MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 9
Liberty University
Question 1
Question
Determine whether the series P∞
n=1 n2+1
3n3−nconverges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series an=n2+1
3n3−nand bn=1
n. We will compare the
given series with the series P∞
n=1 1
nsince limn→∞ an
bn= limn→∞
(n2+1)
(3n3−n)n=
limn→∞
1+ 1
n2
3−1
n2
=1
3which is a positive constant.
Step 1: Check the convergence of P∞
n=1 1
n
The series P∞
n=1 1
nis a harmonic series, which is known to diverge.
Step 2: Apply the limit comparison test
Since limn→∞ an
bn=1
3>0 and the series P∞
n=1 1
ndiverges, by the limit
comparison test, the series P∞
n=1 n2+1
3n3−nalso diverges.
Question 2
Question
Determine whether the series P∞
n=1 n2
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
Step 1: Compute the limit using the ratio test: Let an=n2
2n. We compute
the ratio an+1
an:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 2: Analyze the limit: Since the limit is 1
2<1, the series P∞
n=1 n2
2n
converges by the ratio test.
Therefore, the given series converges.
Question 3
Question
Determine the convergence or divergence of the series P∞
n=1 3n3+n2
4n4+5 .
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test with a series that we know the convergence of.
Step 1: First, we will choose a series to compare with. Let’s consider the
series P∞
n=1 1
np, where p > 0.
Step 2: Next, we will compute the limit of the ratio of the given series to
the chosen series:
lim
n→∞
3n3+n2
4n4+5
1
np
= lim
n→∞
3n3+n2
4n4+ 5 ·np
1
= lim
n→∞
3n3+p+n2+p
4n4+p+ 5np
Step 3: Now, we will find the value of pthat will make the limit above finite
and non-zero: For convergence, we need the exponents of nin the numerator
and denominator to be the same. So we are looking for 3 + p= 4 + p, which
simplifies to 3 = 4, a contradiction. Therefore, the series P∞
n=1 3n3+n2
4n4+5 diverges
by the Limit Comparison Test.
2
Question 4
Question
Determine whether the series P∞
n=1 n2+3n+1
n3+n+5 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n+1
n3+n+5 , we will use the Limit
Comparison Test.
Step 1: Find a suitable series to compare. Let’s consider the series
P∞
n=1 1
n, which is a p-series with p= 1. This series is known to diverge.
Step 2: Compute the limit of the ratio. Calculate the limit limn→∞
n2+3n+1
n3+n+5
1
n
.
Step 3: Simplify the expression.
lim
n→∞
n2+3n+1
n3+n+5
1
n
= lim
n→∞
n2+ 3n+ 1
n3+n+ 5 ·n
= lim
n→∞
n3+ 3n2+n
n3+n+ 5
= lim
n→∞
1 + 3
n+1
n2
1 + 1
n2+5
n3
=1+0+0
1+0+0
= 1
Step 4: Apply the Limit Comparison Test. Since the limit is a finite
positive constant, both series either converge or diverge together. As the series
P∞
n=1 1
ndiverges, by the limit comparison test, the series P∞
n=1 n2+3n+1
n3+n+5 also
diverges. Thus, the given series diverges.
Question 5
Question
Determine the convergence or divergence of the series
∞
X
n=1
2n+ 1
3n2+n+ 5.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series whose convergence or divergence we already know.
3
Consider the series ∞
X
n=1
1
n.
This is the harmonic series, which we know diverges.
Step 2: Calculate the limit of the ratio of the two series.
We will consider the limit
L= lim
n→∞
2n+1
3n2+n+5
1
n
.
Step 3: Evaluate the limit.
L= lim
n→∞
2n2+n
3n2+n+ 5 = lim
n→∞
2 + 1
n
3 + 1
n+5
n2
=2
3.
Step 4: Interpret the result.
Since L > 0 and finite, by the Limit Comparison Test, we conclude that the
given series
∞
X
n=1
2n+ 1
3n2+n+ 5
converges/diverges if and only if the series P∞
n=1 1
nconverges/diverges.
Since the harmonic series diverges, by the Limit Comparison Test, the given
series also diverges.
Question 6
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine if the series P∞
n=1 n!
nnconverges or diverges, we can use the ratio
test.
Step 1: Compute the limit. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n+ 1 ·nn
n!
= lim
n→∞
(n+ 1)n
n+ 1
= lim
n→∞
1 + 1
n
= 1.
Step 2: Analyze the limit. Since the limit is equal to 1, we cannot make
any conclusions using the ratio test. Therefore, we need to consider another
test.
Step 3: Applying the ratio test again. We will consider the sequence
bn=nn
n!. We will compute:
lim
n→∞
bn+1
bn
= lim
n→∞
(n+ 1)n+1/(n+ 1)!
nn/n!
= lim
n→∞
(n+ 1)n+1
nn
= lim
n→∞ (1 + 1/n)n+1=1
e.
4
Step 4: Final conclusion. Since 1
e<1, the series P∞
n=1 n!
nnconverges by
the ratio test.
Question 7
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+n+ 1
n3+ 1 .
Solution
To determine the convergence or divergence of the series ∞
X
n=1
n2+n+ 1
n3+ 1 , we will
use the Limit Comparison Test.
Step 1: Let an=n2+n+ 1
n3+ 1 . We will compare anwith a simpler series
bn=1
n.
Step 2: Find the limit of the ratio lim
n→∞
an
bn
:
lim
n→∞
an
bn
= lim
n→∞
n2+n+ 1
n3+ 1 ·n
1= lim
n→∞
n3+n2+n
n3+ 1 = 1
Step 3: Since lim
n→∞
an
bn
= 1, and ∞
X
n=1
1
nis a harmonic series which diverges,
by the Limit Comparison Test, the series ∞
X
n=1
n2+n+ 1
n3+ 1 also diverges.
Question 8
Question
Determine if the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the ratio of successive terms. Let an=n!
nn. We calculate
the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=n+ 1
(n+ 1)n+1 ·nn
5
Step 2: Simplify the ratio.
an+1
an
=n+ 1
(n+ 1)n+1 ·nn=n+ 1
(n+ 1)n=1
(1 + 1
n)n
Step 3: Consider the limit of the ratio. Taking the limit as napproaches
infinity:
lim
n→∞
1
(1 + 1
n)n=1
e
Step 4: Apply the ratio test. Since the limit of the ratio is less than 1, by
the ratio test, the series P∞
n=1 n!
nnconverges.
Question 9
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Compute the limit for the ratio test: Let an=n!
nn. Then, consider
the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
L= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
Step 3: Further simplify the expression:
L= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1 = 1
Step 4: Analyze the result: Since L= 1, we cannot make a conclusion
about convergence or divergence of the series using the ratio test. Thus, we
have to try a different test.
Step 5: Use Stirling’s approximation: By Stirling’s approximation, we can
estimate n!∼√2πn n
en.
Therefore, n! grows faster than nnwhich implies that n!
nndoes not converge
to 0 as n→ ∞.
Step 6: Conclusion: Since the terms of the series n!
nndo not approach 0,
the series diverges by the Divergence Test.
6
Question 10
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test.
Let an=n!
nn. We compute the ratio
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
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→∞
1
1 + 1
nn
=1
e
Step 3: Analyze the result.
Since r=1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 11
Question
Consider the series P∞
n=1 n!
nn. Determine whether the series converges or di-
verges.
7
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. Then, calculate the limit:
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
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 3: Analyze the limit: Since the limit 1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 12
Question
Determine the convergence or divergence of the series ∞
X
n=1 n3+√n
n4+ 1 using the
limit comparison test.
Solution
To determine the convergence of the series ∞
X
n=1 n3+√n
n4+ 1 , we will use the
limit comparison test. Let’s denote an=n3+√n
n4+1 .
Step 1: Find a comparison series: We want to find a series ∞
X
n=1
bnsuch
that limn→∞ an
bnis a finite positive number.
Let’s choose bn=1
n.
Step 2: Calculate the limit: We will calculate the limit: limn→∞ an
bn.
8
lim
n→∞
an
bn
= lim
n→∞
n3+√n
n4+ 1 ·n
1
= lim
n→∞
n4+n3/2
n4+ 1
= lim
n→∞
1 + n−1/2
1 + n−4
=1+0
1+0
= 1
Step 3: Conclusion: Since limn→∞ an
bn= 1, by the limit comparison test,
the series ∞
X
n=1
n3+√n
n4+ 1 behaves the same as the harmonic series ∞
X
n=1
1
n.
Therefore, ∞
X
n=1
n3+√n
n4+ 1 diverges.
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n3
2n.
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
2n, we will use
the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n3
2n. We compute
the ratio an+1
an:
an+1
an
=(n+ 1)3
2n+1 ·2n
n3
=(n+ 1)3
2n3
=n3+ 3n2+ 3n+ 1
2n3
=1
2+3
2n+3
n2+1
n3.
Step 2: Apply the ratio test. Taking the limit as napproaches infinity, we
have
lim
n→∞ an+1
an= lim
n→∞ 1
2+3
2n+3
n2+1
n3
=1
2.
9
Since the limit is less than 1, by the Ratio Test, the series P∞
n=1 n3
2nconverges.
Therefore, the series P∞
n=1 n3
2nconverges.
Question 14
Question
Consider the series ∞
X
n=1
n2
2n. Determine whether the series converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
=1
2
Step 2: Determine the convergence. Since the limit is 1
2<1, by the ratio
test, the series converges.
