MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 6
Liberty University
Question 1
Question
Determine whether the series P∞
n=1 n3+2n2+1
n5+3 converges or diverges.
Solution
Let’s start by using the Limit Comparison Test to determine the convergence
of the given series. We will compare it to a known series whose convergence is
known.
Step 1: Determine a comparison series
Consider the series P∞
n=1 1
n2. This is a p-series with p= 2, so it converges
(P∞
n=1 1
n2=π2
6). We will compare the given series with this series.
Step 2: Compute the limit for comparison
Let an=n3+2n2+1
n5+3 and bn=1
n2. We will compute the limit of the ratio L
as napproaches infinity:
L= lim
n→∞
an
bn
= lim
n→∞
n3+2n2+1
n5+3
1
n2
Step 3: Simplify the ratio and compute the limit
L= lim
n→∞
n5+ 2n4+n2
n5+ 3 = lim
n→∞
1 + 2
n+1
n3
1 + 3
n5
= 1
Step 4: Conclusion
Since L= 1 and the comparison series P∞
n=1 1
n2converges, by the Limit
Comparison Test, the given series P∞
n=1 n3+2n2+1
n5+3 also converges.
Question 2
Question
Determine the convergence or divergence of the series ∞
X
n=1
n3+ 5n2−3
4n4+n3−2.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let an=n3+5n2−3
4n4+n3−2.
Step 1: Find a simpler series to compare with. We will choose a
simpler series that we know the convergence or divergence of. Let’s consider the
series ∞
X
n=1
1
n.
Step 2: Form the ratio of the two series. Let bn=1
n. We will find the
limit of the ratio lim
n→∞
an
bn
.
Step 3: Calculate the limit and make a conclusion.
lim
n→∞
an
bn
= lim
n→∞
n3+5n2−3
4n4+n3−2
1
n
= lim
n→∞
n4+ 5n3−3n
4n4+n3−2
= lim
n→∞
1+5/n −3/n3
4+1/n2−2/n4
=1+0−0
4+0−0
=1
4.
Since lim
n→∞
an
bn
=1
4>0, the series ∞
X
n=1
n3+ 5n2−3
4n4+n3−2converges or diverges
with the same behavior as ∞
X
n=1
1
n. Therefore, the series ∞
X
n=1
n3+ 5n2−3
4n4+n3−2di-
verges (by divergence of the harmonic series).
Question 3
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
2
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
Step 1: Calculate the limit of the ratio limn→∞
an+1
an.
lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1
n)n
=1
e
Step 2: Determine the convergence based on the limit in Step 1. Since
the limit of the ratio is 1
e, according to the Ratio Test: - If 1
e<1, the series
P∞
n=1 n!
nnconverges. - If 1
e≥1, the series P∞
n=1 n!
nndiverges.
In this case, 1
e<1, so the series P∞
n=1 n!
nnconverges.
Question 4
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n3
n4+ 1
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
n4+1 , we will
use the Limit Comparison Test.
Step 1: Find a series to compare with.
Consider the series P∞
n=1 1
n, which is a well-known divergent p-series.
Step 2: Form the limit of the ratio of the two series.
Let an=n3
n4+1 and bn=1
n. We will consider the limit:
lim
n→∞
an
bn
= lim
n→∞
n3
n4+1
1
n
= lim
n→∞
n4
n4+ 1 = 1
Step 3: Conclude based on the Limit Comparison Test.
Since limn→∞
an
bn= 1 and P∞
n=1 1
nis a divergent series, by the limit com-
parison test, the series P∞
n=1 n3
n4+1 also diverges.
3
Question 5
Question
Determine whether the series P∞
n=1 n2+1
2n3+n+1 converges or diverges.
Solution
To analyze the convergence of the series, we will use the Limit Comparison Test.
Step 1: Let’s consider the series bn=1
n, which we know diverges (it is a
p-series with p= 1). We will compare the given series to bn.
Step 2: Calculate the limit:
lim
n→∞
n2+1
2n3+n+1
1
n
= lim
n→∞
n3+n
2n3+n+ 1 = lim
n→∞
n3(1 + 1
n2)
n3(2 + 1
n2+1
n3)=1
2
Step 3: Since the limit is a finite positive number, by the Limit Compar-
ison Test, the given series P∞
n=1 n2+1
2n3+n+1 must converge if the series P∞
n=1 1
n
converges.
