MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 7
Liberty University
Question 1
Question
Determine the convergence or divergence of the series P∞
n=1 n2+1
n3+1 .
Solution
To determine the convergence or divergence of the given series, we will use the
limit comparison test.
Step 1: Define an=n2+1
n3+1 .
Step 2: Consider the series P∞
n=1 1
n, which is a p-series with p= 1 and
diverges.
Step 3: Calculate the limit:
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1 = 1
Step 4: Since limn→∞
an
bn= 1, by the limit comparison test, the series
P∞
n=1 n2+1
n3+1 has the same convergence/divergence behavior as P∞
n=1 1
n.
Step 5: Since the harmonic series diverges, by the limit comparison test,
the original series P∞
n=1 n2+1
n3+1 also diverges.
Question 2
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2
3n3+ 2.
Solution
Step 1: We can start by simplifying the expression inside the series.
n2
3n3+ 2 =n2
3n3·1
1 + 2
3n3
=1
3n·1
1 + 2
3n3
.
Step 2: We will now find the limit of the series as napproaches infinity.
lim
n→∞
1
3n·1
1 + 2
3n3
= lim
n→∞
1
3n·1
1= lim
n→∞
1
3n= 0.
Step 3: Since the limit of the terms of the series is equal to 0, we can conclude
by the Limit Comparison Test that the series ∞
X
n=1
n2
3n3+ 2 converges.
Question 3
Question
Determine whether the series P∞
n=1 n+2
3n2+1 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 n+2
3n2+1 and the series P∞
n=1 1
n. We will
calculate the limit of the ratio of the two series to apply the Limit Comparison
Test.
Step 1: Find the limit of the ratio.
Let an=n+2
3n2+1 and bn=1
n. We will find the limit of an
bnas napproaches
infinity:
lim
n→∞
an
bn
= lim
n→∞
n+2
3n2+1
1
n
= lim
n→∞
n(n+ 2)
3n2+ 1
Step 2: Simplify the expression.
Simplify the expression inside the limit:
lim
n→∞
n(n+ 2)
3n2+ 1 = lim
n→∞
n2+ 2n
3n2+ 1 = lim
n→∞
1 + 2
n
3 + 1
n2
Step 3: Find the limit.
Taking the limit as napproaches infinity:
lim
n→∞
1 + 2
n
3 + 1
n2
=1+0
3+0 =1
3
Step 4: Apply the Limit Comparison Test.
Since the limit of the ratio an
bnis a finite positive value, we can conclude that
both series P∞
n=1 n+2
3n2+1 and P∞
n=1 1
neither both converge or both diverge.
Since P∞
n=1 1
nis a divergent harmonic series, by the Limit Comparison Test,
we can also conclude that the series P∞
n=1 n+2
3n2+1 diverges.
2
Question 4
Question
Determine the convergence or divergence of the series P∞
n=1 n2+3n
2n3+1 .
Solution
To determine the convergence or divergence of the series P∞
n=1 n2+3n
2n3+1 , we will
use the limit comparison test.
Step 1: Let’s set an=n2+3n
2n3+1 .
Step 2: Find the limit of the ratio of anto a simpler series bn=1
n.
lim
n→∞
an
bn
= lim
n→∞
n2+3n
2n3+1
1
n
= lim
n→∞
n3+ 3n2
2n3+ 1
Step 3: Simplify the limit.
lim
n→∞
n3+ 3n2
2n3+ 1 = lim
n→∞
n3(1 + 3/n)
n3(2 + 1/n2)= lim
n→∞
1+3/n
2+1/n2=1
2
Step 4: Analyze the limit. Since the limit is a finite positive number (i.e.,
not zero or infinite), the series P∞
n=1 n2+3n
2n3+1 behaves similarly to the divergent
series P∞
n=1 1
n.
Step 5: Therefore, by the Limit Comparison Test, the series P∞
n=1 n2+3n
2n3+1
diverges.
Question 5
Question
Let {an}be a sequence such that an=n2+3
n4+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 examine the be-
havior of the sequence anas napproaches infinity.
Step 1: Find the limit of anas napproaches infinity.
lim
n→∞ an= lim
n→∞
n2+ 3
n4+ 1 = 0
Step 2: Use the Limit Comparison Test. Since limn→∞ an= 0, we
will use the Limit Comparison Test with the series P∞
n=1 1
n2, which is a p-series
with p= 2.
3
Step 3: Apply the Limit Comparison Test. Let bn=1
n2. We will
compare anand bn:
lim
n→∞
an
bn
= lim
n→∞
n2+3
n4+1
1
n2
= lim
n→∞
n4+ 3n2
n4+ 1 = 1
Step 4: Conclusion. Since P∞
n=1 1
n2converges by the p-series test with
p= 2, and limn→∞
an
bn= 1, by the Limit Comparison Test, the series P∞
n=1 an
also converges.
Question 6
Question
Determine the convergence or divergence of the series P∞
n=1 n!
(2n)! .
Solution
To determine the convergence of the series, we will use the ratio test. Step 1:
Compute the ratio of consecutive terms: Let an=n!
(2n)! . The ratio of consecutive
terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(2(n+ 1))!
n!/(2n)!
Step 2: Simplify the expression:
= lim
n→∞
(n+ 1)!
(2n+ 2)! ·(2n)!
n!
= lim
n→∞
n+ 1
(2n+ 1)(2n+ 2)
Step 3: Compute the limit:
= lim
n→∞
1
2
=1
2
Step 4: Determine convergence or divergence based on the ratio: Since the
limit is less than 1, by the ratio test, the series P∞
n=1 n!
(2n)! converges.
Question 7
Question
Determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+ 3 .
4
Solution
To determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+ 3 , we will
use the Limit Comparison Test. Let’s consider the series ∞
X
n=1
√n3+ 1
n2+ 3 and the
series ∞
X
n=1
1
n3/2.
Step 1: Find the necessary limit. Let an=√n3+1
n2+3 and bn=1
n3/2.
We will now calculate limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
√n3+ 1
n2+ 3 ·n3/2
1= lim
n→∞
n3+ 1
n2= lim
n→∞ 1 + 1
n2= 1
Step 2: Make a conclusion. Since lim
n→∞
an
bn
= 1 and ∞
X
n=1
1
n3/2is a conver-
gent p-series with p=3
2>1, by the Limit Comparison Test, we can conclude
that ∞
X
n=1
√n3+ 1
n2+ 3 converges.
Question 8
Question
Determine whether the series
∞
X
n=1
2n+ 3n
5n
converges or diverges.
Solution
To determine the convergence of the series, we can rewrite the terms in a simpler
form using the properties of exponents.
∞
X
n=1
2n+ 3n
5n=∞
X
n=1 2n
5n+3n
5n
=∞
X
n=1 2
5n
+3
5n
=∞
X
n=1 2
5n
+∞
X
n=1 3
5n
5
Now, we have the sum of two geometric series. The general form of a ge-
ometric series is P∞
n=1 arn−1where ais the first term and ris the common
ratio.
Step 1: Find the sum of the first geometric series:
∞
X
n=1 2
5n
=2/5
1−2/5=2
5(3/5) = 2
Step 2: Find the sum of the second geometric series:
∞
X
n=1 3
5n
=3/5
1−3/5=3
5(2/5) = 3
Step 3: Add the two sums to find the sum of the original series:
∞
X
n=1
2n+ 3n
5n= 2 + 3 = 5
Since the sum of the series is a finite value (5), the series P∞
n=1 2n+3n
5ncon-
verges.
Question 9
Question
Determine whether the series P∞
n=1 n+2
n3+n+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n+2
n3+n+1 , we will use the limit
comparison test.
Step 1: Find a simpler series to compare with. Consider the series
P∞
n=1 1
n2. This is a p-series with p= 2, and we know that it converges. We will
compare it with the given series by computing the limit of their ratios.
Step 2: Compute the limit of the ratios. Let an=n+2
n3+n+1 and bn=1
n2.
We will find limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n+2
n3+n+1
1
n2
= lim
n→∞
n3+n+ 1
n2(n+ 2) = lim
n→∞
n+ 1/n2+ 1/n3
n+ 2 = 1
Step 3: Apply the limit comparison test. Since the limit is a finite
positive number, by the limit comparison test, we can conclude that the series
P∞
n=1 n+2
n3+n+1 has the same convergence behavior as P∞
n=1 1
n2. Since the latter
series converges, the given series also converges.
Therefore, the series P∞
n=1 n+2
n3+n+1 converges.
6
Question 10
Question
Determine whether the series P∞
n=1 n10
(n+1)! converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms: Let an=n10
(n+1)! . We
compute
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)10
(n+ 2)! ·n!
n10
= lim
n→∞
(n+ 1)10
(n+ 2)(n+ 1)n10
= lim
n→∞
(n+ 1)10
(n+ 2)n10 = lim
n→∞
(1 + 1
n)10
(1 + 2
n)
Step 2: Evaluate the limit: Applying the limit to the above expression, we
get
=1
1= 1
Step 3: Apply the ratio test: Since the limit is equal to 1, we apply the
ratio test. If the limit is less than 1, the series converges; if it is greater than 1,
the series diverges; and if it is equal to 1, the test is inconclusive.
Therefore, since the limit is 1, the series P∞
n=1 n10
(n+1)! diverges by the Ratio
Test.
Question 11
Question
Determine whether the series ∞
X
n=1
n2+ 2n
3n3+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 2n
3n3+ 1 , we will use the Limit
Comparison Test with the series ∞
X
n=1
1
n.
Step 1: Define the two series. Let an=n2+2n
3n3+1 and bn=1
n.