Therefore, the series ∞
X
n=1
n2
2nconverges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
(−1)n√n+ 1
n2+ 1 .
Solution
Step 1: We can use the Alternating Series Test to determine the convergence of
the series. Let an=√n+1
n2+1 .
Step 2: We first need to show that anis positive, decreasing, and limn→∞ an=
0.
Step 3: Since n2+ 1 >0 for all n≥1, we have √n+1
n2+1 >0 for all n≥1.
Step 4: To show that anis decreasing, we examine the derivative of an=
√n+1
n2+1 .
10
Step 5: Let f(x) = √x+1
x2+1 . Then f′(x) = (x2+1)(1
2(x+1)−1/2)−(2x)(√x+1)
(x2+1)2.
Step 6: Simplifying, we get f′(x) = x2+1−x(x+1)
2(x2+1)3/2=1
2(x2+1)3/2>0 for all
x≥1.
Step 7: Since f′(x)>0 for all x≥1, f(x) is decreasing for all x≥1. This
implies that anis decreasing for all n≥1.
Step 8: Finally, we calculate the limit of anas napproaches infinity.
lim
n→∞
√n+ 1
n2+ 1 = lim
n→∞
1
nq1 + 1
n
= lim
n→∞
1
n= 0.
Step 9: By the Alternating Series Test, since anis positive, decreasing, and
limn→∞ an= 0, the series
∞
X
n=1
(−1)n√n+ 1
n2+ 1
converges.
Question 16
Question
Determine whether the series
∞
X
n=1
n2+ sin(n)
n3+ cos(n)
converges or diverges.
Solution
To analyze the convergence of the given series, we can use the Comparison Test.
Step 1: First, note that
n2+ sin(n)
n3+ cos(n)≤n2+ 1
n3−1for n≥1.
Step 2: Now, consider the series
∞
X
n=1
n2+ 1
n3−1.
Step 3: To determine its convergence, we can use the Limit Comparison
Test by comparing it to the harmonic series P∞
n=1 1
n.
Step 4: Let’s calculate the limit of the ratio of the two series:
L= lim
n→∞
(n2+ 1)/(n3−1)
1/n = lim
n→∞
n3+n
n3−1= 1.
11
Step 5: Since L > 0, both series either converge or diverge together.
Since P∞
n=1 1
nis a divergent harmonic series, by the Limit Comparison Test,
P∞
n=1 n2+1
n3−1is also divergent.
Step 6: Therefore, by the Comparison Test, the original series
∞
X
n=1
n2+ sin(n)
n3+ cos(n)
also diverges.
Question 17
Question
Determine whether the series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
and the series ∞
X
n=1
1
n
Step 1: Find the limit of the ratio of the two series. Let
an=n2+ 3n+ 1
n3+ 2n+ 5 and bn=1
n
Then,
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
n3+ 2n+ 5 ·n
1
Step 2: Simplify the expression and find the limit.
lim
n→∞
n2+ 3n+ 1
n3+ 2n+ 5 ·n
1= lim
n→∞
1 + 3
n+1
n2
1 + 2
n2+5
n3·n
12
Solving this limit, we get
lim
n→∞
1 + 3
n+1
n2
1 + 2
n2+5
n3·n= 1
Step 3: Apply the Limit Comparison Test. Since the limit is a finite positive
number, by the Limit Comparison Test, the given series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
converges if and only if the series
∞
X
n=1
1
n
converges. This series is the harmonic series which diverges. Therefore, the
given series also diverges.
Question 18
Question
Determine whether the series ∞
X
n=1
2n+ 3n
5nconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Apply the ratio test. Let an=2n+ 3n
5n. We will compute the limit:
L= lim
n→∞
an+1
an
Step 2: Find an+1 and simplify it. We have:
an+1 =2n+1 + 3n+1
5n+1 =2·2n+ 3 ·3n
5·5n=2
52n+ 3n
5n=2
5an
Step 3: Calculate the limit. Now, we substitute an+1 and aninto the
formula for L:
L= lim
n→∞
2
5an
an
= lim
n→∞
2
5
=2
5
Step 4: Interpret the result If L < 1, then the series converges. If L > 1,
the series diverges. In this case, L=2
5<1, so the series ∞
X
n=1
2n+ 3n
5nconverges
by the ratio test.
13
Question 19
Question
Determine the convergence of the series P∞
n=1 n2+1
n3+1 .
Solution
To analyze the convergence of the series, we can use the Limit Comparison Test.
Step 1: Let’s choose a comparison series. We can consider the series
P∞
n=1 1
n. This is a p-series with p= 1, which diverges.
Step 2: Take the limit of the ratio of the two series:
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1 = 1
Step 3: Since the limit is a positive constant, by the Limit Comparison Test,
the convergence of the original series P∞
n=1 n2+1
n3+1 is the same as the convergence
of the comparison series P∞
n=1 1
n, which diverges.
Therefore, the series P∞
n=1 n2+1
n3+1 also diverges.
Question 20
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms: Let an=n!
nn. Therefore,
the ratio of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio and compute the limit:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
=1
e
Step 3: Apply the ratio test: Since the limit is 1
e, the series P∞
n=1 n!
nn
converges by the ratio test.
Therefore, the series P∞
n=1 n!
nnconverges.
14
Question 21
Question
Determine whether the series ∞
X
n=1
n5
5n
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n5
5n,
we will use the Ratio Test. The Ratio Test states that if
lim
n→∞
an+1
an
<1,
then the series converges absolutely.
Step 1: Let an=n5
5n.
Step 2: Compute the ratio an+1
an
:
an+1
an
=(n+ 1)5
5n+1 ·5n
n5
=(n+ 1)5
5·5n·5n
n5
=(n+ 1)5
5n5
Step 3: Compute the limit of the ratio as napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)5
5n5
= lim
n→∞
(n5+ 5n4+ 10n3+ 10n2+ 5n+ 1)
5n5
= lim
n→∞
1 + 5
n+10
n2+10
n3+5
n4+1
n5
5
=1
5
Step 4: Since limn→∞
an+1
an=1
5<1, by the Ratio Test, the series P∞
n=1 n5
5n
converges absolutely.
15
Question 22
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Apply the ratio test. Let anbe the general term of the series:
an=n!
nn. We need to compute the following limit:
L= lim
n→∞
an+1
an
where
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=(n+ 1)nn
(n+ 1)n+1 =nn
(n+ 1)n=n
n+ 1n
Step 2: Simplify the limit. Then,
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=e−1
Step 3: Analyze the limit. Since L=e−1<1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Therefore, the series converges.
Question 23
Question
Determine whether the series P∞
n=1 n2+1
2n3−nconverges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Find the limit of the ratio. Let an=n2+1
2n3−nand bn=1
n. We will
find the limit of an
bnas napproaches infinity.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 1)/(2n3−n)
1/n = lim
n→∞
n2+ 1
2n3−n·n= lim
n→∞
n3+n
2n3−n= lim
n→∞
n3(1 + 1/n)
n3(2 −1/n)= lim
n→∞
1+1/n
2−1/n =1+0
2−0=1
2
16
Step 2: Interpret the limit. Since limn→∞ an
bn=1
2>0, we can conclude
that either both series Panand Pbnconverge or both diverge.
Step 3: Determine the convergence of Pbn=P1
n. The series P1
nis the
harmonic series, which is a divergent series.
Step 4: Conclude the convergence of Pan. Since the series Pbndiverges
and the limit comparison test gives a positive value for the limit of the ratio,
we can conclude that the series Pan=Pn2+1
2n3−nalso diverges.
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the Ratio Test.
Step 1: Apply the Ratio Test, which states that if limn→∞
an+1
an=L, then
the series converges if L < 1 or diverges if L > 1.
Let an=n!
nn. Then,
L= lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1+1/n
= 1
Step 2: Since L= 1, the Ratio Test is inconclusive. We need to use another
test to determine the convergence of the series.
Step 3: Notice that for large values of n, the factor n! in the numerator
grows much faster than nnin the denominator. Therefore, we can see that the
17
terms of the series do not approach zero, so the series diverges by the Test for
Divergence.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 25
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Calculate the ratio test. Let an=n!
nn. We compute:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1!n
=1
e≈0.3679
Step 2: Apply the ratio test. Since the limit is less than 1, the series
P∞
n=1 n!
nnconverges by the ratio test.
Question 26
Question
Determine whether the series P∞
n=1 n2+1
n4+1 converges or diverges.
18
Solution
To determine the convergence of the series, we can use the limit comparison
test. We will compare the given series to a known series whose convergence is
already known.
Step 1: Find a suitable series to compare with. Let’s compare the given
series with the p-series P∞
n=1 1
n2, since the exponent of nin the denominator is
larger than that of the numerator.
Step 2: Take the limit of the ratio of the two series. Consider the limit
lim
n→∞
n2+1
n4+1
1
n2
= lim
n→∞
n2(n2+ 1)
n4+ 1
Step 3: Simplify the expression.
= lim
n→∞
n4+n2
n4+ 1 = 1
Step 4: Analyze the limit. Since the limit is a finite positive value, by the
limit comparison test, the convergence of the original series P∞
n=1 n2+1
n4+1 is the
same as the convergence of the p-series P∞
n=1 1
n2.