Step 4: Therefore, using the comparison with the divergent series P∞
n=1 1
n,
we can conclude that the given series P∞
n=1 n2+1
2n3+n+1 also diverges.
Question 6
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series, we will use the Ratio Test.
Step 1: Compute the ratio R.
R= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Calculate R.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n+ 1
1 + 1
nn+1
4
=e
where the above limit is a known result in calculus.
Step 3: Analyze the value of R. Since R=e > 1, the series P∞
n=1 n!
nn
diverges by the Ratio Test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 7
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ra-
tio test. Step 1: Let’s apply the ratio test. The ratio test states that if
limn→∞
an+1
an=L, then the series converges if L < 1 and diverges if L > 1.
Consider an=n!
nn. Then,
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!= lim
n→∞
n+ 1
(1 + 1
n)n=1
e<1
Since the limit is less than 1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 8
Question
Determine whether the series P∞
n=1 n2+3n
n3+2 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n
n3+2 , we will use the limit
comparison test.
Step 1: Let’s find a series that we can easily determine the convergence of.
We choose the series P∞
n=1 1
n.
Step 2: Compute the limit:
lim
n→∞
n2+ 3n
n3+ 2 ·n
1= lim
n→∞
n3+ 3n2
n3+ 2 = lim
n→∞
1 + 3
n
1 + 2
n3
= 1
Step 3: Since the limit is a finite positive number (not zero or infinity), by
the limit comparison test, the series P∞
n=1 n2+3n
n3+2 behaves like the series P∞
n=1 1
n.
5
Step 4: Since the harmonic series P∞
n=1 1
ndiverges, our original series
P∞
n=1 n2+3n
n3+2 also diverges by comparison.
Therefore, the series P∞
n=1 n2+3n
n3+2 diverges.
Question 9
Question
Determine if the series ∞
X
n=1
n2
2nconverges or diverges.
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Let an=n2
2n.
Step 2: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 3: Simplify the expression:
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
Step 4: Taking the limit, we get:
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 5: Since the limit is less than 1, by the ratio test, the series ∞
X
n=1
n2
2n
converges.
Therefore, the series converges.
Question 10
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
6
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. We want to
find limn→∞
an+1
an.
Step 2: Compute 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
= lim
n→∞
nn
(n+ 1)n
Step 3: Simplify the expression. Using the properties of limits, we get:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Analyze the result. 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 11
Question
Determine whether the series ∞
X
n=1
n2
2n
converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
Step 1: Apply the ratio test by considering the limit
L= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Find an+1 and
an+1
an.
an+1 =(n+ 1)2
2n+1 and
an+1
an
=(n+ 1)2
2n+1 ·2n
n2
7
Step 3: Simplify the expression
an+1
an.
(n+ 1)2
2n+1 ·2n
n2=(n+ 1)2·2n
2n+1 ·n2=(n+ 1)2
2n2
Step 4: Evaluate the limit L.
L= lim
n→∞
(n+ 1)2
2n2=1
2
Step 5: Analyze the value of L. If L < 1, then the series converges. If
L > 1 or L=∞, then the series diverges. If L= 1, the test is inconclusive.
Since L=1
2<1, by the ratio test, the series P∞
n=1 n2
2nconverges.
Question 12
Question
Determine whether the series P∞
n=1(−1)n√n+1−√n
nconverges or diverges.
Solution
To determine the convergence of the given series, we will analyze it using the
alternating series test.
Step 1: Identify the terms Let an= (−1)n√n+1−√n
nbe the nth term of
the series.
Step 2: Check for absolute convergence First, we will consider the
absolute convergence of the series by finding the absolute value of the terms:
|an|=
√n+ 1 −√n
n
=√n+ 1 −√n
n
Step 3: Simplify the expression Simplify the absolute value of the terms:
|an|=√n+ 1 −√n
n=(√n+ 1 −√n)(√n+1+√n)
n(√n+1+√n)=√n+ 12−√n2
n(√n+1+√n)=1
n(√n+1+√n)
Step 4: Determine the limit Find the limit as napproaches infinity of
|an|to check for absolute convergence:
lim
n→∞ |an|= lim
n→∞
1
n(√n+1+√n)= lim
n→∞
1
n√n(1 + 1
n)= lim
n→∞
1
n3/2(1 + 1
n)= 0
Step 5: Apply the alternating series test Since the series P∞
n=1(−1)n√n+1−√n
n
satisfies the conditions of the alternating series test (terms are decreasing in ab-
solute value and approach 0), the series converges.