Step 2: Find the limit of the ratio. We will find limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+2n
3n3+1
1
n
7
= lim
n→∞
n3+ 2n2
3n3+ 1 =1
3
Step 3: Apply the Limit Comparison Test. Since limn→∞
an
bn=1
3>0, and
∞
X
n=1
1
nis a divergent p-series with p= 1, by the Limit Comparison Test, the
series ∞
X
n=1
n2+ 2n
3n3+ 1 also diverges.
Therefore, the series ∞
X
n=1
n2+ 2n
3n3+ 1 diverges.
Question 12
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.
Step 1: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Simplify the expression:
lim
n→∞
(n+ 1)2
2n2·2n
2n+1
= lim
n→∞
(n+ 1)2
2n2·1
2
lim
n→∞
(n2+ 2n+ 1)
2n2·1
2
= lim
n→∞
1
2+1
n+1
2n2=1
2
Step 2: Analyze the result of the limit: If the limit is less than 1, the series
will converge. If it is greater than 1 or equal to 1, the series will diverge. Since
the limit is 1
2, the series ∞
X
n=1
n2
2nconverges by the ratio test.
Question 13
Question
Let (an) be a sequence such that an=n2
2nfor all n≥1. Determine whether the
series P∞
n=1 anconverges or diverges.
8
Solution
To analyze the convergence of the series P∞
n=1 an=P∞
n=1 n2
2n, we will use the
Ratio Test.
Step 1: Find the limit of the ratio of consecutive terms. Let’s
calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 2: Simplify the expression. Simplify the ratio:
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Apply the Ratio Test. Since the limit is 1
2<1, by the Ratio
Test, the series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 an=P∞
n=1 n2
2nconverges.
Question 14
Question
Let {an}be a sequence of positive real numbers such that limn→∞
an+1
an=L.
Determine the convergence of the series P∞
n=1 anin terms of L.
Solution
Given that limn→∞
an+1
an=L, we consider the following cases:
If L < 1, then we have an+1
an<1 for sufficiently large n, which implies
that an+1 < anfor sufficiently large n. This means that the terms of
the sequence eventually decrease, and by the Comparison Test, the series
P∞
n=1 anconverges.
If L > 1, then we have an+1
an>1 for sufficiently large n, which implies
that an+1 > anfor sufficiently large n. This means that the terms of the
sequence eventually increase, and the sequence {an}does not tend to zero.
Therefore, the series P∞
n=1 andiverges.
If L= 1, the ratio test is inconclusive, and we may need to consider
other tests or techniques to determine the convergence or divergence of
the series.
In summary:
If limn→∞
an+1
an<1, then P∞
n=1 anconverges.
If limn→∞
an+1
an>1, then P∞
n=1 andiverges.
9
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3
n!
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio R.
R= lim
n→∞
an+1
an
where an=n3
n!.
Step 2: Evaluate the ratio R.
R= lim
n→∞
(n+ 1)3/(n+ 1)!
n3/n!
= lim
n→∞
(n+ 1)3·n!
(n+ 1)! ·n3
= lim
n→∞
(n+ 1)3
(n+ 1)(n+ 1)(n+ 1)
= lim
n→∞
(n+ 1)2
(n+ 1)(n+ 1)
= 0
Step 3: Decide the convergence of the series based on the value of R. Since
R < 1, by the Ratio Test, the series P∞
n=1 n3
n!converges.
Question 16
Question
Prove or disprove the convergence of the series
∞
X
n=1
n2
2n.
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
10
Step 1: Compute the ratio test Using the ratio test, we consider the
limit
L= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Calculate the limit
L= lim
n→∞
(n+ 1)2/2n+1
n2/2n
= lim
n→∞
(n+ 1)2
2(n2)
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2.
Step 3: Analyze the limit Since the limit L=1
2is less than 1, by the
ratio test, the series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 17
Question
Determine the convergence or divergence of the series P∞
n=1 1
nln n.
Solution
Step 1: Let’s use the limit comparison test to determine the convergence of the
series.
Step 2: Consider the series P∞
n=1 1
n, which is a p-series with p= 1. This
series diverges (harmonic series).
Step 3: Now, we calculate the following limit:
lim
n→∞
1
nln n
1
n
= lim
n→∞
n
nln n= lim
n→∞
1
nln n−1
Step 4: Since the exponent ln n−1 grows larger than 1 as napproaches
infinity, the limit of the ratio is 0.
Step 5: By the limit comparison test, since the limit is a finite positive
number, we conclude that the series P∞
n=1 1
nln nconverges.
11
Question 18
Question
Determine whether the series ∞
X
n=1
n3+ 3
2n5+ 7 converges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n3+ 3
2n5+ 7, we can use the Limit
Comparison Test or Direct Comparison Test.
Step 1: Let’s first simplify the series:
n3+ 3
2n5+ 7 =n31 + 3
n3
n52
n2+7
n5=1 + 3
n3
2n2+ 7/n3
Step 2: The series ∞
X
n=1
1 + 3
n3
2n2+ 7/n3can be simplified further by applying
the Direct Comparison Test:
0≤1 + 3
n3
2n2+ 7/n3≤1 + 3
n3
2n2=1
2n2+3
2n5
Step 3: Now, consider the series ∞
X
n=1 1
2n2+3
2n5for comparison. Since
∞
X
n=1
1
2n2is a convergent p-series with p= 2 >1, and ∞
X
n=1
3
2n5is also a convergent
p-series with p= 5 >1, both series converge.
Step 4: By the Direct Comparison Test, since ∞
X
n=1
1 + 3
n3
2n2+ 7/n3is bounded
above by a convergent series, ∞
X
n=1
n3+ 3
2n5+ 7 converges by comparison.
Question 19
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
12
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 consider 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 inside the absolute value:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
(n+ 1)
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Use the limit definition of e as napproaches infinity:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Analyze the limit: Since 1
e<1, by the ratio test, the series ∞
X
n=1
n!
nn
converges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 20
Question
Determine whether the series ∞
X
n=1
n3
2nconverges or diverges.
Solution
Step 1: We will use the Ratio Test to determine the convergence of the series.
The Ratio Test states that for a series ∞
X
n=1
an, if limn→∞
an+1
an
<1, then the
series converges. If limn→∞
an+1
an
>1, then the series diverges.
Step 2: Calculate the ratio r= lim
n→∞
an+1
an
for our series:
r= lim
n→∞
(n+1)3
2n+1
n3
2n
13
= lim
n→∞
(n+ 1)3·2n
2n+1 ·n3
= lim
n→∞
(n+ 1)3
2n3
Step 3: Simplify the limit:
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1 + 3
n+3
n2+1
n3
2
=1
2
Step 4: Since r=1
2<1, by the Ratio Test, the series ∞
X
n=1
n3
2nconverges.
Question 21
Question
Determine whether the series P∞
n=1 n4
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n4
2n, we will use the ratio test.
Step 1: Compute the ratio test. Consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)42n
2n+1n4
= lim
n→∞
(n+ 1)4
2n4
Step 2: Simplify the expression inside the limit.
lim
n→∞
(n+ 1)4
2n4
= lim
n→∞
n4+ 4n3+ 6n2+ 4n+ 1
2n4
= lim
n→∞
1
2+2
n+3
n2+2
n3+1
n4
=1
2
Step 3: Check the limit. Since the limit is 1
2<1, by the ratio test, the
series P∞
n=1 n4
2nconverges.
Therefore, the series P∞
n=1 n4
2nconverges.
Question 22
Question
Let (an) be a sequence of nonnegative real numbers such that the series P∞
n=1 an
converges. Prove that the series P∞
n=1 √anan+1 also converges.
14
Solution
Let’s prove the convergence of the series P∞
n=1 √anan+1 using the Comparison
Test.
Step 1: Relationship between anand √anan+1
Note that for any nonnegative real numbers xand y, we have the inequality
2√xy ≤x+y. Applying this inequality to anand an+1, we get:
2√anan+1 ≤an+an+1
Step 2: Convergence of the series
Since the series P∞
n=1 anconverges, the sequence (an) converges to zero as
napproaches infinity.
Let’s denote Sas the sum of the series P∞
n=1 an. Thus, we have:
an→0 as n→ ∞
Step 3: Comparison Test
Now, we will compare the series P∞
n=1 √anan+1 to the series P∞
n=1
an+an+1
2.
Since P∞
n=1 anconverges, we have:
lim
n→∞
an+an+1
2= 0
Therefore, by the Comparison Test, and based on the inequality 2√anan+1 ≤
an+an+1, we conclude that the series P∞
n=1 √anan+1 converges.
Question 23
Question
Consider the series P∞
n=1 3n+1
n2+2 n. Determine whether the series converges or
diverges.
Solution
To determine the convergence or divergence of the series, we can use the root
test.
Step 1: Apply the root test Let’s define an=3n+1
n2+2 n. We will compute
the limit of the n-th root of the absolute value of an.
lim
n→∞
n
p|an|= lim
n→∞
3n+ 1
n2+ 2
Step 2: Evaluate the limit We can simplify the expression under the absolute
value:
lim
n→∞
3n+ 1
n2+ 2
= lim
n→∞
3 + 1
n
n+2
n
= lim
n→∞
3
n
= 0
15
Step 3: Analyze the limit result Since the limit of the n-th root of |an|is
less than 1, by the root test, the series P∞
n=1 3n+1
n2+2 nconverges.
Therefore, the given series converges.
Question 24
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. Let’s compare it to the series P∞
n=1 1
n.
Step 1: Find the limit. Consider the limit:
lim
n→∞
n2
n3+1
1
n
= lim
n→∞
n3
n3+ 1 = 1
Step 2: Apply the Limit Comparison Test. Since the limit is finite and
positive, by the Limit Comparison Test, P∞
n=1 n2
n3+1 and P∞
n=1 1
neither both
converge or both diverge.
Step 3: Evaluate the comparison series. The series P∞
n=1 1
nis a well-
known divergent series (harmonic series) as p-series with p= 1.