Step 5: Conclude the convergence. The p-series P∞
n=1 1
n2converges as p=
2>1. Therefore, by the limit comparison test, the original series P∞
n=1 n2+1
n4+1
also converges.
Question 27
Question
Determine whether the series P∞
n=1 n3+3n+1
n4+4 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
already known.
Step 1: Let’s consider the series P∞
n=1 1
n, which is a p-series with p= 1.
This series is known to diverge.
Step 2: Let an=n3+3n+1
n4+4 be the general term of the given series.
Step 3: We will calculate the limit of the ratio between the general term of
the given series and the general term of the series P∞
n=1 1
n:
lim
n→∞
an
1/n = lim
n→∞
n3+ 3n+ 1
n4+ 4 ·n
1= lim
n→∞
n4+ 3n2+n
n4+ 4 = 1
Step 4: Since the limit is a positive finite number, by the Limit Comparison
Test, we conclude that both series either converge or diverge.
19
Step 5: Therefore, by the Limit Comparison Test, the series P∞
n=1 n3+3n+1
n4+4
has the same convergence behavior as the harmonic series P∞
n=1 1
n. Since the
harmonic series diverges, we can conclude that the given series diverges as well.
Question 28
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 an=n!
nn. We will calculate the limit of
the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/(nn)
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
=1
e<1
Step 2: Determine the convergence. Since the limit of the ratio is less than
1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 29
Question
Determine whether the series ∞
X
n=1
n2+ 1
n4+ 3
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s find a suitable series to compare with. We can rewrite the
given series as
∞
X
n=1
n2+ 1
n4+ 3 =∞
X
n=1
1 + 1
n2
n2+3
n2
20
Step 2: Consider the series
∞
X
n=1
1
n2
This is a p-series with p= 2, which we know to converge.
Step 3: Now, let’s evaluate the limit
lim
n→∞
1 + 1
n2
n2+3
n2
Step 4: We can simplify the limit as follows:
lim
n→∞
1 + 1
n2
n2+3
n2
= lim
n→∞
1/n2+1
n4
n2/n2+3
n2n2
= lim
n→∞
0+0
1+0
= 0
Step 5: Since the limit is a positive finite number, by the Limit Comparison
Test, the given series
∞
X
n=1
n2+ 1
n4+ 3
converges.
Question 30
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn, then
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)! ·nn
(n+ 1)n+1 ·n!=(n+ 1) ·nn
(n+ 1)n+1 =nn
(n+ 1)n=1 + 1
n−n
Step 2: Apply the limit of the ratio. We will compute the limit of the ratio
as napproaches infinity: limn→∞ 1 + 1
n−n.
Using the limit form (1+ 1
n)nfor e, we rewrite the expression as limn→∞ 1
(1+ 1
n)n.
21
Step 3: Determine the limit. As napproaches infinity, 1 + 1
nnapproaches
e. Thus, limn→∞ 1 + 1
n−n=1
e= 0.
Step 4: Conclusion. Since the limit of the ratio is not 0, the series P∞
n=1 n!
nn
diverges by the ratio test.
Question 31
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test:
lim
n→∞
an+1
an
,
where an=n!
nn.
Step 1: Find the ratio of consecutive terms The ratio of consecutive
terms is:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!
=n+ 1
(n+ 1)n+1 ·nn
=n+ 1
(n+ 1)n+1 ·nn
=n
(n+ 1)n.
Step 2: Find the limit of the ratio Taking the limit as napproaches
22
infinity:
lim
n→∞
an+1
an
= lim
n→∞
n
(n+ 1)n
= lim
n→∞
n1
(1 + 1
n)n
= lim
n→∞
n1
e
=∞.
Step 3: Evaluate the ratio test Since the limit is infinity, the ratio test
tells us that the series ∞
X
n=1
n!
nndiverges.
Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 32
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Then, we
compute
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
.
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n.
Step 3: Apply limit properties.
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e.
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
23
Question 33
Question
Let {an}be a sequence such that an=n2+1
3n3+4n+5 . Determine if the series
P∞
n=1 anconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 an, we will use the limit com-
parison test. We will compare the given series with a series whose behavior we
know.
Step 1: Find a series for comparison Consider the series P∞
n=1 1
n. This
is the harmonic series, which is known to diverge.
Step 2: Form the ratio limn→∞ an
1/n We calculate the limit:
lim
n→∞
an
1/n = lim
n→∞
(n2+ 1)/(3n3+ 4n+ 5)
1/n .
Step 3: Simplify the expression Simplify the expression to get:
lim
n→∞
n2+ 1
3n3+ 4n+ 5 ·n
1= lim
n→∞
n3+n
3n3+ 4n+ 5.
Step 4: Evaluate the limit Divide through by the highest power of nin
the denominator to get:
lim
n→∞
1+1/n2
3+4/n2+ 5/n3.
As n→ ∞, the terms with 1/n2and 1/n3approach 0, leaving:
1
3.
Step 5: Interpret the result Since the limit limn→∞ an
1/n is a finite positive
number, by the limit comparison test, we conclude that the series P∞
n=1 anhas
the same convergence behavior as the harmonic series. Thus, the series P∞
n=1 an
diverges.
Question 34
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Show that the series P∞
n=1
nan
1+n2anconverges.
24
Solution
To show that the series P∞
n=1
nan
1+n2anconverges, we will use the Limit Compar-
ison Test.
Step 1: Find a suitable series to compare to. Consider the series
P∞
n=1 1
n2. This series is known to converge since it is a p-series with p= 2.
Step 2: Compute the limit of the ratio of the two series. We consider
the limit:
lim
n→∞
nan
1+n2an
1
n2
= lim
n→∞
n3an
1 + n2an
.
Step 3: Simplify the expression to find the limit. Rewrite the limit
as:
lim
n→∞
n3an
1 + n2an
= lim
n→∞
n2an
1
n2+an
.
Step 4: Use the comparison test. Since P∞
n=1 anconverges, it follows
that an→0 as n→ ∞. Thus, for large n, we have an<1
n2. Therefore, the
series P∞
n=1
nan
1+n2anconverges by the Limit Comparison Test with P∞
n=1 1
n2.
Question 35
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test. The ratio
test states that for a series P∞
n=1 an, if limn→∞
an+1
an=L, then the series
converges if L < 1 and diverges if L > 1.
Step 1: Calculate the ratio Compute the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
n!·nn
(n+ 1)n+1
Simplify the expression:
an+1
an
= (n+ 1) ·nn
(n+ 1)n+1 =n+ 1
(1 + 1
n)n
Step 2: Find the limit Calculate the limit as napproaches infinity:
lim
n→∞
n+ 1
(1 + 1
n)n
This limit can be evaluated using L’Hˆopital’s Rule or recognizing it as the
form of the limit definition of e:
25
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
Step 1: Compute the limit using the ratio test: Let an=n2
2n. We compute
the ratio an+1
an:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 2: Analyze the limit: Since the limit is 1
2<1, the series P∞
n=1 n2
2n
converges by the ratio test.
Therefore, the given series converges.
Question 3
Question
Determine the convergence or divergence of the series P∞
n=1 3n3+n2
4n4+5 .
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test with a series that we know the convergence of.
Step 1: First, we will choose a series to compare with. Let’s consider the
series P∞
n=1 1
np, where p > 0.
Step 2: Next, we will compute the limit of the ratio of the given series to
the chosen series:
lim
n→∞
3n3+n2
4n4+5
1
np
= lim
n→∞
3n3+n2
4n4+ 5 ·np
1
= lim
n→∞
3n3+p+n2+p
4n4+p+ 5np
Step 3: Now, we will find the value of pthat will make the limit above finite
and non-zero: For convergence, we need the exponents of nin the numerator
and denominator to be the same. So we are looking for 3 + p= 4 + p, which
simplifies to 3 = 4, a contradiction. Therefore, the series P∞
n=1 3n3+n2
4n4+5 diverges
by the Limit Comparison Test.
2
Question 4
Question
Determine whether the series P∞
n=1 n2+3n+1
n3+n+5 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n+1
n3+n+5 , we will use the Limit
Comparison Test.
Step 1: Find a suitable series to compare. Let’s consider the series
P∞
n=1 1
n, which is a p-series with p= 1. This series is known to diverge.
Step 2: Compute the limit of the ratio. Calculate the limit limn→∞
n2+3n+1
n3+n+5
1
n
.
Step 3: Simplify the expression.
lim
n→∞
n2+3n+1
n3+n+5
1
n
= lim
n→∞
n2+ 3n+ 1
n3+n+ 5 ·n
= lim
n→∞
n3+ 3n2+n
n3+n+ 5
= lim
n→∞
1 + 3
n+1
n2
1 + 1
n2+5
n3
=1+0+0
1+0+0
= 1
Step 4: Apply the Limit Comparison Test. Since the limit is a finite
positive constant, both series either converge or diverge together. As the series
P∞
n=1 1
ndiverges, by the limit comparison test, the series P∞
n=1 n2+3n+1
n3+n+5 also
diverges. Thus, the given series diverges.
Question 5
Question
Determine the convergence or divergence of the series
∞
X
n=1
2n+ 1
3n2+n+ 5.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series whose convergence or divergence we already know.
3
Consider the series ∞
X
n=1
1
n.
This is the harmonic series, which we know diverges.
Step 2: Calculate the limit of the ratio of the two series.
We will consider the limit
L= lim
n→∞
2n+1
3n2+n+5
1
n
.
Step 3: Evaluate the limit.
