the series P∞
n=1 1
n. We
will find the limit limn→∞
an
bn, where an=n2+5n+1
n3+2 and bn=1
n.
lim
n→∞
an
bn
= lim
n→∞ n·n2+ 5n+ 1
n3+ 2 = lim
n→∞
n3+ 5n2+n
n3+ 2 = lim
n→∞
1 + 5
n+1
n2
1 + 2
n3
= 1
Since the limit is a nonzero finite number, we can conclude that the given series
and the harmonic series either both converge or both diverge.
Step 2: Since the harmonic series diverges, by the limit comparison test,
the original series P∞
n=1 n2+5n+1
n3+2 also diverges. Thus, the given series diverges.
Question 28
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 calculate
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n
Step 2: Simplify the limit.
lim
n→∞
(n+ 1) ·nn
(n+ 1)n= lim
n→∞
nn
(n+ 1)n−1= lim
n→∞ n
n+ 1n−1
=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 given series converges.
19
Question 29
Question
Let an=n2
2n. Determine whether the series P∞
n=1 anconverges or diverges.
Solution
To determine whether the series P∞
n=1 anconverges or diverges, we will use the
Ratio Test.
Step 1: Calculate the limit of the absolute ratio.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the limit.
= lim
n→∞
(n+ 1)2
2n2·2n
2n+1
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n·2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Analyze the limit. Since the limit is 1
2<1, by the Ratio Test, the
series P∞
n=1 anconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 30
Question
Determine whether the series P∞
n=1 n2
n3+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
n3+1 , we will use the Limit
Comparison Test.
Step 1: Let an=n2
n3+1 . We will find a series bnthat we know the conver-
gence of, and then calculate the following limit:
lim
n→∞
an
bn
20
Step 2: Let’s choose bn=1
n. Now, calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n2
n3+ 1 ·n
1= lim
n→∞
n3
n3+ 1
Step 3: To evaluate this limit, we divide the numerator and denominator
by n3:
lim
n→∞
n3
n3+ 1 = lim
n→∞
1
1 + 1
n3
= 1
Step 4: Since the limit is a finite positive number, we can conclude that
the series P∞
n=1 n2
n3+1 converges by the Limit Comparison Test, as it behaves
similarly to the convergent series P∞
n=1 1
n.
Question 31
Question
Determine whether the series P∞
n=1 n3+2n
n4+3 converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. We will compare the given series to a known series whose convergence is
known.
Step 1: Find a known series to compare Let’s consider the series
P∞
n=1 1
n. This series is known to diverge (Harmonic series).
Step 2: Calculate the limit We will calculate the limit of the ratio of the
terms of the two series.
lim
n→∞
n3+2n
n4+3
1
n
Step 3: Simplify the limit expression Simplify the expression and find
the limit.
lim
n→∞
n4+ 2n2
n4+ 3 = lim
n→∞
1 + 2
n2
1 + 3
n4
= 1
Step 4: Conclusion Since the limit is a finite positive number, by the
limit comparison test, the given series P∞
n=1 n3+2n
n4+3 has the same convergence
behavior as the divergent harmonic series P∞
n=1 1
n. Therefore, the given series
diverges.
Question 32
Question
Determine whether the series P∞
n=1 n2+n+1
2n3+3 converges or diverges.
21
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s choose a series to compare with. Consider the series P∞
n=1 1
n
which is known to be a p-series with p= 1. This will be our benchmark series.
Step 2: We will calculate the following limit:
lim
n→∞
n2+n+1
2n3+3
1
n
= lim
n→∞
n3+n2+n
2n2+ 3/n = lim
n→∞
1+1/n + 1/n2
2+3/n2=1+0+0
2+0 =1
2
Step 3: Since the limit is a positive finite value, by the Limit Compari-
son Test, the series P∞
n=1 n2+n+1
2n3+3 converges if and only if the series P∞
n=1 1
n
converges.
Step 4: Since the series P∞
n=1 1
nis a harmonic series and diverges, we
conclude that the given series P∞
n=1 n2+n+1
2n3+3 also diverges.
Question 33
Question
Determine whether the series P∞
n=1 n2+n
n4+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+n
n4+1 , we can use the compar-
ison test.
Step 1: Find a suitable series for comparison Notice that for n≥1,
we have n2+n
n4+ 1 ≤n2+n
n4=1
n2.
Step 2: Determine the convergence of the series P∞
n=1 1
n2The series
P∞
n=1 1
n2is a known convergent series (it is a p-series with p= 2 >1).
Step 3: Apply the comparison test Since n2+n
n4+1 ≤1
n2for all n≥1
and P∞
n=1 1
n2converges, by the comparison test, the series P∞
n=1 n2+n
n4+1 also
converges.
Therefore, the series P∞
n=1 n2+n
n4+1 converges.
Question 34
Question
Determine the convergence or divergence of the series P∞
n=1
(−1)n
√n+(−1)n.
22
Solution
To determine the convergence or divergence of the series, we will use the Alter-
nating Series Test.
Step 1: Identify the terms of the series.
The terms of the series are an=(−1)n
√n+(−1)n.
Step 2: Check the conditions of the Alternating Series Test.
We need to verify if the sequence {an}is (i) decreasing and (ii) approaching
zero as napproaches infinity.
Step 3: Show that the sequence {an}is decreasing.
Consider the difference between consecutive terms:
an+1 −an=(−1)n+1
√n+ 1 + (−1)n+1 −(−1)n
√n+ (−1)n
=(−1)n+1√n+ (−1)2n−(−1)n(√n+ 1 + (−1)n+1)
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
=−√n−1 + √n+ 1 + 1
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
=√n+ 1 −√n
(√n+ (−1)n)(√n+ 1 + (−1)n+1)
The numerator is positive, so an+1 > an, which means the sequence {an}is
decreasing.
Step 4: Show that the sequence {an}approaches zero.
We need to take the limit as napproaches infinity of anto show that it ap-
proaches zero.
lim
n→∞
(−1)n
√n+ (−1)n= 0
Therefore, the sequence {an}approaches zero.
Step 5: Conclude by the Alternating Series Test.
Since the sequence {an}is decreasing and approaches zero, and all conditions
of the Alternating Series Test are satisfied, the series P∞
n=1 anis convergent.
Question 35
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+ 3n
2n3+n+ 1.
23
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: Consider the series ∞
X
n=1
n2+ 3n
2n3+n+ 1 and a known series ∞
X
n=1
1
np
which diverges for p≤1.
Step 2: Let’s find the limit of the ratio of the terms of the given series and
the known series:
L= lim
n→∞
n2+3n
2n3+n+1
1
n
Step 3: Simplifying the expression by dividing each term by the leading
term in the denominator:
L= lim
n→∞
n2+3n
n(2n2+1/n+1/n)
1/n = lim
n→∞
n2+ 3n
n(2n2+ 2) = lim
n→∞
n+ 3
2n2+ 2
Step 4: The limit Lcan be simplified by dividing each term by the highest
power in the denominator:
L= lim
n→∞
1 + 3
n
2 + 2
n2
=1+0
2+0 =1
2
Step 5: Since 0 <1
2<∞, by the limit comparison test, as p= 1 and L > 0,
the series ∞
X
n=1
n2+ 3n
2n3+n+ 1 diverges.
24