Step 4: Conclusion. Since the comparison series diverges, by the Limit
Comparison Test, the original series P∞
n=1 n2
n3+1 also diverges. Therefore, the
series P∞
n=1 n2
n3+1 diverges.
Question 25
Question
Determine whether the series P∞
n=1 n!
1·3·5···(2n−1) converges or diverges.
Solution
To determine the convergence of the given series P∞
n=1 n!
1·3·5···(2n−1) , we will use
the Ratio Test.
Step 1: Compute the ratio Rusing the formula for the Ratio Test:
R= lim
n→∞
an+1
an
where an=n!
1·3·5···(2n−1) .
16
Step 2: Find an+1 by plugging n+ 1 into the expression for an:
an+1 =(n+ 1)!
1·3·5···(2(n+ 1) −1)
Step 3: Compute the limit L:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!
1·3·5···(2(n+ 1) −1) ·1·3·5···(2n−1)
n!
Step 4: Simplify the expression:
L= lim
n→∞
(n+ 1)(2n)(2n−1)
2n+ 1
Step 5: Calculate the limit:
L= lim
n→∞
4n3+O(n2)
2n+ 1
= lim
n→∞
4n3
2n= lim
n→∞ 2n2=∞
Step 6: Analyze the limit L: Since the limit L=∞, the series P∞
n=1 n!
1·3·5···(2n−1)
diverges by the Ratio Test.
Question 26
Question
Determine whether the series P∞
n=1 n2
4n3+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=n2
4n3+5 .
Step 1: Find limn→∞
an
1
n
lim
n→∞
an
1
n
= lim
n→∞
n2
4n3+ 5 ·n= lim
n→∞
n3
4n3+ 5 = lim
n→∞
1
4 + 5
n3
=1
4
Step 2: Interpret the result Since the limit is a positive finite number, we
can conclude that P∞
n=1 n2
4n3+5 has the same convergence behavior as P∞
n=1 1
n.
Step 3: Apply the Limit Comparison Test We know that P∞
n=1 1
nis
the harmonic series, which diverges. Therefore, by the Limit Comparison Test,
the series P∞
n=1 n2
4n3+5 also diverges.
Question 27
Question
Show whether the series P∞
n=1 n2+1
n3+2 converges or diverges.
17
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series P∞
n=1 an=P∞
n=1 n2+1
n3+2 .
Step 1: Calculate the limit Find the limit of the ratio n2+1
n3+2 as nap-
proaches infinity.
lim
n→∞
n2+ 1
n3+ 2 = lim
n→∞
1 + 1
n2
n+2
n3
= 0
Step 2: Choose a sequence to compare Let’s choose a sequence bnsuch
that limn→∞
an
bnexists and is a finite positive number. A good choice in this
case is bn=1
n.
Step 3: Calculate the new limit Now, compute the limit of the ratio an
bn.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 1)/(n3+ 2)
1/n = lim
n→∞
n3+n
n3+ 2 = 1
Step 4: Apply the Limit Comparison Test Since the limit is a finite
positive number, by the Limit Comparison Test, if P∞
n=1 bnconverges, then
P∞
n=1 analso converges. Since P∞
n=1 1
ndiverges (Harmonic series), then the
series P∞
n=1 n2+1
n3+2 also diverges.
Question 28
Question
Consider the series P∞
n=1 n!
nn. Determine if the series converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=e−1
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
18
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
Step 1: Apply the ratio test: Let’s calculate the limit:
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
Using the properties of factorials, we simplify this expression:
= lim
n→∞
n+ 1
(n+ 1)n+1 = lim
n→∞
1
(1 + 1/n)n=1
e
Step 2: Analyze the limit: Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 30
Question
Let (an) be a sequence of positive numbers such that P∞
n=1 anconverges. Define
bn=nan
1+n2. Determine whether the series P∞
n=1 bnconverges or diverges.
Solution
Step 1: We will use the limit comparison test to determine convergence. Let
cn=an
n2. Since P∞
n=1 anconverges, we have that limn→∞ cn= 0.
Step 2: Now, compute the limit of the ratio of bnand cnas napproaches
infinity:
lim
n→∞
bn
cn
= lim
n→∞
nan
1+n2
an
n2
= lim
n→∞
n3
1 + n2= lim
n→∞
n3
n2= lim
n→∞ n=∞.
Step 3: Since the limit above diverges, we conclude by the limit comparison
test that P∞
n=1 bnalso diverges.
Question 31
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
19
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Compute
the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(n+ 1)n+1
.
Step 3: Simplify further and find the limit.
R= lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n= 0.
Step 4: Apply the ratio test. Since R < 1, the series P∞
n=1 n!
nnis absolutely
convergent.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 32
Question
Determine whether the series P∞
n=1 n2+1
n3−2converges or diverges.
Solution
To investigate the convergence of the series P∞
n=1 n2+1
n3−2, we can use the Limit
Comparison Test. Let’s denote an=n2+1
n3−2.
Step 1: Find the limit of the ratio of terms Consider bn=1
n, a known
divergent series. We will compare anand bnby computing the limit of their
ratio:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3−2·n
1= lim
n→∞
n3+n
n3−2= lim
n→∞
1 + 1
n2
1−2
n3
= 1.
Step 2: Apply the Limit Comparison Test Since the limit is a fi-
nite positive value, by the Limit Comparison Test, if P∞
n=1 bnconverges, then
P∞
n=1 anconverges. However, since P∞
n=1 1
ndiverges (by the p-series test with
p= 1), then P∞
n=1 n2+1
n3−2must also diverge.
Therefore, the series P∞
n=1 n2+1
n3−2diverges.
20
Question 33
Question
Determine whether the series ∞
X
n=1
1
n(ln n)pconverges or diverges, where pis a
positive constant.
Solution
To investigate the convergence of the given series, we will apply the Integral
Test.
Step 1: Express the series as an integral Consider the function f(x) =
1
x(ln x)pfor x≥2. This function is continuous, positive, and decreasing for
x≥2.
The integral of f(x) over the interval [2,∞) can be expressed as:
Z∞
2
1
x(ln x)pdx
Step 2: Evaluate the integral To evaluate this integral, let u= ln x.
Then du =1
xdx.
The integral becomes:
Z1
updu =1
(1 −p)up−1=1
(1 −p)(ln x)p−1
Step 3: Apply the Integral Test Since the integral R∞
2
1
x(ln x)pdx con-
verges if and only if the series ∞
X
n=2
1
n(ln n)pconverges, we must determine the
convergence of the improper integral:
Z∞
2
1
x(ln x)pdx = lim
b→∞ Zb
2
1
x(ln x)pdx = lim
b→∞
1
(1 −p)(ln b)p−1−1
(1 −p)(ln 2)p−1
Step 4: Conclude the convergence The integral converges if and only if
the limit is finite.
Since pis a positive constant, the limit is finite if p > 1. Thus, the series
∞
X
n=1
1
n(ln n)pconverges if p > 1 and diverges if p≤1.
21
Question 34
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Define bn=P2n
k=nak. Prove that limn→∞ bn= 0.
Solution
Suppose P∞
n=1 anconverges. We want to show that limn→∞ bn= 0 where
bn=P2n
k=nak.
Step 1: Decompose bn.Define sn=Pn
k=1 ak. Then we can express bnin
terms of sn:
bn=s2n−sn−1
Step 2: Consider the difference s2n−sn.Since P∞
n=1 anconverges, sn
is a Cauchy sequence. This implies that given ε > 0, there exists N∈Nsuch
that for all m, n > N:
|sm−sn|< ε
Step 3: Bound |s2n−sn|.Using the result from Step 2, we have for n>N:
|s2n−sn|< ε
Step 4: Bound |bn|.Substitute bn=s2n−sn−1:
|bn|=|s2n−sn−1|
=|s2n−sn+sn−sn−1|
≤ |s2n−sn|+|sn−sn−1|
<2ε
Step 5: Conclusion. Since εis arbitrary and |bn|<2ε, we have shown
that limn→∞ bn= 0.
Question 35
Question
Determine whether the series P∞
n=1 n2+3
2n3+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s choose a suitable series to compare with. Consider the series
P∞
n=1 1
n. This is a known divergent series.
22
Question 4
Question
Determine the convergence or divergence of the series P∞
n=1 n2+3n
2n3+1 .
Solution
To determine the convergence or divergence of the series P∞
n=1 n2+3n
2n3+1 , we will
use the limit comparison test.
Step 1: Let’s set an=n2+3n
2n3+1 .
Step 2: Find the limit of the ratio of anto a simpler series bn=1
n.
lim
n→∞
an
bn
= lim
n→∞
n2+3n
2n3+1
1
n
= lim
n→∞
n3+ 3n2
2n3+ 1
Step 3: Simplify the limit.
lim
n→∞
n3+ 3n2
2n3+ 1 = lim
n→∞
n3(1 + 3/n)
n3(2 + 1/n2)= lim
n→∞
1+3/n
2+1/n2=1
2
Step 4: Analyze the limit. Since the limit is a finite positive number (i.e.,
not zero or infinite), the series P∞
n=1 n2+3n
2n3+1 behaves similarly to the divergent
series P∞
n=1 1
n.
Step 5: Therefore, by the Limit Comparison Test, the series P∞
n=1 n2+3n
2n3+1
diverges.
Question 5
Question
Let {an}be a sequence such that an=n2+3
n4+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 examine the be-
havior of the sequence anas napproaches infinity.
Step 1: Find the limit of anas napproaches infinity.
lim
n→∞ an= lim
n→∞
n2+ 3
n4+ 1 = 0
Step 2: Use the Limit Comparison Test. Since limn→∞ an= 0, we
will use the Limit Comparison Test with the series P∞
n=1 1
n2, which is a p-series
with p= 2.