L= lim
n→∞
2n2+n
3n2+n+ 5 = lim
n→∞
2 + 1
n
3 + 1
n+5
n2
=2
3.
Step 4: Interpret the result.
Since L > 0 and finite, by the Limit Comparison Test, we conclude that the
given series
∞
X
n=1
2n+ 1
3n2+n+ 5
converges/diverges if and only if the series P∞
n=1 1
nconverges/diverges.
Since the harmonic series diverges, by the Limit Comparison Test, the given
series also diverges.
Question 6
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine if the series P∞
n=1 n!
nnconverges or diverges, we can use the ratio
test.
Step 1: Compute the limit. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n+ 1 ·nn
n!
= lim
n→∞
(n+ 1)n
n+ 1
= lim
n→∞
1 + 1
n
= 1.
Step 2: Analyze the limit. Since the limit is equal to 1, we cannot make
any conclusions using the ratio test. Therefore, we need to consider another
test.
Step 3: Applying the ratio test again. We will consider the sequence
bn=nn
n!. We will compute:
lim
n→∞
bn+1
bn
= lim
n→∞
(n+ 1)n+1/(n+ 1)!
nn/n!
= lim
n→∞
(n+ 1)n+1
nn
= lim
n→∞ (1 + 1/n)n+1=1
e.
4
Step 4: Final conclusion. Since 1
e<1, the series P∞
n=1 n!
nnconverges by
the ratio test.
Question 7
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+n+ 1
n3+ 1 .
Solution
To determine the convergence or divergence of the series ∞
X
n=1
n2+n+ 1
n3+ 1 , we will
use the Limit Comparison Test.
Step 1: Let an=n2+n+ 1
n3+ 1 . We will compare anwith a simpler series
bn=1
n.
Step 2: Find the limit of the ratio lim
n→∞
an
bn
:
lim
n→∞
an
bn
= lim
n→∞
n2+n+ 1
n3+ 1 ·n
1= lim
n→∞
n3+n2+n
n3+ 1 = 1
Step 3: Since lim
n→∞
an
bn
= 1, and ∞
X
n=1
1
nis a harmonic series which diverges,
by the Limit Comparison Test, the series ∞
X
n=1
n2+n+ 1
n3+ 1 also diverges.
Question 8
Question
Determine if the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the ratio of successive terms. Let an=n!
nn. We calculate
the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=n+ 1
(n+ 1)n+1 ·nn
5
Step 2: Simplify the ratio.
an+1
an
=n+ 1
(n+ 1)n+1 ·nn=n+ 1
(n+ 1)n=1
(1 + 1
n)n
Step 3: Consider the limit of the ratio. Taking the limit as napproaches
infinity:
lim
n→∞
1
(1 + 1
n)n=1
e
Step 4: Apply the ratio test. Since the limit of the ratio is less than 1, by
the ratio test, the series P∞
n=1 n!
nnconverges.
Question 9
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Compute the limit for the ratio test: Let an=n!
nn. Then, consider
the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
L= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
Step 3: Further simplify the expression:
L= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1 = 1
Step 4: Analyze the result: Since L= 1, we cannot make a conclusion
about convergence or divergence of the series using the ratio test. Thus, we
have to try a different test.
Step 5: Use Stirling’s approximation: By Stirling’s approximation, we can
estimate n!∼√2πn n
en.
Therefore, n! grows faster than nnwhich implies that n!
nndoes not converge
to 0 as n→ ∞.
Step 6: Conclusion: Since the terms of the series n!
nndo not approach 0,
the series diverges by the Divergence Test.
6
Question 10
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test.
Let an=n!
nn. We compute the ratio
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
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→∞
1
1 + 1
nn
=1
e
Step 3: Analyze the result.
Since r=1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 11
Question
Consider the series P∞
n=1 n!
nn. Determine whether the series converges or di-
verges.
7
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. Then, calculate the limit:
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
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 3: Analyze the limit: Since the limit 1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 12
Question
Determine the convergence or divergence of the series ∞
X
n=1 n3+√n
n4+ 1 using the
limit comparison test.
Solution
To determine the convergence of the series ∞
X
n=1 n3+√n
n4+ 1 , we will use the
limit comparison test. Let’s denote an=n3+√n
n4+1 .
Step 1: Find a comparison series: We want to find a series ∞
X
n=1
bnsuch
that limn→∞ an
bnis a finite positive number.
Let’s choose bn=1
n.
Step 2: Calculate the limit: We will calculate the limit: limn→∞ an
bn.
8
lim
n→∞
an
bn
= lim
n→∞
n3+√n
n4+ 1 ·n
1
= lim
n→∞
n4+n3/2
n4+ 1
= lim
n→∞
1 + n−1/2
1 + n−4
=1+0
1+0
= 1
Step 3: Conclusion: Since limn→∞ an
bn= 1, by the limit comparison test,
the series ∞
X
n=1
n3+√n
n4+ 1 behaves the same as the harmonic series ∞
X
n=1
1
n.
Therefore, ∞
X
n=1
n3+√n
n4+ 1 diverges.
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n3
2n.
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
2n, we will use
the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n3
2n. We compute
the ratio an+1
an:
an+1
an
=(n+ 1)3
2n+1 ·2n
n3
=(n+ 1)3
2n3
=n3+ 3n2+ 3n+ 1
2n3
=1
2+3
2n+3
n2+1
n3.
Step 2: Apply the ratio test. Taking the limit as napproaches infinity, we
have
lim
n→∞ an+1
an= lim
n→∞ 1
2+3
2n+3
n2+1
n3
=1
2.
9
Since the limit is less than 1, by the Ratio Test, the series P∞
n=1 n3
2nconverges.
Therefore, the series P∞
n=1 n3
2nconverges.
Question 14
Question
Consider the series ∞
X
n=1
n2
2n. Determine whether the series converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
=1
2
Step 2: Determine the convergence. Since the limit is 1
2<1, by the ratio
test, the series converges.
Therefore, the series ∞
X
n=1
n2
2nconverges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
(−1)n√n+ 1
n2+ 1 .
Solution
Step 1: We can use the Alternating Series Test to determine the convergence of
the series. Let an=√n+1
n2+1 .
Step 2: We first need to show that anis positive, decreasing, and limn→∞ an=
0.
Step 3: Since n2+ 1 >0 for all n≥1, we have √n+1
n2+1 >0 for all n≥1.
Step 4: To show that anis decreasing, we examine the derivative of an=
√n+1
n2+1 .
10
Step 5: Let f(x) = √x+1
x2+1 . Then f′(x) = (x2+1)(1
2(x+1)−1/2)−(2x)(√x+1)
(x2+1)2.
Step 6: Simplifying, we get f′(x) = x2+1−x(x+1)
2(x2+1)3/2=1
2(x2+1)3/2>0 for all
x≥1.
Step 7: Since f′(x)>0 for all x≥1, f(x) is decreasing for all x≥1. This
implies that anis decreasing for all n≥1.
Step 8: Finally, we calculate the limit of anas napproaches infinity.
lim
n→∞
√n+ 1
n2+ 1 = lim
n→∞
1
nq1 + 1
n
= lim
n→∞
1
n= 0.
Step 9: By the Alternating Series Test, since anis positive, decreasing, and
limn→∞ an= 0, the series
∞
X
n=1
(−1)n√n+ 1
n2+ 1
converges.
Question 16
Question
Determine whether the series
∞
X
n=1
n2+ sin(n)
n3+ cos(n)
converges or diverges.
Solution
To analyze the convergence of the given series, we can use the Comparison Test.
Step 1: First, note that
n2+ sin(n)
n3+ cos(n)≤n2+ 1
n3−1for n≥1.
Step 2: Now, consider the series
∞
X
n=1
n2+ 1
n3−1.
Step 3: To determine its convergence, we can use the Limit Comparison
Test by comparing it to the harmonic series P∞
n=1 1
n.
Step 4: Let’s calculate the limit of the ratio of the two series:
L= lim
n→∞
(n2+ 1)/(n3−1)
1/n = lim
n→∞
n3+n
n3−1= 1.
11
Step 5: Since L > 0, both series either converge or diverge together.
Since P∞
n=1 1
nis a divergent harmonic series, by the Limit Comparison Test,
P∞
n=1 n2+1
n3−1is also divergent.
Step 6: Therefore, by the Comparison Test, the original series
∞
X
n=1
n2+ sin(n)
n3+ cos(n)
also diverges.
Question 17
Question
Determine whether the series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
and the series ∞
X
n=1
1
n
Step 1: Find the limit of the ratio of the two series. Let
an=n2+ 3n+ 1
n3+ 2n+ 5 and bn=1
n
Then,
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
n3+ 2n+ 5 ·n
1
Step 2: Simplify the expression and find the limit.
lim
n→∞
n2+ 3n+ 1
n3+ 2n+ 5 ·n
1= lim
n→∞
1 + 3
n+1
n2
1 + 2
n2+5
n3·n
12
Solving this limit, we get
lim
n→∞
1 + 3
n+1
n2
1 + 2
n2+5
n3·n= 1
Step 3: Apply the Limit Comparison Test. Since the limit is a finite positive
number, by the Limit Comparison Test, the given series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
converges if and only if the series
∞
X
n=1
1
n
converges. This series is the harmonic series which diverges. Therefore, the
given series also diverges.