Therefore, the series P∞
n=1(−1)n√n+1−√n
nconverges.
8
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n2
2n.
Solution
To determine the convergence or divergence of the series P∞
n=1 n2
2n, we will use
the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the limit expression.
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: Apply the ratio test. - If the limit is less than 1, the series converges
absolutely. - If the limit is greater than 1 or does not exist, the series diverges.
- If the limit is equal to 1, the test is inconclusive.
Since the limit is 1
2which is less than 1, by the ratio test, the series P∞
n=1 n2
2n
converges absolutely.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
9
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n(n+ 1)
= lim
n→∞
nn
(n+ 1)n+1
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
n
= 1
Since the limit is equal to 1, the ratio test is inconclusive. Let’s try another
test, the root test.
lim
n→∞
n
p|an|= lim
n→∞
n
s
n!
nn
= lim
n→∞
n
√n!
n
=limn→∞
n
√n!
limn→∞ n
=1
∞
= 0
Since the limit is less than 1, the root test tells us that the series P∞
n=1 n!
nn
converges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
10
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. The ratio test states that if
L= lim
n→∞
an+1
an
,
then: - If L < 1, the series converges absolutely. - If L > 1, the series diverges.
- If L= 1, the test is inconclusive.
Step 1: Compute the limit involved in the ratio test:
L= 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
= lim
n→∞ 1
1+1/nn
=1
e,
where the last step follows from the limit limn→∞ 1 + 1
nn=e.
Step 2: Analyze the value of L: Since L=1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges absolutely.
Therefore, the series converges.
Question 16
Question
Determine whether the series ∞
X
n=1
n2+ 3
3n3+ 4 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 3
3n3+ 4, we can use the limit
comparison test.
11
Step 1: Find the limit for comparison Let an=n2+3
3n3+4 be the general
term of the series. We will find limn→∞
an
1
n
to compare.
lim
n→∞
an
1
n
= lim
n→∞
n2+ 3
3n3+ 4 ·n= lim
n→∞
n3+ 3n
3n3+ 4 =1
3
Step 2: Interpret the limit Since limn→∞
an
1
n
=1
3>0, and ∞
X
n=1
1
nis
ap-series with p= 1 (which diverges), we can conclude that ∞
X
n=1
n2+ 3
3n3+ 4 also
diverges by the limit comparison test.
Question 17
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n2+ 1
n3+ 1
Solution
To analyze the convergence of the given series, we can use the limit comparison
test. We will compare it to the series P∞
n=1 1
n.
Step 1: Find the limit of the ratio: Let an=n2+1
n3+1 and bn=1
n. We
want to find:
lim
n→∞
an
bn
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1 = 1
Step 2: Conclude with the limit comparison test: Since the limit is
a finite nonzero number, we can conclude by the limit comparison test that the
series P∞
n=1 n2+1
n3+1 behaves the same as the divergent harmonic series P∞
n=1 1
n.
Therefore, by the limit comparison test, the series P∞
n=1 n2+1
n3+1 diverges.
Question 18
Question
Determine the convergence or divergence of the series P∞
n=1 n2
2n.
12
Solution
To determine the convergence or divergence of the series P∞
n=1 n2
2n, we will use
the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n2
2n. We will consider the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the ratio.
L= lim
n→∞
(n+ 1)2
2n2
Step 3: Calculate the limit.
L= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1
2+1
n+1
2n2
=1
2
Step 4: Determine convergence. Since L=1
2<1, by the Ratio Test, the
series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 19
Question
Determine whether the series P∞
n=1 n+1
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n+1
2n, we will use the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms: Let an=n+1
2n.
We compute:
lim
n→∞
an+1
an
= lim
n→∞
n+2
2n+1
n+1
2n
= lim
n→∞
n+ 2
2(n+ 1)
= lim
n→∞
n
2n
= lim
n→∞
1
2
=1
2
Step 2: Apply the ratio test: - If limn→∞
an+1
an<1, the series converges
absolutely. - If limn→∞
an+1
an>1 or the limit is infinite, the series diverges. -
If limn→∞
an+1
an= 1, the test is inconclusive.
Since the limit we found is 1
2<1, by the ratio test, the series P∞
n=1 n+1
2n
converges absolutely.