3
Step 3: Apply the Limit Comparison Test. Let bn=1
n2. We will
compare anand bn:
lim
n→∞
an
bn
= lim
n→∞
n2+3
n4+1
1
n2
= lim
n→∞
n4+ 3n2
n4+ 1 = 1
Step 4: Conclusion. Since P∞
n=1 1
n2converges by the p-series test with
p= 2, and limn→∞
an
bn= 1, by the Limit Comparison Test, the series P∞
n=1 an
also converges.
Question 6
Question
Determine the convergence or divergence of the series P∞
n=1 n!
(2n)! .
Solution
To determine the convergence of the series, we will use the ratio test. Step 1:
Compute the ratio of consecutive terms: Let an=n!
(2n)! . The ratio of consecutive
terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(2(n+ 1))!
n!/(2n)!
Step 2: Simplify the expression:
= lim
n→∞
(n+ 1)!
(2n+ 2)! ·(2n)!
n!
= lim
n→∞
n+ 1
(2n+ 1)(2n+ 2)
Step 3: Compute the limit:
= lim
n→∞
1
2
=1
2
Step 4: Determine convergence or divergence based on the ratio: Since the
limit is less than 1, by the ratio test, the series P∞
n=1 n!
(2n)! converges.
Question 7
Question
Determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+ 3 .
4
Solution
To determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+ 3 , we will
use the Limit Comparison Test. Let’s consider the series ∞
X
n=1
√n3+ 1
n2+ 3 and the
series ∞
X
n=1
1
n3/2.
Step 1: Find the necessary limit. Let an=√n3+1
n2+3 and bn=1
n3/2.
We will now calculate limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
√n3+ 1
n2+ 3 ·n3/2
1= lim
n→∞
n3+ 1
n2= lim
n→∞ 1 + 1
n2= 1
Step 2: Make a conclusion. Since lim
n→∞
an
bn
= 1 and ∞
X
n=1
1
n3/2is a conver-
gent p-series with p=3
2>1, by the Limit Comparison Test, we can conclude
that ∞
X
n=1
√n3+ 1
n2+ 3 converges.
Question 8
Question
Determine whether the series
∞
X
n=1
2n+ 3n
5n
converges or diverges.
Solution
To determine the convergence of the series, we can rewrite the terms in a simpler
form using the properties of exponents.
∞
X
n=1
2n+ 3n
5n=∞
X
n=1 2n
5n+3n
5n
=∞
X
n=1 2
5n
+3
5n
=∞
X
n=1 2
5n
+∞
X
n=1 3
5n
5
Now, we have the sum of two geometric series. The general form of a ge-
ometric series is P∞
n=1 arn−1where ais the first term and ris the common
ratio.
Step 1: Find the sum of the first geometric series:
∞
X
n=1 2
5n
=2/5
1−2/5=2
5(3/5) = 2
Step 2: Find the sum of the second geometric series:
∞
X
n=1 3
5n
=3/5
1−3/5=3
5(2/5) = 3
Step 3: Add the two sums to find the sum of the original series:
∞
X
n=1
2n+ 3n
5n= 2 + 3 = 5
Since the sum of the series is a finite value (5), the series P∞
n=1 2n+3n
5ncon-
verges.
Question 9
Question
Determine whether the series P∞
n=1 n+2
n3+n+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n+2
n3+n+1 , we will use the limit
comparison test.
Step 1: Find a simpler series to compare with. Consider the series
P∞
n=1 1
n2. This is a p-series with p= 2, and we know that it converges. We will
compare it with the given series by computing the limit of their ratios.
Step 2: Compute the limit of the ratios. Let an=n+2
n3+n+1 and bn=1
n2.
We will find limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n+2
n3+n+1
1
n2
= lim
n→∞
n3+n+ 1
n2(n+ 2) = lim
n→∞
n+ 1/n2+ 1/n3
n+ 2 = 1
Step 3: Apply the limit comparison test. Since the limit is a finite
positive number, by the limit comparison test, we can conclude that the series
P∞
n=1 n+2
n3+n+1 has the same convergence behavior as P∞
n=1 1
n2. Since the latter
series converges, the given series also converges.
Therefore, the series P∞
n=1 n+2
n3+n+1 converges.
6
Question 10
Question
Determine whether the series P∞
n=1 n10
(n+1)! converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms: Let an=n10
(n+1)! . We
compute
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)10
(n+ 2)! ·n!
n10
= lim
n→∞
(n+ 1)10
(n+ 2)(n+ 1)n10
= lim
n→∞
(n+ 1)10
(n+ 2)n10 = lim
n→∞
(1 + 1
n)10
(1 + 2
n)
Step 2: Evaluate the limit: Applying the limit to the above expression, we
get
=1
1= 1
Step 3: Apply the ratio test: Since the limit is equal to 1, we apply the
ratio test. If the limit is less than 1, the series converges; if it is greater than 1,
the series diverges; and if it is equal to 1, the test is inconclusive.
Therefore, since the limit is 1, the series P∞
n=1 n10
(n+1)! diverges by the Ratio
Test.
Question 11
Question
Determine whether the series ∞
X
n=1
n2+ 2n
3n3+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 2n
3n3+ 1 , we will use the Limit
Comparison Test with the series ∞
X
n=1
1
n.
Step 1: Define the two series. Let an=n2+2n
3n3+1 and bn=1
n.
Step 2: Find the limit of the ratio. We will find limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+2n
3n3+1
1
n
7
= lim
n→∞
n3+ 2n2
3n3+ 1 =1
3
Step 3: Apply the Limit Comparison Test. Since limn→∞
an
bn=1
3>0, and
∞
X
n=1
1
nis a divergent p-series with p= 1, by the Limit Comparison Test, the
series ∞
X
n=1
n2+ 2n
3n3+ 1 also diverges.
Therefore, the series ∞
X
n=1
n2+ 2n
3n3+ 1 diverges.
Question 12
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.
Step 1: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Simplify the expression:
lim
n→∞
(n+ 1)2
2n2·2n
2n+1
= lim
n→∞
(n+ 1)2
2n2·1
2
lim
n→∞
(n2+ 2n+ 1)
2n2·1
2
= lim
n→∞
1
2+1
n+1
2n2=1
2
Step 2: Analyze the result of the limit: If the limit is less than 1, the series
will converge. If it is greater than 1 or equal to 1, the series will diverge. Since
the limit is 1
2, the series ∞
X
n=1
n2
2nconverges by the ratio test.
Question 13
Question
Let (an) be a sequence such that an=n2
2nfor all n≥1. Determine whether the
series P∞
n=1 anconverges or diverges.
8
Solution
To analyze the convergence of the series P∞
n=1 an=P∞
n=1 n2
2n, we will use the
Ratio Test.
Step 1: Find the limit of the ratio of consecutive terms. Let’s
calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 2: Simplify the expression. Simplify the ratio:
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Apply the Ratio Test. Since the limit is 1
2<1, by the Ratio
Test, the series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 an=P∞
n=1 n2
2nconverges.
Question 14
Question
Let {an}be a sequence of positive real numbers such that limn→∞
an+1
an=L.
Determine the convergence of the series P∞
n=1 anin terms of L.
Solution
Given that limn→∞
an+1
an=L, we consider the following cases:
If L < 1, then we have an+1
an<1 for sufficiently large n, which implies
that an+1 < anfor sufficiently large n. This means that the terms of
the sequence eventually decrease, and by the Comparison Test, the series
P∞
n=1 anconverges.
If L > 1, then we have an+1
an>1 for sufficiently large n, which implies
that an+1 > anfor sufficiently large n. This means that the terms of the
sequence eventually increase, and the sequence {an}does not tend to zero.
Therefore, the series P∞
n=1 andiverges.
If L= 1, the ratio test is inconclusive, and we may need to consider
other tests or techniques to determine the convergence or divergence of
the series.
In summary:
If limn→∞
an+1
an<1, then P∞
n=1 anconverges.
If limn→∞
an+1
an>1, then P∞
n=1 andiverges.
9
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3
n!
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio R.
R= lim
n→∞
an+1
an
where an=n3
n!.
Step 2: Evaluate the ratio R.
R= lim
n→∞
(n+ 1)3/(n+ 1)!
n3/n!
= lim
n→∞
(n+ 1)3·n!
(n+ 1)! ·n3
= lim
n→∞
(n+ 1)3
(n+ 1)(n+ 1)(n+ 1)
= lim
n→∞
(n+ 1)2
(n+ 1)(n+ 1)
= 0
Step 3: Decide the convergence of the series based on the value of R. Since
R < 1, by the Ratio Test, the series P∞
n=1 n3
n!converges.
Question 16
Question
Prove or disprove the convergence of the series
∞
X
n=1
n2
2n.
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
10
Step 1: Compute the ratio test Using the ratio test, we consider the
limit
L= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Calculate the limit
L= lim
n→∞
(n+ 1)2/2n+1
n2/2n
= lim
n→∞
(n+ 1)2
2(n2)
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2.
Step 3: Analyze the limit Since the limit L=1
2is less than 1, by the
ratio test, the series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 17
Question
Determine the convergence or divergence of the series P∞
n=1 1
nln n.
Solution
Step 1: Let’s use the limit comparison test to determine the convergence of the
series.
Step 2: Consider the series P∞
n=1 1
n, which is a p-series with p= 1. This
series diverges (harmonic series).
Step 3: Now, we calculate the following limit:
lim
n→∞
1
nln n
1
n
= lim
n→∞
n
nln n= lim
n→∞
1
nln n−1
Step 4: Since the exponent ln n−1 grows larger than 1 as napproaches
infinity, the limit of the ratio is 0.
Step 5: By the limit comparison test, since the limit is a finite positive
number, we conclude that the series P∞
n=1 1
nln nconverges.
11
Question 18
Question
Determine whether the series ∞
X
n=1
n3+ 3
2n5+ 7 converges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n3+ 3
2n5+ 7, we can use the Limit
Comparison Test or Direct Comparison Test.