Question 18
Question
Determine whether the series ∞
X
n=1
2n+ 3n
5nconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Apply the ratio test. Let an=2n+ 3n
5n. We will compute the limit:
L= lim
n→∞
an+1
an
Step 2: Find an+1 and simplify it. We have:
an+1 =2n+1 + 3n+1
5n+1 =2·2n+ 3 ·3n
5·5n=2
52n+ 3n
5n=2
5an
Step 3: Calculate the limit. Now, we substitute an+1 and aninto the
formula for L:
L= lim
n→∞
2
5an
an
= lim
n→∞
2
5
=2
5
Step 4: Interpret the result If L < 1, then the series converges. If L > 1,
the series diverges. In this case, L=2
5<1, so the series ∞
X
n=1
2n+ 3n
5nconverges
by the ratio test.
13
Question 19
Question
Determine the convergence of the series P∞
n=1 n2+1
n3+1 .
Solution
To analyze the convergence of the series, we can use the Limit Comparison Test.
Step 1: Let’s choose a comparison series. We can consider the series
P∞
n=1 1
n. This is a p-series with p= 1, which diverges.
Step 2: Take the limit of the ratio of the two series:
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1 = 1
Step 3: Since the limit is a positive constant, by the Limit Comparison Test,
the convergence of the original series P∞
n=1 n2+1
n3+1 is the same as the convergence
of the comparison series P∞
n=1 1
n, which diverges.
Therefore, the series P∞
n=1 n2+1
n3+1 also diverges.
Question 20
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms: Let an=n!
nn. Therefore,
the ratio of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio and compute the limit:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
=1
e
Step 3: Apply the ratio test: Since the limit is 1
e, the series P∞
n=1 n!
nn
converges by the ratio test.
Therefore, the series P∞
n=1 n!
nnconverges.
14
Question 21
Question
Determine whether the series ∞
X
n=1
n5
5n
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n5
5n,
we will use the Ratio Test. The Ratio Test states that if
lim
n→∞
an+1
an
<1,
then the series converges absolutely.
Step 1: Let an=n5
5n.
Step 2: Compute the ratio an+1
an
:
an+1
an
=(n+ 1)5
5n+1 ·5n
n5
=(n+ 1)5
5·5n·5n
n5
=(n+ 1)5
5n5
Step 3: Compute the limit of the ratio as napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)5
5n5
= lim
n→∞
(n5+ 5n4+ 10n3+ 10n2+ 5n+ 1)
5n5
= lim
n→∞
1 + 5
n+10
n2+10
n3+5
n4+1
n5
5
=1
5
Step 4: Since limn→∞
an+1
an=1
5<1, by the Ratio Test, the series P∞
n=1 n5
5n
converges absolutely.
15
Question 22
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Apply the ratio test. Let anbe the general term of the series:
an=n!
nn. We need to compute the following limit:
L= lim
n→∞
an+1
an
where
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=(n+ 1)nn
(n+ 1)n+1 =nn
(n+ 1)n=n
n+ 1n
Step 2: Simplify the limit. Then,
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=e−1
Step 3: Analyze the limit. Since L=e−1<1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Therefore, the series converges.
Question 23
Question
Determine whether the series P∞
n=1 n2+1
2n3−nconverges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Find the limit of the ratio. Let an=n2+1
2n3−nand bn=1
n. We will
find the limit of an
bnas napproaches infinity.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 1)/(2n3−n)
1/n = lim
n→∞
n2+ 1
2n3−n·n= lim
n→∞
n3+n
2n3−n= lim
n→∞
n3(1 + 1/n)
n3(2 −1/n)= lim
n→∞
1+1/n
2−1/n =1+0
2−0=1
2
16
Step 2: Interpret the limit. Since limn→∞ an
bn=1
2>0, we can conclude
that either both series Panand Pbnconverge or both diverge.
Step 3: Determine the convergence of Pbn=P1
n. The series P1
nis the
harmonic series, which is a divergent series.
Step 4: Conclude the convergence of Pan. Since the series Pbndiverges
and the limit comparison test gives a positive value for the limit of the ratio,
we can conclude that the series Pan=Pn2+1
2n3−nalso diverges.
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the Ratio Test.
Step 1: Apply the Ratio Test, which states that if limn→∞
an+1
an=L, then
the series converges if L < 1 or diverges if L > 1.
Let an=n!
nn. Then,
L= lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1+1/n
= 1
Step 2: Since L= 1, the Ratio Test is inconclusive. We need to use another
test to determine the convergence of the series.
Step 3: Notice that for large values of n, the factor n! in the numerator
grows much faster than nnin the denominator. Therefore, we can see that the
17
terms of the series do not approach zero, so the series diverges by the Test for
Divergence.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 25
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Calculate the ratio test. Let an=n!
nn. We compute:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1!n
=1
e≈0.3679
Step 2: Apply the ratio test. Since the limit is less than 1, the series
P∞
n=1 n!
nnconverges by the ratio test.
Question 26
Question
Determine whether the series P∞
n=1 n2+1
n4+1 converges or diverges.
18
Solution
To determine the convergence of the series, we can use the limit comparison
test. We will compare the given series to a known series whose convergence is
already known.
Step 1: Find a suitable series to compare with. Let’s compare the given
series with the p-series P∞
n=1 1
n2, since the exponent of nin the denominator is
larger than that of the numerator.
Step 2: Take the limit of the ratio of the two series. Consider the limit
lim
n→∞
n2+1
n4+1
1
n2
= lim
n→∞
n2(n2+ 1)
n4+ 1
Step 3: Simplify the expression.
= lim
n→∞
n4+n2
n4+ 1 = 1
Step 4: Analyze the limit. Since the limit is a finite positive value, by the
limit comparison test, the convergence of the original series P∞
n=1 n2+1
n4+1 is the
same as the convergence of the p-series P∞
n=1 1
n2.
Step 5: Conclude the convergence. The p-series P∞
n=1 1
n2converges as p=
2>1. Therefore, by the limit comparison test, the original series P∞
n=1 n2+1
n4+1
also converges.
Question 27
Question
Determine whether the series P∞
n=1 n3+3n+1
n4+4 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
already known.
Step 1: Let’s consider the series P∞
n=1 1
n, which is a p-series with p= 1.
This series is known to diverge.
Step 2: Let an=n3+3n+1
n4+4 be the general term of the given series.
Step 3: We will calculate the limit of the ratio between the general term of
the given series and the general term of the series P∞
n=1 1
n:
lim
n→∞
an
1/n = lim
n→∞
n3+ 3n+ 1
n4+ 4 ·n
1= lim
n→∞
n4+ 3n2+n
n4+ 4 = 1
Step 4: Since the limit is a positive finite number, by the Limit Comparison
Test, we conclude that both series either converge or diverge.
19
Step 5: Therefore, by the Limit Comparison Test, the series P∞
n=1 n3+3n+1
n4+4
has the same convergence behavior as the harmonic series P∞
n=1 1
n. Since the
harmonic series diverges, we can conclude that the given series diverges as well.
Question 28
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 an=n!
nn. We will calculate the limit of
the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/(nn)
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
=1
e<1
Step 2: Determine the convergence. Since the limit of the ratio is less than
1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 29
Question
Determine whether the series ∞
X
n=1
n2+ 1
n4+ 3
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s find a suitable series to compare with. We can rewrite the
given series as
∞
X
n=1
n2+ 1
n4+ 3 =∞
X
n=1
1 + 1
n2
n2+3
n2
20
Step 2: Consider the series
∞
X
n=1
1
n2
This is a p-series with p= 2, which we know to converge.
Step 3: Now, let’s evaluate the limit
lim
n→∞
1 + 1
n2
n2+3
n2
Step 4: We can simplify the limit as follows:
lim
n→∞
1 + 1
n2
n2+3
n2
= lim
n→∞
1/n2+1
n4
n2/n2+3
n2n2
= lim
n→∞
0+0
1+0
= 0
Step 5: Since the limit is a positive finite number, by the Limit Comparison
Test, the given series
∞
X
n=1
n2+ 1
n4+ 3
converges.
Question 30
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn, then
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)! ·nn
(n+ 1)n+1 ·n!=(n+ 1) ·nn
(n+ 1)n+1 =nn
(n+ 1)n=1 + 1
n−n
Step 2: Apply the limit of the ratio. We will compute the limit of the ratio
as napproaches infinity: limn→∞ 1 + 1
n−n.
Using the limit form (1+ 1
n)nfor e, we rewrite the expression as limn→∞ 1
(1+ 1
n)n.
21
Step 3: Determine the limit. As napproaches infinity, 1 + 1
nnapproaches
e. Thus, limn→∞ 1 + 1
n−n=1
e= 0.
Step 4: Conclusion. Since the limit of the ratio is not 0, the series P∞
n=1 n!
nn
diverges by the ratio test.
Question 31
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test:
lim
n→∞
an+1
an
,
where an=n!
nn.
Step 1: Find the ratio of consecutive terms The ratio of consecutive
terms is:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!
=n+ 1
(n+ 1)n+1 ·nn
=n+ 1
(n+ 1)n+1 ·nn
=n
(n+ 1)n.
Step 2: Find the limit of the ratio Taking the limit as napproaches
22
infinity:
lim
n→∞
an+1
an
= lim
n→∞
n
(n+ 1)n
= lim
n→∞
n1
(1 + 1
n)n
= lim
n→∞
n1
e
=∞.