13
Question 20
Question
Let P∞
n=1 anbe a series such that an=1
n2for n= 2kand an= 0 otherwise.
Determine whether the series converges or diverges.
Solution
Step 1: We will first rewrite the series using only the terms where anis nonzero.
Step 2: Notice that the terms where anis nonzero are when nis a power
of 2, i.e., n= 2,4,8,16, . . .. Step 3: Therefore, we can rewrite the series as
P∞
k=1 1
(2k)2=P∞
k=1 1
4k. Step 4: We recognize this series as a geometric series
with first term a= 1/4 and common ratio r= 1/4. Step 5: The geometric
series converges if |r|<1, which is true in this case since |1/4|<1. Step 6:
Therefore, the series P∞
n=1 anconverges.
Question 21
Question
Let P∞
n=1 anbe a series such that limn→∞ nan=c= 0. Determine whether
the series converges, diverges, or cannot be determined. Justify your answer.
Solution
Given: limn→∞ nan=c= 0
It is known that if limn→∞ nanis a non-zero finite value, then the series
P∞
n=1 andiverges.
Step 1: Let’s prove this result using the Limit Comparison Test.
Since limn→∞ nan=c= 0, we can consider the series bn=1
n.
Step 2: Now, we calculate the limit of the ratio limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
nan
n= lim
n→∞ an=c
Step 3: According to the Limit Comparison Test, if limn→∞
an
bnexists and is
a positive finite number, then either both series P∞
n=1 anand P∞
n=1 bnconverge
or both diverge.
Since the Harmonic Series P∞
n=1 1
nis divergent, and we have shown that
limn→∞
an
bn=c= 0, it follows that the series P∞
n=1 analso diverges.
Hence, the given series P∞
n=1 andiverges.
14
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test.
Let an=n!
nn. We will calculate the following limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the limit.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
= lim
n→∞
1
e
=1
e
Step 3: Analyze the limit.
Since the limit of the absolute value of the ratio is less than 1 (specifically,
it is 1
e<1), by the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 23
Question
Let P∞
n=1 anbe a convergent series such that an>0 for all n≥1. Prove or
disprove the following statement: If limn→∞ an= 0, then the series P∞
n=1 √an
converges.
Solution
To prove or disprove the statement, we will provide a counterexample.
Counterexample: Consider the series where an=1
n2. We know that
P∞
n=1 1
n2is a convergent series by the p-series test.
Now, let’s calculate bn=√an=q1
n2=1
n. We know that the harmonic
series P∞
n=1 1
nis a divergent series.
15
Since the series P∞
n=1 1
ndiverges, this provides a counterexample to the
statement. Therefore, the statement is disproved.
In conclusion, the statement ”If limn→∞ an= 0, then the series P∞
n=1 √an
converges” is false, as shown by the counterexample provided.
Question 24
Question
Determine whether the series ∞
X
k=1
3k2−1
4k3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
k=1
3k2−1
4k3+ 2, we will first simplify
the terms and then apply the limit comparison test.
Step 1: Simplifying the terms Consider the term 3k2−1
4k3+2 . We notice that
the degree of the numerator and denominator is the same, so we can simplify
by dividing each term by the highest power of kpresent in the denominator:
3k2−1
4k3+ 2 =k2(3 −1
k2)
k3(4 + 2
k3)=3−1
k2
4 + 2
k3
Step 2: Applying the limit comparison test Let’s find the limit of
the absolute value of the ratio of the given series term and a known convergent
series term. We will compare it to the alternating harmonic series ∞
X
k=1
(−1)k
k.
Consider the limit:
lim
k→∞
3−1
k2
4+ 2
k3
(−1)k
k
Simplifying the expression inside the limit further, we get:
lim
k→∞
3k+ 2
4k3(−1)k+k(−1)k+ 2k4(−1)k
Since we have both positive and negative terms in the denominator, the
absolute values will alternate.
For the limit comparison test, we consider the positive terms only and find
that the above series is comparable to ∞
X
k=1
1
k2.
16
Step 3: Conclusion Since ∞
X
k=1
1
k2is a convergent p-series with p= 2 >1,
by the limit comparison test, the original series ∞
X
k=1
3k2−1
4k3+ 2 also converges.
Question 25
Question
Determine whether the series
∞
X
n=1
3n+ (−2)n
5n
converges or diverges.