Step 1: Let’s first simplify the series:
n3+ 3
2n5+ 7 =n31 + 3
n3
n52
n2+7
n5=1 + 3
n3
2n2+ 7/n3
Step 2: The series ∞
X
n=1
1 + 3
n3
2n2+ 7/n3can be simplified further by applying
the Direct Comparison Test:
0≤1 + 3
n3
2n2+ 7/n3≤1 + 3
n3
2n2=1
2n2+3
2n5
Step 3: Now, consider the series ∞
X
n=1 1
2n2+3
2n5for comparison. Since
∞
X
n=1
1
2n2is a convergent p-series with p= 2 >1, and ∞
X
n=1
3
2n5is also a convergent
p-series with p= 5 >1, both series converge.
Step 4: By the Direct Comparison Test, since ∞
X
n=1
1 + 3
n3
2n2+ 7/n3is bounded
above by a convergent series, ∞
X
n=1
n3+ 3
2n5+ 7 converges by comparison.
Question 19
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
12
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 consider 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 inside the absolute value:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
(n+ 1)
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Use the limit definition of e as napproaches infinity:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Analyze the limit: Since 1
e<1, by the ratio test, the series ∞
X
n=1
n!
nn
converges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 20
Question
Determine whether the series ∞
X
n=1
n3
2nconverges or diverges.
Solution
Step 1: We will use the Ratio Test to determine the convergence of the series.
The Ratio Test states that for a series ∞
X
n=1
an, if limn→∞
an+1
an
<1, then the
series converges. If limn→∞
an+1
an
>1, then the series diverges.
Step 2: Calculate the ratio r= lim
n→∞
an+1
an
for our series:
r= lim
n→∞
(n+1)3
2n+1
n3
2n
13
= lim
n→∞
(n+ 1)3·2n
2n+1 ·n3
= lim
n→∞
(n+ 1)3
2n3
Step 3: Simplify the limit:
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1 + 3
n+3
n2+1
n3
2
=1
2
Step 4: Since r=1
2<1, by the Ratio Test, the series ∞
X
n=1
n3
2nconverges.
Question 21
Question
Determine whether the series P∞
n=1 n4
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n4
2n, we will use the ratio test.
Step 1: Compute the ratio test. Consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)42n
2n+1n4
= lim
n→∞
(n+ 1)4
2n4
Step 2: Simplify the expression inside the limit.
lim
n→∞
(n+ 1)4
2n4
= lim
n→∞
n4+ 4n3+ 6n2+ 4n+ 1
2n4
= lim
n→∞
1
2+2
n+3
n2+2
n3+1
n4
=1
2
Step 3: Check the limit. Since the limit is 1
2<1, by the ratio test, the
series P∞
n=1 n4
2nconverges.
Therefore, the series P∞
n=1 n4
2nconverges.
Question 22
Question
Let (an) be a sequence of nonnegative real numbers such that the series P∞
n=1 an
converges. Prove that the series P∞
n=1 √anan+1 also converges.
14
Solution
Let’s prove the convergence of the series P∞
n=1 √anan+1 using the Comparison
Test.
Step 1: Relationship between anand √anan+1
Note that for any nonnegative real numbers xand y, we have the inequality
2√xy ≤x+y. Applying this inequality to anand an+1, we get:
2√anan+1 ≤an+an+1
Step 2: Convergence of the series
Since the series P∞
n=1 anconverges, the sequence (an) converges to zero as
napproaches infinity.
Let’s denote Sas the sum of the series P∞
n=1 an. Thus, we have:
an→0 as n→ ∞
Step 3: Comparison Test
Now, we will compare the series P∞
n=1 √anan+1 to the series P∞
n=1
an+an+1
2.
Since P∞
n=1 anconverges, we have:
lim
n→∞
an+an+1
2= 0
Therefore, by the Comparison Test, and based on the inequality 2√anan+1 ≤
an+an+1, we conclude that the series P∞
n=1 √anan+1 converges.
Question 23
Question
Consider the series P∞
n=1 3n+1
n2+2 n. Determine whether the series converges or
diverges.
Solution
To determine the convergence or divergence of the series, we can use the root
test.
Step 1: Apply the root test Let’s define an=3n+1
n2+2 n. We will compute
the limit of the n-th root of the absolute value of an.
lim
n→∞
n
p|an|= lim
n→∞
3n+ 1
n2+ 2
Step 2: Evaluate the limit We can simplify the expression under the absolute
value:
lim
n→∞
3n+ 1
n2+ 2
= lim
n→∞
3 + 1
n
n+2
n
= lim
n→∞
3
n
= 0
15
Step 3: Analyze the limit result Since the limit of the n-th root of |an|is
less than 1, by the root test, the series P∞
n=1 3n+1
n2+2 nconverges.
Therefore, the given series converges.
Question 24
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. Let’s compare it to the series P∞
n=1 1
n.
Step 1: Find the limit. Consider the limit:
lim
n→∞
n2
n3+1
1
n
= lim
n→∞
n3
n3+ 1 = 1
Step 2: Apply the Limit Comparison Test. Since the limit is finite and
positive, by the Limit Comparison Test, P∞
n=1 n2
n3+1 and P∞
n=1 1
neither both
converge or both diverge.
Step 3: Evaluate the comparison series. The series P∞
n=1 1
nis a well-
known divergent series (harmonic series) as p-series with p= 1.
Step 4: Conclusion. Since the comparison series diverges, by the Limit
Comparison Test, the original series P∞
n=1 n2
n3+1 also diverges. Therefore, the
series P∞
n=1 n2
n3+1 diverges.
Question 25
Question
Determine whether the series P∞
n=1 n!
1·3·5···(2n−1) converges or diverges.
Solution
To determine the convergence of the given series P∞
n=1 n!
1·3·5···(2n−1) , we will use
the Ratio Test.
Step 1: Compute the ratio Rusing the formula for the Ratio Test:
R= lim
n→∞
an+1
an
where an=n!
1·3·5···(2n−1) .
16
Step 2: Find an+1 by plugging n+ 1 into the expression for an:
an+1 =(n+ 1)!
1·3·5···(2(n+ 1) −1)
Step 3: Compute the limit L:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!
1·3·5···(2(n+ 1) −1) ·1·3·5···(2n−1)
n!
Step 4: Simplify the expression:
L= lim
n→∞
(n+ 1)(2n)(2n−1)
2n+ 1
Step 5: Calculate the limit:
L= lim
n→∞
4n3+O(n2)
2n+ 1
= lim
n→∞
4n3
2n= lim
n→∞ 2n2=∞
Step 6: Analyze the limit L: Since the limit L=∞, the series P∞
n=1 n!
1·3·5···(2n−1)
diverges by the Ratio Test.
Question 26
Question
Determine whether the series P∞
n=1 n2
4n3+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=n2
4n3+5 .
Step 1: Find limn→∞
an
1
n
lim
n→∞
an
1
n
= lim
n→∞
n2
4n3+ 5 ·n= lim
n→∞
n3
4n3+ 5 = lim
n→∞
1
4 + 5
n3
=1
4
Step 2: Interpret the result Since the limit is a positive finite number, we
can conclude that P∞
n=1 n2
4n3+5 has the same convergence behavior as P∞
n=1 1
n.
Step 3: Apply the Limit Comparison Test We know that P∞
n=1 1
nis
the harmonic series, which diverges. Therefore, by the Limit Comparison Test,
the series P∞
n=1 n2
4n3+5 also diverges.
Question 27
Question
Show whether the series P∞
n=1 n2+1
n3+2 converges or diverges.
17
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series P∞
n=1 an=P∞
n=1 n2+1
n3+2 .
Step 1: Calculate the limit Find the limit of the ratio n2+1
n3+2 as nap-
proaches infinity.
lim
n→∞
n2+ 1
n3+ 2 = lim
n→∞
1 + 1
n2
n+2
n3
= 0
Step 2: Choose a sequence to compare Let’s choose a sequence bnsuch
that limn→∞
an
bnexists and is a finite positive number. A good choice in this
case is bn=1
n.
Step 3: Calculate the new limit Now, compute the limit of the ratio an
bn.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 1)/(n3+ 2)
1/n = lim
n→∞
n3+n
n3+ 2 = 1
Step 4: Apply the Limit Comparison Test Since the limit is a finite
positive number, by the Limit Comparison Test, if P∞
n=1 bnconverges, then
P∞
n=1 analso converges. Since P∞
n=1 1
ndiverges (Harmonic series), then the
series P∞
n=1 n2+1
n3+2 also diverges.
Question 28
Question
Consider the series P∞
n=1 n!
nn. Determine if the series converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=e−1
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
18
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
Step 1: Apply the ratio test: Let’s calculate the limit:
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
Using the properties of factorials, we simplify this expression:
= lim
n→∞
n+ 1
(n+ 1)n+1 = lim
n→∞
1
(1 + 1/n)n=1
e
Step 2: Analyze the limit: Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 30
Question
Let (an) be a sequence of positive numbers such that P∞
n=1 anconverges. Define
bn=nan
1+n2. Determine whether the series P∞
n=1 bnconverges or diverges.
Solution
Step 1: We will use the limit comparison test to determine convergence. Let
cn=an
n2. Since P∞
n=1 anconverges, we have that limn→∞ cn= 0.
Step 2: Now, compute the limit of the ratio of bnand cnas napproaches
infinity:
lim
n→∞
bn
cn
= lim
n→∞
nan
1+n2
an
n2
= lim
n→∞
n3
1 + n2= lim
n→∞
n3
n2= lim
n→∞ n=∞.
Step 3: Since the limit above diverges, we conclude by the limit comparison
test that P∞
n=1 bnalso diverges.
Question 31
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
19
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Compute
the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(n+ 1)n+1
.
Step 3: Simplify further and find the limit.
R= lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n= 0.