Step 3: Evaluate the ratio test Since the limit is infinity, the ratio test
tells us that the series ∞
X
n=1
n!
nndiverges.
Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 32
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Then, we
compute
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
.
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n.
Step 3: Apply limit properties.
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e.
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
23
Question 33
Question
Let {an}be a sequence such that an=n2+1
3n3+4n+5 . Determine if the series
P∞
n=1 anconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 an, we will use the limit com-
parison test. We will compare the given series with a series whose behavior we
know.
Step 1: Find a series for comparison Consider the series P∞
n=1 1
n. This
is the harmonic series, which is known to diverge.
Step 2: Form the ratio limn→∞ an
1/n We calculate the limit:
lim
n→∞
an
1/n = lim
n→∞
(n2+ 1)/(3n3+ 4n+ 5)
1/n .
Step 3: Simplify the expression Simplify the expression to get:
lim
n→∞
n2+ 1
3n3+ 4n+ 5 ·n
1= lim
n→∞
n3+n
3n3+ 4n+ 5.
Step 4: Evaluate the limit Divide through by the highest power of nin
the denominator to get:
lim
n→∞
1+1/n2
3+4/n2+ 5/n3.
As n→ ∞, the terms with 1/n2and 1/n3approach 0, leaving:
1
3.
Step 5: Interpret the result Since the limit limn→∞ an
1/n is a finite positive
number, by the limit comparison test, we conclude that the series P∞
n=1 anhas
the same convergence behavior as the harmonic series. Thus, the series P∞
n=1 an
diverges.
Question 34
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Show that the series P∞
n=1
nan
1+n2anconverges.
24
Solution
To show that the series P∞
n=1
nan
1+n2anconverges, we will use the Limit Compar-
ison Test.
Step 1: Find a suitable series to compare to. Consider the series
P∞
n=1 1
n2. This series is known to converge since it is a p-series with p= 2.
Step 2: Compute the limit of the ratio of the two series. We consider
the limit:
lim
n→∞
nan
1+n2an
1
n2
= lim
n→∞
n3an
1 + n2an
.
Step 3: Simplify the expression to find the limit. Rewrite the limit
as:
lim
n→∞
n3an
1 + n2an
= lim
n→∞
n2an
1
n2+an
.
Step 4: Use the comparison test. Since P∞
n=1 anconverges, it follows
that an→0 as n→ ∞. Thus, for large n, we have an<1
n2. Therefore, the
series P∞
n=1
nan
1+n2anconverges by the Limit Comparison Test with P∞
n=1 1
n2.
Question 35
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test. The ratio
test states that for a series P∞
n=1 an, if limn→∞
an+1
an=L, then the series
converges if L < 1 and diverges if L > 1.
Step 1: Calculate the ratio Compute the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
n!·nn
(n+ 1)n+1
Simplify the expression:
an+1
an
= (n+ 1) ·nn
(n+ 1)n+1 =n+ 1
(1 + 1
n)n
Step 2: Find the limit Calculate the limit as napproaches infinity:
lim
n→∞
n+ 1
(1 + 1
n)n
This limit can be evaluated using L’Hˆopital’s Rule or recognizing it as the
form of the limit definition of e:
25
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
Step 1: Compute the limit using the ratio test: Let an=n2
2n. We compute
the ratio an+1
an:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 2: Analyze the limit: Since the limit is 1
2<1, the series P∞
n=1 n2
2n
converges by the ratio test.
Therefore, the given series converges.
Question 3
Question
Determine the convergence or divergence of the series P∞
n=1 3n3+n2
4n4+5 .
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test with a series that we know the convergence of.
Step 1: First, we will choose a series to compare with. Let’s consider the
series P∞
n=1 1
np, where p > 0.
Step 2: Next, we will compute the limit of the ratio of the given series to
the chosen series:
lim
n→∞
3n3+n2
4n4+5
1
np
= lim
n→∞
3n3+n2
4n4+ 5 ·np
1
= lim
n→∞
3n3+p+n2+p
4n4+p+ 5np
Step 3: Now, we will find the value of pthat will make the limit above finite
and non-zero: For convergence, we need the exponents of nin the numerator
and denominator to be the same. So we are looking for 3 + p= 4 + p, which
simplifies to 3 = 4, a contradiction. Therefore, the series P∞
n=1 3n3+n2
4n4+5 diverges
by the Limit Comparison Test.
2
Question 4
Question
Determine whether the series P∞
n=1 n2+3n+1
n3+n+5 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n+1
n3+n+5 , we will use the Limit
Comparison Test.
Step 1: Find a suitable series to compare. Let’s consider the series
P∞
n=1 1
n, which is a p-series with p= 1. This series is known to diverge.
Step 2: Compute the limit of the ratio. Calculate the limit limn→∞
n2+3n+1
n3+n+5
1
n
.
Step 3: Simplify the expression.
lim
n→∞
n2+3n+1
n3+n+5
1
n
= lim
n→∞
n2+ 3n+ 1
n3+n+ 5 ·n
= lim
n→∞
n3+ 3n2+n
n3+n+ 5
= lim
n→∞
1 + 3
n+1
n2
1 + 1
n2+5
n3
=1+0+0
1+0+0
= 1
Step 4: Apply the Limit Comparison Test. Since the limit is a finite
positive constant, both series either converge or diverge together. As the series
P∞
n=1 1
ndiverges, by the limit comparison test, the series P∞
n=1 n2+3n+1
n3+n+5 also
diverges. Thus, the given series diverges.
Question 5
Question
Determine the convergence or divergence of the series
∞
X
n=1
2n+ 1
3n2+n+ 5.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series whose convergence or divergence we already know.
3
Consider the series ∞
X
n=1
1
n.
This is the harmonic series, which we know diverges.
Step 2: Calculate the limit of the ratio of the two series.
We will consider the limit
L= lim
n→∞
2n+1
3n2+n+5
1
n
.
Step 3: Evaluate the limit.
L= lim
n→∞
2n2+n
3n2+n+ 5 = lim
n→∞
2 + 1
n
3 + 1
n+5
n2
=2
3.
Step 4: Interpret the result.
Since L > 0 and finite, by the Limit Comparison Test, we conclude that the
given series
∞
X
n=1
2n+ 1
3n2+n+ 5
converges/diverges if and only if the series P∞
n=1 1
nconverges/diverges.
Since the harmonic series diverges, by the Limit Comparison Test, the given
series also diverges.
Question 6
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine if the series P∞
n=1 n!
nnconverges or diverges, we can use the ratio
test.
Step 1: Compute the limit. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n+ 1 ·nn
n!
= lim
n→∞
(n+ 1)n
n+ 1
= lim
n→∞
1 + 1
n
= 1.
Step 2: Analyze the limit. Since the limit is equal to 1, we cannot make
any conclusions using the ratio test. Therefore, we need to consider another
test.
Step 3: Applying the ratio test again. We will consider the sequence
bn=nn
n!. We will compute:
lim
n→∞
bn+1
bn
= lim
n→∞
(n+ 1)n+1/(n+ 1)!
nn/n!
= lim
n→∞
(n+ 1)n+1
nn
= lim
n→∞ (1 + 1/n)n+1=1
e.
4
Step 4: Final conclusion. Since 1
e<1, the series P∞
n=1 n!
nnconverges by
the ratio test.
Question 7
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+n+ 1
n3+ 1 .
Solution
To determine the convergence or divergence of the series ∞
X
n=1
n2+n+ 1
n3+ 1 , we will
use the Limit Comparison Test.
Step 1: Let an=n2+n+ 1
n3+ 1 . We will compare anwith a simpler series
bn=1
n.
Step 2: Find the limit of the ratio lim
n→∞
an
bn
:
lim
n→∞
an
bn
= lim
n→∞
n2+n+ 1
n3+ 1 ·n
1= lim
n→∞
n3+n2+n
n3+ 1 = 1
Step 3: Since lim
n→∞
an
bn
= 1, and ∞
X
n=1
1
nis a harmonic series which diverges,
by the Limit Comparison Test, the series ∞
X
n=1
n2+n+ 1
n3+ 1 also diverges.
Question 8
Question
Determine if the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the ratio of successive terms. Let an=n!
nn. We calculate
the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=n+ 1
(n+ 1)n+1 ·nn
5
Step 2: Simplify the ratio.
an+1
an
=n+ 1
(n+ 1)n+1 ·nn=n+ 1
(n+ 1)n=1
(1 + 1
n)n
Step 3: Consider the limit of the ratio. Taking the limit as napproaches
infinity:
lim
n→∞
1
(1 + 1
n)n=1
e
Step 4: Apply the ratio test. Since the limit of the ratio is less than 1, by
the ratio test, the series P∞
n=1 n!
nnconverges.
Question 9
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Compute the limit for the ratio test: Let an=n!
nn. Then, consider
the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression:
L= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
Step 3: Further simplify the expression:
L= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1 = 1
Step 4: Analyze the result: Since L= 1, we cannot make a conclusion
about convergence or divergence of the series using the ratio test. Thus, we
have to try a different test.
Step 5: Use Stirling’s approximation: By Stirling’s approximation, we can
estimate n!∼√2πn n
en.
Therefore, n! grows faster than nnwhich implies that n!
nndoes not converge
to 0 as n→ ∞.
Step 6: Conclusion: Since the terms of the series n!
nndo not approach 0,
the series diverges by the Divergence Test.
6
Question 10
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test.