Solution
To determine the convergence of the series, we will split it into two separate
series and analyze each one individually.
Let’s denote the two series as an=3n
5nand bn=(−2)n
5n.
Step 1: Analyzing the series anFor the series an:
an=3n
5n=3
5n
This is a geometric series with r=3
5. The series converges if |r|<1, so:
|r|=
3
5
=3
5<1
Therefore, the series P∞
n=1 3n
5nconverges.
Step 2: Analyzing the series bnFor the series bn:
bn=(−2)n
5n=−2
5n
This is also a geometric series with r=−2
5. The series converges if |r|<1, so:
|r|=
−2
5
=2
5<1
Therefore, the series P∞
n=1
(−2)n
5nconverges.
Step 3: Combining the series Since both series anand bnconverge, and
the original series is the sum of these two convergent series, the series
∞
X
n=1
3n+ (−2)n
5n
also converges by the Comparison Test.
17
Question 26
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: Calculate the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 2: Simplify the expression and calculate the limit.
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 3: Analyze the limit. Since 1
e<1, the series converges by the ratio
test.
Therefore, the series
∞
X
n=1
n!
nn
converges.
Question 27
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series
∞
X
n=1
n!
nn,
18
we can use the ratio test. Let’s compute the limit of the ratio test.
Step 1: Compute the ratio test. Consider the ratio rn=an+1
an, where
an=n!
nn. Then,
rn=(n+ 1)!/(n+ 1)n+1
n!/nn
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!
=nn
(n+ 1)n
=n
n+ 1n
=1−1
n+ 1n
.
Step 2: Find the limit of the ratio. As napproaches infinity, we have
lim
n→∞ 1−1
n+ 1n
=1
e,
where eis the base of the natural logarithm.
Step 3: Determine the convergence of the series. Since the limit of the ratio
test is 1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 28
Question
Determine whether the series
∞
X
n=1
2n2+ 1
3n3−1
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 an=P∞
n=1 2n2+1
3n3−1and the series P∞
n=1 bn=
P∞
n=1 1
n.
19
Step 1: Find limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
2n2+1
3n3−1
1
n
= lim
n→∞
(2n2+ 1)n
3n3−1
= lim
n→∞
2n3+n
3n3−1
= lim
n→∞
2 + 1
n2
3−1
n3
=2
3.
Step 2: Analyze the limit. Since limn→∞
an
bn=2
3<∞, by the Limit
Comparison Test, either both series P∞
n=1 anand P∞
n=1 bnconverge or both
diverge.
Step 3: Determine the convergence of P∞
n=1 bn. The series P∞
n=1 1
nis the
harmonic series which is known to diverge.
Step 4: Conclusion. Since P∞
n=1 1
ndiverges and limn→∞
an
bn=2
3<∞, by
the Limit Comparison Test, the series P∞
n=1 2n2+1
3n3−1also diverges.
Question 29
Question
Let (an)∞
n=1 be a sequence such that an=n2+5
n3+3 . Determine whether the series
P∞
n=1 anconverges or diverges.
Solution
To investigate the convergence of the series P∞
n=1 an, we will apply the Limit
Comparison Test.
Step 1: Find a reference series. Let’s consider the series P∞
n=1 1
n, which
is known to be a divergent p-series with p= 1.
Step 2: Compute the limit. We will calculate the limit of the ratio
limn→∞
an
1/n :
lim
n→∞
an
1/n = lim
n→∞
n2+ 5
n3+ 3 ×n
1= lim
n→∞
n3+ 5n
n3+ 3 = 1
Step 3: Apply the Limit Comparison Test. Since the limit from Step 2
is a positive finite value, the series P∞
n=1 anhas the same convergence behavior
as the reference series P∞
n=1 1
n. Thus, P∞
n=1 andiverges.
20
Question 30
Question
Determine the convergence or divergence of the series P∞
n=1 3n+n2
n!.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. The ratio test states that if limn→∞
an+1
an<1, then the series Pan
converges. If limn→∞
an+1
an>1 or the limit does not exist, then the series
diverges.
Step 1: Compute the ratio an+1
an.
an+1
an
=
3n+1+(n+1)2
(n+1)!
3n+n2
n!
an+1
an
=3n+1 + (n+ 1)2
(n+ 1)! ·n!
3n+n2
an+1
an
=3·3n+ (n+ 1)2
(n+ 1)! ·n!