Step 4: Apply the ratio test. Since R < 1, the series P∞
n=1 n!
nnis absolutely
convergent.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 32
Question
Determine whether the series P∞
n=1 n2+1
n3−2converges or diverges.
Solution
To investigate the convergence of the series P∞
n=1 n2+1
n3−2, we can use the Limit
Comparison Test. Let’s denote an=n2+1
n3−2.
Step 1: Find the limit of the ratio of terms Consider bn=1
n, a known
divergent series. We will compare anand bnby computing the limit of their
ratio:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3−2·n
1= lim
n→∞
n3+n
n3−2= lim
n→∞
1 + 1
n2
1−2
n3
= 1.
Step 2: Apply the Limit Comparison Test Since the limit is a fi-
nite positive value, by the Limit Comparison Test, if P∞
n=1 bnconverges, then
P∞
n=1 anconverges. However, since P∞
n=1 1
ndiverges (by the p-series test with
p= 1), then P∞
n=1 n2+1
n3−2must also diverge.
Therefore, the series P∞
n=1 n2+1
n3−2diverges.
20
Question 33
Question
Determine whether the series ∞
X
n=1
1
n(ln n)pconverges or diverges, where pis a
positive constant.
Solution
To investigate the convergence of the given series, we will apply the Integral
Test.
Step 1: Express the series as an integral Consider the function f(x) =
1
x(ln x)pfor x≥2. This function is continuous, positive, and decreasing for
x≥2.
The integral of f(x) over the interval [2,∞) can be expressed as:
Z∞
2
1
x(ln x)pdx
Step 2: Evaluate the integral To evaluate this integral, let u= ln x.
Then du =1
xdx.
The integral becomes:
Z1
updu =1
(1 −p)up−1=1
(1 −p)(ln x)p−1
Step 3: Apply the Integral Test Since the integral R∞
2
1
x(ln x)pdx con-
verges if and only if the series ∞
X
n=2
1
n(ln n)pconverges, we must determine the
convergence of the improper integral:
Z∞
2
1
x(ln x)pdx = lim
b→∞ Zb
2
1
x(ln x)pdx = lim
b→∞
1
(1 −p)(ln b)p−1−1
(1 −p)(ln 2)p−1
Step 4: Conclude the convergence The integral converges if and only if
the limit is finite.
Since pis a positive constant, the limit is finite if p > 1. Thus, the series
∞
X
n=1
1
n(ln n)pconverges if p > 1 and diverges if p≤1.
21
Question 34
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Define bn=P2n
k=nak. Prove that limn→∞ bn= 0.
Solution
Suppose P∞
n=1 anconverges. We want to show that limn→∞ bn= 0 where
bn=P2n
k=nak.
Step 1: Decompose bn.Define sn=Pn
k=1 ak. Then we can express bnin
terms of sn:
bn=s2n−sn−1
Step 2: Consider the difference s2n−sn.Since P∞
n=1 anconverges, sn
is a Cauchy sequence. This implies that given ε > 0, there exists N∈Nsuch
that for all m, n > N:
|sm−sn|< ε
Step 3: Bound |s2n−sn|.Using the result from Step 2, we have for n>N:
|s2n−sn|< ε
Step 4: Bound |bn|.Substitute bn=s2n−sn−1:
|bn|=|s2n−sn−1|
=|s2n−sn+sn−sn−1|
≤ |s2n−sn|+|sn−sn−1|
<2ε
Step 5: Conclusion. Since εis arbitrary and |bn|<2ε, we have shown
that limn→∞ bn= 0.
Question 35
Question
Determine whether the series P∞
n=1 n2+3
2n3+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s choose a suitable series to compare with. Consider the series
P∞
n=1 1
n. This is a known divergent series.
22
Question 4
Question
Determine the convergence or divergence of the series P∞
n=1 n2+3n
2n3+1 .
Solution
To determine the convergence or divergence of the series P∞
n=1 n2+3n
2n3+1 , we will
use the limit comparison test.
Step 1: Let’s set an=n2+3n
2n3+1 .
Step 2: Find the limit of the ratio of anto a simpler series bn=1
n.
lim
n→∞
an
bn
= lim
n→∞
n2+3n
2n3+1
1
n
= lim
n→∞
n3+ 3n2
2n3+ 1
Step 3: Simplify the limit.
lim
n→∞
n3+ 3n2
2n3+ 1 = lim
n→∞
n3(1 + 3/n)
n3(2 + 1/n2)= lim
n→∞
1+3/n
2+1/n2=1
2
Step 4: Analyze the limit. Since the limit is a finite positive number (i.e.,
not zero or infinite), the series P∞
n=1 n2+3n
2n3+1 behaves similarly to the divergent
series P∞
n=1 1
n.
Step 5: Therefore, by the Limit Comparison Test, the series P∞
n=1 n2+3n
2n3+1
diverges.
Question 5
Question
Let {an}be a sequence such that an=n2+3
n4+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 examine the be-
havior of the sequence anas napproaches infinity.
Step 1: Find the limit of anas napproaches infinity.
lim
n→∞ an= lim
n→∞
n2+ 3
n4+ 1 = 0
Step 2: Use the Limit Comparison Test. Since limn→∞ an= 0, we
will use the Limit Comparison Test with the series P∞
n=1 1
n2, which is a p-series
with p= 2.
3
Step 3: Apply the Limit Comparison Test. Let bn=1
n2. We will
compare anand bn:
lim
n→∞
an
bn
= lim
n→∞
n2+3
n4+1
1
n2
= lim
n→∞
n4+ 3n2
n4+ 1 = 1
Step 4: Conclusion. Since P∞
n=1 1
n2converges by the p-series test with
p= 2, and limn→∞
an
bn= 1, by the Limit Comparison Test, the series P∞
n=1 an
also converges.
Question 6
Question
Determine the convergence or divergence of the series P∞
n=1 n!
(2n)! .
Solution
To determine the convergence of the series, we will use the ratio test. Step 1:
Compute the ratio of consecutive terms: Let an=n!
(2n)! . The ratio of consecutive
terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(2(n+ 1))!
n!/(2n)!
Step 2: Simplify the expression:
= lim
n→∞
(n+ 1)!
(2n+ 2)! ·(2n)!
n!
= lim
n→∞
n+ 1
(2n+ 1)(2n+ 2)
Step 3: Compute the limit:
= lim
n→∞
1
2
=1
2
Step 4: Determine convergence or divergence based on the ratio: Since the
limit is less than 1, by the ratio test, the series P∞
n=1 n!
(2n)! converges.
Question 7
Question
Determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+ 3 .
4
Solution
To determine the convergence or divergence of the series ∞
X
n=1
√n3+ 1
n2+ 3 , we will
use the Limit Comparison Test. Let’s consider the series ∞
X
n=1
√n3+ 1
n2+ 3 and the
series ∞
X
n=1
1
n3/2.
Step 1: Find the necessary limit. Let an=√n3+1
n2+3 and bn=1
n3/2.
We will now calculate limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
√n3+ 1
n2+ 3 ·n3/2
1= lim
n→∞
n3+ 1
n2= lim
n→∞ 1 + 1
n2= 1
Step 2: Make a conclusion. Since lim
n→∞
an
bn
= 1 and ∞
X
n=1
1
n3/2is a conver-
gent p-series with p=3
2>1, by the Limit Comparison Test, we can conclude
that ∞
X
n=1
√n3+ 1
n2+ 3 converges.
Question 8
Question
Determine whether the series
∞
X
n=1
2n+ 3n
5n
converges or diverges.
Solution
To determine the convergence of the series, we can rewrite the terms in a simpler
form using the properties of exponents.
∞
X
n=1
2n+ 3n
5n=∞
X
n=1 2n
5n+3n
5n
=∞
X
n=1 2
5n
+3
5n
=∞
X
n=1 2
5n
+∞
X
n=1 3
5n
5
Now, we have the sum of two geometric series. The general form of a ge-
ometric series is P∞
n=1 arn−1where ais the first term and ris the common
ratio.
Step 1: Find the sum of the first geometric series:
∞
X
n=1 2
5n
=2/5
1−2/5=2
5(3/5) = 2
Step 2: Find the sum of the second geometric series:
∞
X
n=1 3
5n
=3/5
1−3/5=3
5(2/5) = 3
Step 3: Add the two sums to find the sum of the original series:
∞
X
n=1
2n+ 3n
5n= 2 + 3 = 5
Since the sum of the series is a finite value (5), the series P∞
n=1 2n+3n
5ncon-
verges.
Question 9
Question
Determine whether the series P∞
n=1 n+2
n3+n+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n+2
n3+n+1 , we will use the limit
comparison test.
Step 1: Find a simpler series to compare with. Consider the series
P∞
n=1 1
n2. This is a p-series with p= 2, and we know that it converges. We will
compare it with the given series by computing the limit of their ratios.
Step 2: Compute the limit of the ratios. Let an=n+2
n3+n+1 and bn=1
n2.
We will find limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n+2
n3+n+1
1
n2
= lim
n→∞
n3+n+ 1
n2(n+ 2) = lim
n→∞
n+ 1/n2+ 1/n3
n+ 2 = 1
Step 3: Apply the limit comparison test. Since the limit is a finite
positive number, by the limit comparison test, we can conclude that the series
P∞
n=1 n+2
n3+n+1 has the same convergence behavior as P∞
n=1 1
n2. Since the latter
series converges, the given series also converges.
Therefore, the series P∞
n=1 n+2
n3+n+1 converges.