Let an=n!
nn. We compute the ratio
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
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→∞
1
1 + 1
nn
=1
e
Step 3: Analyze the result.
Since r=1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 11
Question
Consider the series P∞
n=1 n!
nn. Determine whether the series converges or di-
verges.
7
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. Then, calculate the limit:
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
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 3: Analyze the limit: Since the limit 1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 12
Question
Determine the convergence or divergence of the series ∞
X
n=1 n3+√n
n4+ 1 using the
limit comparison test.
Solution
To determine the convergence of the series ∞
X
n=1 n3+√n
n4+ 1 , we will use the
limit comparison test. Let’s denote an=n3+√n
n4+1 .
Step 1: Find a comparison series: We want to find a series ∞
X
n=1
bnsuch
that limn→∞ an
bnis a finite positive number.
Let’s choose bn=1
n.
Step 2: Calculate the limit: We will calculate the limit: limn→∞ an
bn.
8
lim
n→∞
an
bn
= lim
n→∞
n3+√n
n4+ 1 ·n
1
= lim
n→∞
n4+n3/2
n4+ 1
= lim
n→∞
1 + n−1/2
1 + n−4
=1+0
1+0
= 1
Step 3: Conclusion: Since limn→∞ an
bn= 1, by the limit comparison test,
the series ∞
X
n=1
n3+√n
n4+ 1 behaves the same as the harmonic series ∞
X
n=1
1
n.
Therefore, ∞
X
n=1
n3+√n
n4+ 1 diverges.
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n3
2n.
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
2n, we will use
the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n3
2n. We compute
the ratio an+1
an:
an+1
an
=(n+ 1)3
2n+1 ·2n
n3
=(n+ 1)3
2n3
=n3+ 3n2+ 3n+ 1
2n3
=1
2+3
2n+3
n2+1
n3.
Step 2: Apply the ratio test. Taking the limit as napproaches infinity, we
have
lim
n→∞ an+1
an= lim
n→∞ 1
2+3
2n+3
n2+1
n3
=1
2.
9
Since the limit is less than 1, by the Ratio Test, the series P∞
n=1 n3
2nconverges.
Therefore, the series P∞
n=1 n3
2nconverges.
Question 14
Question
Consider the series ∞
X
n=1
n2
2n. Determine whether the series converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
=1
2
Step 2: Determine the convergence. Since the limit is 1
2<1, by the ratio
test, the series converges.
Therefore, the series ∞
X
n=1
n2
2nconverges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
(−1)n√n+ 1
n2+ 1 .
Solution
Step 1: We can use the Alternating Series Test to determine the convergence of
the series. Let an=√n+1
n2+1 .
Step 2: We first need to show that anis positive, decreasing, and limn→∞ an=
0.
Step 3: Since n2+ 1 >0 for all n≥1, we have √n+1
n2+1 >0 for all n≥1.
Step 4: To show that anis decreasing, we examine the derivative of an=
√n+1
n2+1 .
10
Step 5: Let f(x) = √x+1
x2+1 . Then f′(x) = (x2+1)(1
2(x+1)−1/2)−(2x)(√x+1)
(x2+1)2.
Step 6: Simplifying, we get f′(x) = x2+1−x(x+1)
2(x2+1)3/2=1
2(x2+1)3/2>0 for all
x≥1.
Step 7: Since f′(x)>0 for all x≥1, f(x) is decreasing for all x≥1. This
implies that anis decreasing for all n≥1.
Step 8: Finally, we calculate the limit of anas napproaches infinity.
lim
n→∞
√n+ 1
n2+ 1 = lim
n→∞
1
nq1 + 1
n
= lim
n→∞
1
n= 0.
Step 9: By the Alternating Series Test, since anis positive, decreasing, and
limn→∞ an= 0, the series
∞
X
n=1
(−1)n√n+ 1
n2+ 1
converges.
Question 16
Question
Determine whether the series
∞
X
n=1
n2+ sin(n)
n3+ cos(n)
converges or diverges.
Solution
To analyze the convergence of the given series, we can use the Comparison Test.
Step 1: First, note that
n2+ sin(n)
n3+ cos(n)≤n2+ 1
n3−1for n≥1.
Step 2: Now, consider the series
∞
X
n=1
n2+ 1
n3−1.
Step 3: To determine its convergence, we can use the Limit Comparison
Test by comparing it to the harmonic series P∞
n=1 1
n.
Step 4: Let’s calculate the limit of the ratio of the two series:
L= lim
n→∞
(n2+ 1)/(n3−1)
1/n = lim
n→∞
n3+n
n3−1= 1.
11
Step 5: Since L > 0, both series either converge or diverge together.
Since P∞
n=1 1
nis a divergent harmonic series, by the Limit Comparison Test,
P∞
n=1 n2+1
n3−1is also divergent.
Step 6: Therefore, by the Comparison Test, the original series
∞
X
n=1
n2+ sin(n)
n3+ cos(n)
also diverges.
Question 17
Question
Determine whether the series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
and the series ∞
X
n=1
1
n
Step 1: Find the limit of the ratio of the two series. Let
an=n2+ 3n+ 1
n3+ 2n+ 5 and bn=1
n
Then,
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
n3+ 2n+ 5 ·n
1
Step 2: Simplify the expression and find the limit.
lim
n→∞
n2+ 3n+ 1
n3+ 2n+ 5 ·n
1= lim
n→∞
1 + 3
n+1
n2
1 + 2
n2+5
n3·n
12
Solving this limit, we get
lim
n→∞
1 + 3
n+1
n2
1 + 2
n2+5
n3·n= 1
Step 3: Apply the Limit Comparison Test. Since the limit is a finite positive
number, by the Limit Comparison Test, the given series
∞
X
n=1
n2+ 3n+ 1
n3+ 2n+ 5
converges if and only if the series
∞
X
n=1
1
n
converges. This series is the harmonic series which diverges. Therefore, the
given series also diverges.
Question 18
Question
Determine whether the series ∞
X
n=1
2n+ 3n
5nconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Apply the ratio test. Let an=2n+ 3n
5n. We will compute the limit:
L= lim
n→∞
an+1
an
Step 2: Find an+1 and simplify it. We have:
an+1 =2n+1 + 3n+1
5n+1 =2·2n+ 3 ·3n
5·5n=2
52n+ 3n
5n=2
5an
Step 3: Calculate the limit. Now, we substitute an+1 and aninto the
formula for L:
L= lim
n→∞
2
5an
an
= lim
n→∞
2
5
=2
5
Step 4: Interpret the result If L < 1, then the series converges. If L > 1,
the series diverges. In this case, L=2
5<1, so the series ∞
X
n=1
2n+ 3n
5nconverges
by the ratio test.
13
Question 19
Question
Determine the convergence of the series P∞
n=1 n2+1
n3+1 .
Solution
To analyze the convergence of the series, we can use the Limit Comparison Test.
Step 1: Let’s choose a comparison series. We can consider the series
P∞
n=1 1
n. This is a p-series with p= 1, which diverges.
Step 2: Take the limit of the ratio of the two series:
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1 = 1
Step 3: Since the limit is a positive constant, by the Limit Comparison Test,
the convergence of the original series P∞
n=1 n2+1
n3+1 is the same as the convergence
of the comparison series P∞
n=1 1
n, which diverges.
Therefore, the series P∞
n=1 n2+1
n3+1 also diverges.
Question 20
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms: Let an=n!
nn. Therefore,
the ratio of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio and compute the limit:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
=1
e
Step 3: Apply the ratio test: Since the limit is 1
e, the series P∞
n=1 n!
nn
converges by the ratio test.
Therefore, the series P∞
n=1 n!
nnconverges.
14
Question 21
Question
Determine whether the series ∞
X
n=1
n5
5n
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n5
5n,
we will use the Ratio Test. The Ratio Test states that if
lim
n→∞
an+1
an
<1,
then the series converges absolutely.
Step 1: Let an=n5
5n.
Step 2: Compute the ratio an+1
an
:
an+1
an
=(n+ 1)5
5n+1 ·5n
n5
=(n+ 1)5
5·5n·5n
n5
=(n+ 1)5
5n5
Step 3: Compute the limit of the ratio as napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)5
5n5
= lim
n→∞
(n5+ 5n4+ 10n3+ 10n2+ 5n+ 1)
5n5
= lim
n→∞
1 + 5
n+10
n2+10
n3+5
n4+1
n5
5
=1
5
Step 4: Since limn→∞
an+1
an=1
5<1, by the Ratio Test, the series P∞
n=1 n5
5n
converges absolutely.
15
Question 22
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Apply the ratio test. Let anbe the general term of the series:
an=n!
nn. We need to compute the following limit:
L= lim
n→∞
an+1
an
where
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
(n+ 1)n+1 ·nn
n!=(n+ 1)nn
(n+ 1)n+1 =nn
(n+ 1)n=n
n+ 1n
Step 2: Simplify the limit. Then,
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=e−1
Step 3: Analyze the limit. Since L=e−1<1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Therefore, the series converges.
Question 23
Question
Determine whether the series P∞
n=1 n2+1
2n3−nconverges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Find the limit of the ratio. Let an=n2+1
2n3−nand bn=1
n. We will
find the limit of an
bnas napproaches infinity.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 1)/(2n3−n)
1/n = lim
n→∞
n2+ 1
2n3−n·n= lim
n→∞
n3+n
2n3−n= lim
n→∞
n3(1 + 1/n)
n3(2 −1/n)= lim
n→∞
1+1/n
2−1/n =1+0
2−0=1
2
16
Step 2: Interpret the limit. Since limn→∞ an
bn=1
2>0, we can conclude
that either both series Panand Pbnconverge or both diverge.