6
Question 10
Question
Determine whether the series P∞
n=1 n10
(n+1)! converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms: Let an=n10
(n+1)! . We
compute
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)10
(n+ 2)! ·n!
n10
= lim
n→∞
(n+ 1)10
(n+ 2)(n+ 1)n10
= lim
n→∞
(n+ 1)10
(n+ 2)n10 = lim
n→∞
(1 + 1
n)10
(1 + 2
n)
Step 2: Evaluate the limit: Applying the limit to the above expression, we
get
=1
1= 1
Step 3: Apply the ratio test: Since the limit is equal to 1, we apply the
ratio test. If the limit is less than 1, the series converges; if it is greater than 1,
the series diverges; and if it is equal to 1, the test is inconclusive.
Therefore, since the limit is 1, the series P∞
n=1 n10
(n+1)! diverges by the Ratio
Test.
Question 11
Question
Determine whether the series ∞
X
n=1
n2+ 2n
3n3+ 1 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 2n
3n3+ 1 , we will use the Limit
Comparison Test with the series ∞
X
n=1
1
n.
Step 1: Define the two series. Let an=n2+2n
3n3+1 and bn=1
n.
Step 2: Find the limit of the ratio. We will find limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+2n
3n3+1
1
n
7
= lim
n→∞
n3+ 2n2
3n3+ 1 =1
3
Step 3: Apply the Limit Comparison Test. Since limn→∞
an
bn=1
3>0, and
∞
X
n=1
1
nis a divergent p-series with p= 1, by the Limit Comparison Test, the
series ∞
X
n=1
n2+ 2n
3n3+ 1 also diverges.
Therefore, the series ∞
X
n=1
n2+ 2n
3n3+ 1 diverges.
Question 12
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.
Step 1: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Simplify the expression:
lim
n→∞
(n+ 1)2
2n2·2n
2n+1
= lim
n→∞
(n+ 1)2
2n2·1
2
lim
n→∞
(n2+ 2n+ 1)
2n2·1
2
= lim
n→∞
1
2+1
n+1
2n2=1
2
Step 2: Analyze the result of the limit: If the limit is less than 1, the series
will converge. If it is greater than 1 or equal to 1, the series will diverge. Since
the limit is 1
2, the series ∞
X
n=1
n2
2nconverges by the ratio test.
Question 13
Question
Let (an) be a sequence such that an=n2
2nfor all n≥1. Determine whether the
series P∞
n=1 anconverges or diverges.
8
Solution
To analyze the convergence of the series P∞
n=1 an=P∞
n=1 n2
2n, we will use the
Ratio Test.
Step 1: Find the limit of the ratio of consecutive terms. Let’s
calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 2: Simplify the expression. Simplify the ratio:
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 3: Apply the Ratio Test. Since the limit is 1
2<1, by the Ratio
Test, the series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 an=P∞
n=1 n2
2nconverges.
Question 14
Question
Let {an}be a sequence of positive real numbers such that limn→∞
an+1
an=L.
Determine the convergence of the series P∞
n=1 anin terms of L.
Solution
Given that limn→∞
an+1
an=L, we consider the following cases:
If L < 1, then we have an+1
an<1 for sufficiently large n, which implies
that an+1 < anfor sufficiently large n. This means that the terms of
the sequence eventually decrease, and by the Comparison Test, the series
P∞
n=1 anconverges.
If L > 1, then we have an+1
an>1 for sufficiently large n, which implies
that an+1 > anfor sufficiently large n. This means that the terms of the
sequence eventually increase, and the sequence {an}does not tend to zero.
Therefore, the series P∞
n=1 andiverges.
If L= 1, the ratio test is inconclusive, and we may need to consider
other tests or techniques to determine the convergence or divergence of
the series.
In summary:
If limn→∞
an+1
an<1, then P∞
n=1 anconverges.
If limn→∞
an+1
an>1, then P∞
n=1 andiverges.
9
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n3
n!
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio R.
R= lim
n→∞
an+1
an
where an=n3
n!.
Step 2: Evaluate the ratio R.
R= lim
n→∞
(n+ 1)3/(n+ 1)!
n3/n!
= lim
n→∞
(n+ 1)3·n!
(n+ 1)! ·n3
= lim
n→∞
(n+ 1)3
(n+ 1)(n+ 1)(n+ 1)
= lim
n→∞
(n+ 1)2
(n+ 1)(n+ 1)
= 0
Step 3: Decide the convergence of the series based on the value of R. Since
R < 1, by the Ratio Test, the series P∞
n=1 n3
n!converges.
Question 16
Question
Prove or disprove the convergence of the series
∞
X
n=1
n2
2n.
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
10
Step 1: Compute the ratio test Using the ratio test, we consider the
limit
L= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Calculate the limit
L= lim
n→∞
(n+ 1)2/2n+1
n2/2n
= lim
n→∞
(n+ 1)2
2(n2)
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2.
Step 3: Analyze the limit Since the limit L=1
2is less than 1, by the
ratio test, the series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 17
Question
Determine the convergence or divergence of the series P∞
n=1 1
nln n.
Solution
Step 1: Let’s use the limit comparison test to determine the convergence of the
series.
Step 2: Consider the series P∞
n=1 1
n, which is a p-series with p= 1. This
series diverges (harmonic series).
Step 3: Now, we calculate the following limit:
lim
n→∞
1
nln n
1
n
= lim
n→∞
n
nln n= lim
n→∞
1
nln n−1
Step 4: Since the exponent ln n−1 grows larger than 1 as napproaches
infinity, the limit of the ratio is 0.
Step 5: By the limit comparison test, since the limit is a finite positive
number, we conclude that the series P∞
n=1 1
nln nconverges.
11
Question 18
Question
Determine whether the series ∞
X
n=1
n3+ 3
2n5+ 7 converges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n3+ 3
2n5+ 7, we can use the Limit
Comparison Test or Direct Comparison Test.
Step 1: Let’s first simplify the series:
n3+ 3
2n5+ 7 =n31 + 3
n3
n52
n2+7
n5=1 + 3
n3
2n2+ 7/n3
Step 2: The series ∞
X
n=1
1 + 3
n3
2n2+ 7/n3can be simplified further by applying
the Direct Comparison Test:
0≤1 + 3
n3
2n2+ 7/n3≤1 + 3
n3
2n2=1
2n2+3
2n5
Step 3: Now, consider the series ∞
X
n=1 1
2n2+3
2n5for comparison. Since
∞
X
n=1
1
2n2is a convergent p-series with p= 2 >1, and ∞
X
n=1
3
2n5is also a convergent
p-series with p= 5 >1, both series converge.
Step 4: By the Direct Comparison Test, since ∞
X
n=1
1 + 3
n3
2n2+ 7/n3is bounded
above by a convergent series, ∞
X
n=1
n3+ 3
2n5+ 7 converges by comparison.
Question 19
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
12
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 consider 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 inside the absolute value:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
(n+ 1)
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
Step 3: Use the limit definition of e as napproaches infinity:
lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Analyze the limit: Since 1
e<1, by the ratio test, the series ∞
X
n=1
n!
nn
converges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 20
Question
Determine whether the series ∞
X
n=1
n3
2nconverges or diverges.
Solution
Step 1: We will use the Ratio Test to determine the convergence of the series.
The Ratio Test states that for a series ∞
X
n=1
an, if limn→∞
an+1
an
<1, then the
series converges. If limn→∞
an+1
an
>1, then the series diverges.
Step 2: Calculate the ratio r= lim
n→∞
an+1
an
for our series:
r= lim
n→∞
(n+1)3
2n+1
n3
2n
13
= lim
n→∞
(n+ 1)3·2n
2n+1 ·n3
= lim
n→∞
(n+ 1)3
2n3
Step 3: Simplify the limit:
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1 + 3
n+3
n2+1
n3
2
=1
2
Step 4: Since r=1
2<1, by the Ratio Test, the series ∞
X
n=1
n3
2nconverges.
Question 21
Question
Determine whether the series P∞
n=1 n4
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n4
2n, we will use the ratio test.
Step 1: Compute the ratio test. Consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)42n
2n+1n4
= lim
n→∞
(n+ 1)4
2n4
Step 2: Simplify the expression inside the limit.
lim
n→∞
(n+ 1)4
2n4
= lim
n→∞
n4+ 4n3+ 6n2+ 4n+ 1
2n4
= lim
n→∞
1
2+2
n+3
n2+2
n3+1
n4
=1
2
Step 3: Check the limit. Since the limit is 1
2<1, by the ratio test, the
series P∞
n=1 n4
2nconverges.
Therefore, the series P∞
n=1 n4
2nconverges.
Question 22
Question
Let (an) be a sequence of nonnegative real numbers such that the series P∞
n=1 an
converges. Prove that the series P∞
n=1 √anan+1 also converges.
14
Solution
Let’s prove the convergence of the series P∞
n=1 √anan+1 using the Comparison
Test.
Step 1: Relationship between anand √anan+1
Note that for any nonnegative real numbers xand y, we have the inequality
2√xy ≤x+y. Applying this inequality to anand an+1, we get:
2√anan+1 ≤an+an+1
Step 2: Convergence of the series
Since the series P∞
n=1 anconverges, the sequence (an) converges to zero as
napproaches infinity.
Let’s denote Sas the sum of the series P∞
n=1 an. Thus, we have:
an→0 as n→ ∞
Step 3: Comparison Test
Now, we will compare the series P∞
n=1 √anan+1 to the series P∞
n=1
an+an+1
2.
Since P∞
n=1 anconverges, we have:
lim
n→∞
an+an+1
2= 0
Therefore, by the Comparison Test, and based on the inequality 2√anan+1 ≤
an+an+1, we conclude that the series P∞
n=1 √anan+1 converges.
Question 23
Question
Consider the series P∞
n=1 3n+1
n2+2 n. Determine whether the series converges or
diverges.
Solution
To determine the convergence or divergence of the series, we can use the root
test.
Step 1: Apply the root test Let’s define an=3n+1
n2+2 n. We will compute
the limit of the n-th root of the absolute value of an.
lim
n→∞
n
p|an|= lim
n→∞
3n+ 1
n2+ 2
Step 2: Evaluate the limit We can simplify the expression under the absolute
value:
lim
n→∞
3n+ 1
n2+ 2
= lim
n→∞
3 + 1
n
n+2
n
= lim
n→∞
3
n
= 0
15
Step 3: Analyze the limit result Since the limit of the n-th root of |an|is
less than 1, by the root test, the series P∞
n=1 3n+1
n2+2 nconverges.
Therefore, the given series converges.
Question 24
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. Let’s compare it to the series P∞
n=1 1
n.
Step 1: Find the limit. Consider the limit:
lim
n→∞
n2
n3+1
1
n
= lim
n→∞
n3
n3+ 1 = 1
Step 2: Apply the Limit Comparison Test. Since the limit is finite and
positive, by the Limit Comparison Test, P∞
n=1 n2
n3+1 and P∞
n=1 1
neither both
converge or both diverge.
Step 3: Evaluate the comparison series. The series P∞
n=1 1
nis a well-
known divergent series (harmonic series) as p-series with p= 1.
Step 4: Conclusion. Since the comparison series diverges, by the Limit
Comparison Test, the original series P∞
n=1 n2
n3+1 also diverges. Therefore, the
series P∞
n=1 n2
n3+1 diverges.
Question 25
Question
Determine whether the series P∞
n=1 n!
1·3·5···(2n−1) converges or diverges.
Solution
To determine the convergence of the given series P∞
n=1 n!
1·3·5···(2n−1) , we will use
the Ratio Test.
Step 1: Compute the ratio Rusing the formula for the Ratio Test:
R= lim
n→∞
an+1
an
where an=n!
1·3·5···(2n−1) .
16
Step 2: Find an+1 by plugging n+ 1 into the expression for an:
an+1 =(n+ 1)!
1·3·5···(2(n+ 1) −1)
Step 3: Compute the limit L:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!
1·3·5···(2(n+ 1) −1) ·1·3·5···(2n−1)
n!
Step 4: Simplify the expression:
L= lim
n→∞
(n+ 1)(2n)(2n−1)
2n+ 1
Step 5: Calculate the limit:
L= lim
n→∞
4n3+O(n2)
2n+ 1
= lim
n→∞
4n3
2n= lim
n→∞ 2n2=∞
Step 6: Analyze the limit L: Since the limit L=∞, the series P∞
n=1 n!
1·3·5···(2n−1)
diverges by the Ratio Test.
Question 26
Question
Determine whether the series P∞
n=1 n2
4n3+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=n2
4n3+5 .
Step 1: Find limn→∞
an
1
n
lim
n→∞
an
1
n
= lim
n→∞
n2
4n3+ 5 ·n= lim
n→∞
n3
4n3+ 5 = lim
n→∞
1
4 + 5
n3
=1
4
Step 2: Interpret the result Since the limit is a positive finite number, we
can conclude that P∞
n=1 n2
4n3+5 has the same convergence behavior as P∞
n=1 1
n.
Step 3: Apply the Limit Comparison Test We know that P∞
n=1 1
nis
the harmonic series, which diverges. Therefore, by the Limit Comparison Test,
the series P∞
n=1 n2
4n3+5 also diverges.
Question 27
Question
Show whether the series P∞
n=1 n2+1
n3+2 converges or diverges.
17
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series P∞
n=1 an=P∞
n=1 n2+1
n3+2 .
Step 1: Calculate the limit Find the limit of the ratio n2+1
n3+2 as nap-
proaches infinity.
lim
n→∞
n2+ 1
n3+ 2 = lim
n→∞
1 + 1
n2
n+2
n3
= 0
Step 2: Choose a sequence to compare Let’s choose a sequence bnsuch
that limn→∞
an
bnexists and is a finite positive number. A good choice in this
case is bn=1
n.
Step 3: Calculate the new limit Now, compute the limit of the ratio an
bn.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 1)/(n3+ 2)
1/n = lim
n→∞
n3+n
n3+ 2 = 1
Step 4: Apply the Limit Comparison Test Since the limit is a finite
positive number, by the Limit Comparison Test, if P∞
n=1 bnconverges, then
P∞
n=1 analso converges. Since P∞
n=1 1
ndiverges (Harmonic series), then the
series P∞
n=1 n2+1
n3+2 also diverges.
Question 28
Question
Consider the series P∞
n=1 n!
nn. Determine if the series converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=e−1
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
18
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we can use
the ratio test.
Step 1: Apply the ratio test: Let’s calculate the limit:
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
Using the properties of factorials, we simplify this expression:
= lim
n→∞
n+ 1
(n+ 1)n+1 = lim
n→∞
1
(1 + 1/n)n=1
e
Step 2: Analyze the limit: Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 30
Question
Let (an) be a sequence of positive numbers such that P∞
n=1 anconverges. Define
bn=nan
1+n2. Determine whether the series P∞
n=1 bnconverges or diverges.
Solution
Step 1: We will use the limit comparison test to determine convergence. Let
cn=an
n2. Since P∞
n=1 anconverges, we have that limn→∞ cn= 0.
Step 2: Now, compute the limit of the ratio of bnand cnas napproaches
infinity:
lim
n→∞
bn
cn
= lim
n→∞
nan
1+n2
an
n2
= lim
n→∞
n3
1 + n2= lim
n→∞
n3
n2= lim
n→∞ n=∞.
Step 3: Since the limit above diverges, we conclude by the limit comparison
test that P∞
n=1 bnalso diverges.
Question 31
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
19
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Compute
the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(n+ 1)n+1
.
Step 3: Simplify further and find the limit.
R= lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n= 0.
Step 4: Apply the ratio test. Since R < 1, the series P∞
n=1 n!
nnis absolutely
convergent.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 32
Question
Determine whether the series P∞
n=1 n2+1
n3−2converges or diverges.
Solution
To investigate the convergence of the series P∞
n=1 n2+1
n3−2, we can use the Limit
Comparison Test. Let’s denote an=n2+1
n3−2.
Step 1: Find the limit of the ratio of terms Consider bn=1
n, a known
divergent series. We will compare anand bnby computing the limit of their
ratio:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3−2·n
1= lim
n→∞
n3+n
n3−2= lim
n→∞
1 + 1
n2
1−2
n3
= 1.
Step 2: Apply the Limit Comparison Test Since the limit is a fi-
nite positive value, by the Limit Comparison Test, if P∞
n=1 bnconverges, then
P∞
n=1 anconverges. However, since P∞
n=1 1
ndiverges (by the p-series test with
p= 1), then P∞
n=1 n2+1
n3−2must also diverge.
Therefore, the series P∞
n=1 n2+1
n3−2diverges.
20
Question 33
Question
Determine whether the series ∞
X
n=1
1
n(ln n)pconverges or diverges, where pis a
positive constant.
Solution
To investigate the convergence of the given series, we will apply the Integral
Test.
Step 1: Express the series as an integral Consider the function f(x) =
1
x(ln x)pfor x≥2. This function is continuous, positive, and decreasing for
x≥2.
The integral of f(x) over the interval [2,∞) can be expressed as:
Z∞
2
1
x(ln x)pdx
Step 2: Evaluate the integral To evaluate this integral, let u= ln x.
Then du =1
xdx.
The integral becomes:
Z1
updu =1
(1 −p)up−1=1
(1 −p)(ln x)p−1
Step 3: Apply the Integral Test Since the integral R∞
2
1
x(ln x)pdx con-
verges if and only if the series ∞
X
n=2
1
n(ln n)pconverges, we must determine the
convergence of the improper integral:
Z∞
2
1
x(ln x)pdx = lim
b→∞ Zb
2
1
x(ln x)pdx = lim
b→∞
1
(1 −p)(ln b)p−1−1
(1 −p)(ln 2)p−1
Step 4: Conclude the convergence The integral converges if and only if
the limit is finite.
Since pis a positive constant, the limit is finite if p > 1. Thus, the series
∞
X
n=1
1
n(ln n)pconverges if p > 1 and diverges if p≤1.
21
Question 34
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Define bn=P2n
k=nak. Prove that limn→∞ bn= 0.
Solution
Suppose P∞
n=1 anconverges. We want to show that limn→∞ bn= 0 where
bn=P2n
k=nak.
Step 1: Decompose bn.Define sn=Pn
k=1 ak. Then we can express bnin
terms of sn:
bn=s2n−sn−1
Step 2: Consider the difference s2n−sn.Since P∞
n=1 anconverges, sn
is a Cauchy sequence. This implies that given ε > 0, there exists N∈Nsuch
that for all m, n > N:
|sm−sn|< ε
Step 3: Bound |s2n−sn|.Using the result from Step 2, we have for n>N:
|s2n−sn|< ε
Step 4: Bound |bn|.Substitute bn=s2n−sn−1:
|bn|=|s2n−sn−1|
=|s2n−sn+sn−sn−1|
≤ |s2n−sn|+|sn−sn−1|
<2ε
Step 5: Conclusion. Since εis arbitrary and |bn|<2ε, we have shown
that limn→∞ bn= 0.
Question 35
Question
Determine whether the series P∞
n=1 n2+3
2n3+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s choose a suitable series to compare with. Consider the series
P∞
n=1 1
n. This is a known divergent series.
22
Step 2: Calculate the limit limn→∞
n2+3
2n3+5
1
n
.
lim
n→∞
n2+3
2n3+5
1
n
= lim
n→∞
n3+n
2n3+ 5 ·n= lim
n→∞
1 + 1
n2
2 + 5
n3
=1
2
Therefore, the limit is a finite positive value.
Step 3: Since the limit is finite and positive, by the Limit Comparison Test,
the given series P∞
n=1 n2+3
2n3+5 converges/diverges in the same way as the series
P∞
n=1 1
n. Since P∞
n=1 1
ndiverges (by the p-series test with p= 1), the given
series P∞
n=1 n2+3
2n3+5 also diverges.
23