Step 3: Determine the convergence of Pbn=P1
n. The series P1
nis the
harmonic series, which is a divergent series.
Step 4: Conclude the convergence of Pan. Since the series Pbndiverges
and the limit comparison test gives a positive value for the limit of the ratio,
we can conclude that the series Pan=Pn2+1
2n3−nalso diverges.
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the Ratio Test.
Step 1: Apply the Ratio Test, which states that if limn→∞
an+1
an=L, then
the series converges if L < 1 or diverges if L > 1.
Let an=n!
nn. Then,
L= lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1+1/n
= 1
Step 2: Since L= 1, the Ratio Test is inconclusive. We need to use another
test to determine the convergence of the series.
Step 3: Notice that for large values of n, the factor n! in the numerator
grows much faster than nnin the denominator. Therefore, we can see that the
17
terms of the series do not approach zero, so the series diverges by the Test for
Divergence.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 25
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Calculate the ratio test. Let an=n!
nn. We compute:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1!n
=1
e≈0.3679
Step 2: Apply the ratio test. Since the limit is less than 1, the series
P∞
n=1 n!
nnconverges by the ratio test.
Question 26
Question
Determine whether the series P∞
n=1 n2+1
n4+1 converges or diverges.
18
Solution
To determine the convergence of the series, we can use the limit comparison
test. We will compare the given series to a known series whose convergence is
already known.
Step 1: Find a suitable series to compare with. Let’s compare the given
series with the p-series P∞
n=1 1
n2, since the exponent of nin the denominator is
larger than that of the numerator.
Step 2: Take the limit of the ratio of the two series. Consider the limit
lim
n→∞
n2+1
n4+1
1
n2
= lim
n→∞
n2(n2+ 1)
n4+ 1
Step 3: Simplify the expression.
= lim
n→∞
n4+n2
n4+ 1 = 1
Step 4: Analyze the limit. Since the limit is a finite positive value, by the
limit comparison test, the convergence of the original series P∞
n=1 n2+1
n4+1 is the
same as the convergence of the p-series P∞
n=1 1
n2.
Step 5: Conclude the convergence. The p-series P∞
n=1 1
n2converges as p=
2>1. Therefore, by the limit comparison test, the original series P∞
n=1 n2+1
n4+1
also converges.
Question 27
Question
Determine whether the series P∞
n=1 n3+3n+1
n4+4 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
already known.
Step 1: Let’s consider the series P∞
n=1 1
n, which is a p-series with p= 1.
This series is known to diverge.
Step 2: Let an=n3+3n+1
n4+4 be the general term of the given series.
Step 3: We will calculate the limit of the ratio between the general term of
the given series and the general term of the series P∞
n=1 1
n:
lim
n→∞
an
1/n = lim
n→∞
n3+ 3n+ 1
n4+ 4 ·n
1= lim
n→∞
n4+ 3n2+n
n4+ 4 = 1
Step 4: Since the limit is a positive finite number, by the Limit Comparison
Test, we conclude that both series either converge or diverge.
19
Step 5: Therefore, by the Limit Comparison Test, the series P∞
n=1 n3+3n+1
n4+4
has the same convergence behavior as the harmonic series P∞
n=1 1
n. Since the
harmonic series diverges, we can conclude that the given series diverges as well.
Question 28
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 an=n!
nn. We will calculate the limit of
the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/(nn)
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
=1
e<1
Step 2: Determine the convergence. Since the limit of the ratio is less than
1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 29
Question
Determine whether the series ∞
X
n=1
n2+ 1
n4+ 3
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s find a suitable series to compare with. We can rewrite the
given series as
∞
X
n=1
n2+ 1
n4+ 3 =∞
X
n=1
1 + 1
n2
n2+3
n2
20
Step 2: Consider the series
∞
X
n=1
1
n2
This is a p-series with p= 2, which we know to converge.
Step 3: Now, let’s evaluate the limit
lim
n→∞
1 + 1
n2
n2+3
n2
Step 4: We can simplify the limit as follows:
lim
n→∞
1 + 1
n2
n2+3
n2
= lim
n→∞
1/n2+1
n4
n2/n2+3
n2n2
= lim
n→∞
0+0
1+0
= 0
Step 5: Since the limit is a positive finite number, by the Limit Comparison
Test, the given series
∞
X
n=1
n2+ 1
n4+ 3
converges.
Question 30
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn, then
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)! ·nn
(n+ 1)n+1 ·n!=(n+ 1) ·nn
(n+ 1)n+1 =nn
(n+ 1)n=1 + 1
n−n
Step 2: Apply the limit of the ratio. We will compute the limit of the ratio
as napproaches infinity: limn→∞ 1 + 1
n−n.
Using the limit form (1+ 1
n)nfor e, we rewrite the expression as limn→∞ 1
(1+ 1
n)n.
21
Step 3: Determine the limit. As napproaches infinity, 1 + 1
nnapproaches
e. Thus, limn→∞ 1 + 1
n−n=1
e= 0.
Step 4: Conclusion. Since the limit of the ratio is not 0, the series P∞
n=1 n!
nn
diverges by the ratio test.
Question 31
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test:
lim
n→∞
an+1
an
,
where an=n!
nn.
Step 1: Find the ratio of consecutive terms The ratio of consecutive
terms is:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!
=n+ 1
(n+ 1)n+1 ·nn
=n+ 1
(n+ 1)n+1 ·nn
=n
(n+ 1)n.
Step 2: Find the limit of the ratio Taking the limit as napproaches
22
infinity:
lim
n→∞
an+1
an
= lim
n→∞
n
(n+ 1)n
= lim
n→∞
n1
(1 + 1
n)n
= lim
n→∞
n1
e
=∞.
Step 3: Evaluate the ratio test Since the limit is infinity, the ratio test
tells us that the series ∞
X
n=1
n!
nndiverges.
Therefore, the series ∞
X
n=1
n!
nndiverges.
Question 32
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. Then, we
compute
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
.
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n.
Step 3: Apply limit properties.
lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e.
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
23
Question 33
Question
Let {an}be a sequence such that an=n2+1
3n3+4n+5 . Determine if the series
P∞
n=1 anconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 an, we will use the limit com-
parison test. We will compare the given series with a series whose behavior we
know.
Step 1: Find a series for comparison Consider the series P∞
n=1 1
n. This
is the harmonic series, which is known to diverge.
Step 2: Form the ratio limn→∞ an
1/n We calculate the limit:
lim
n→∞
an
1/n = lim
n→∞
(n2+ 1)/(3n3+ 4n+ 5)
1/n .
Step 3: Simplify the expression Simplify the expression to get:
lim
n→∞
n2+ 1
3n3+ 4n+ 5 ·n
1= lim
n→∞
n3+n
3n3+ 4n+ 5.
Step 4: Evaluate the limit Divide through by the highest power of nin
the denominator to get:
lim
n→∞
1+1/n2
3+4/n2+ 5/n3.
As n→ ∞, the terms with 1/n2and 1/n3approach 0, leaving:
1
3.
Step 5: Interpret the result Since the limit limn→∞ an
1/n is a finite positive
number, by the limit comparison test, we conclude that the series P∞
n=1 anhas
the same convergence behavior as the harmonic series. Thus, the series P∞
n=1 an
diverges.
Question 34
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Show that the series P∞
n=1
nan
1+n2anconverges.
24
Solution
To show that the series P∞
n=1
nan
1+n2anconverges, we will use the Limit Compar-
ison Test.
Step 1: Find a suitable series to compare to. Consider the series
P∞
n=1 1
n2. This series is known to converge since it is a p-series with p= 2.
Step 2: Compute the limit of the ratio of the two series. We consider
the limit:
lim
n→∞
nan
1+n2an
1
n2
= lim
n→∞
n3an
1 + n2an
.
Step 3: Simplify the expression to find the limit. Rewrite the limit
as:
lim
n→∞
n3an
1 + n2an
= lim
n→∞
n2an
1
n2+an
.
Step 4: Use the comparison test. Since P∞
n=1 anconverges, it follows
that an→0 as n→ ∞. Thus, for large n, we have an<1
n2. Therefore, the
series P∞
n=1
nan
1+n2anconverges by the Limit Comparison Test with P∞
n=1 1
n2.
Question 35
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test. The ratio
test states that for a series P∞
n=1 an, if limn→∞
an+1
an=L, then the series
converges if L < 1 and diverges if L > 1.
Step 1: Calculate the ratio Compute the ratio an+1
an:
an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
n!·nn
(n+ 1)n+1
Simplify the expression:
an+1
an
= (n+ 1) ·nn
(n+ 1)n+1 =n+ 1
(1 + 1
n)n
Step 2: Find the limit Calculate the limit as napproaches infinity:
lim
n→∞
n+ 1
(1 + 1
n)n
This limit can be evaluated using L’Hˆopital’s Rule or recognizing it as the
form of the limit definition of e:
25
lim
n→∞
n+ 1
(1 + 1
n)n=e
Step 3: Determine the convergence Since limn→∞
an+1
an=e > 1, the
series P∞
n=1 n!
nndiverges by the ratio test.
26