3n+n2
an+1
an
=3n+1 + 3(n+ 1)2
(n+ 1)! ·(3n+n2)
Step 2: Simplify the expression and find the limit.
lim
n→∞
an+1
an
= lim
n→∞
3n+1 + 3(n+ 1)2
(n+ 1)! ·(3n+n2)
lim
n→∞
3n+1 + 3n2+ 6n+ 3
(n+ 1)! ·(3n+n2)
As napproaches infinity, the terms with highest power dominate, so the limit
simplifies to:
lim
n→∞
3n
nn
= 0
Step 3: Determine convergence or divergence. Since the limit is less than
1, by the ratio test, the series P∞
n=1 3n+n2
n!converges.
Question 31
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n3+ 2.
21
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s consider the series P∞
n=1 n2+1
n3+2 and the series P∞
n=1 1
n.
Step 2: We will find the limit of their ratio:
lim
n→∞
n2+1
n3+2
1
n
= lim
n→∞
n(n−1+n−3)
1+2n−3= lim
n→∞
1 + n−2
1= 1
Step 3: Since the limit is a finite positive value, we can conclude that both
series either converge or diverge together.
Step 4: Since P∞
n=1 1
nis a divergent p-series with p= 1, by the Limit
Comparison Test, we can assert that P∞
n=1 n2+1
n3+2 also diverges. Hence, the
series diverges.
Question 32
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the limit limn→∞
an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 2: Analyze the limit. Since the limit limn→∞
an+1
an= 0 <1, by the
ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
22
Question 33
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 test. Let an=n!
nn. We will calculate the
limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio. Simplifying the above expression, we get:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
n
n+ 1
n
= lim
n→∞
1
1+1/n
= 1
Step 3: Analyze the result. Since the limit of the absolute value of the
ratio is equal to 1, the ratio test is inconclusive.
Step 4: Apply another test. Let’s try to use the root test. Consider the
auxiliary series bn=n
p|an|=n
qn!
nn.
Step 5: Calculate the root test. We will calculate the limit:
lim
n→∞ bn= lim
n→∞
n
rn!
nn
Step 6: Simplify the root. Simplifying the above expression, we get:
lim
n→∞
n
rn!
nn= lim
n→∞
n
√n!
n
Step 7: Apply the root test. As n→ ∞, we have n
√n!≈n/e, where e
is the base of the natural logarithm. Therefore,
lim
n→∞
n
√n!
n= lim
n→∞
n/e
n=1
e<1
Step 8: Conclusion. Since the limit of n
p|an|is less than 1, by the root
test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
23
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n+ 1
n4+ 1 .
Solution
To analyze the convergence of the series, we will use the limit comparison test.
Step 1: Find a suitable series for comparison. Consider the series
P∞
n=1 1
n2. This series is a p-series with p= 2, which converges.
Step 2: Establish a comparison criterion. We will compare the given
series to P∞
n=1 1
n2.
Step 3: Calculate the limit. Let an=n2+3n+1
n4+1 and bn=1
n2. We need
to evaluate the following limit:
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
n4+ 1 ·n2
1= lim
n→∞
n4+ 3n3+n2
n4+ 1 = 1.
Step 4: Apply the comparison test. Since the limit in Step 3 is a positive
finite number, the series P∞
n=1 n2+3n+1
n4+1 and P∞
n=1 1
n2either both converge or
both diverge.
Step 5: Final conclusion. Since P∞
n=1 1
n2converges, by the limit com-
parison test, the series P∞
n=1 n2+3n+1
n4+1 also converges.
Question 35
Question
Determine whether the series P∞
n=1
log(n+1)
n2converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series an=log(n+1)
n2.
Step 1: Find the limit of the ratio of the terms We will find the limit
of the ratio limn→∞
an
1
n2
.
lim
n→∞
an
1
n2
= lim
n→∞
log(n+1)
n2
1
n2
= lim
n→∞ log(n+ 1) = ∞
Step 2: Interpret the limit Since the limit is ∞, we cannot conclude
anything yet. Therefore, let’s use the Limit Comparison Test.
24
Question 2
Question
Determine the convergence or divergence of the series ∞
X
n=1
n3+ 5n2−3
4n4+n3−2.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let an=n3+5n2−3
4n4+n3−2.
Step 1: Find a simpler series to compare with. We will choose a
simpler series that we know the convergence or divergence of. Let’s consider the
series ∞
X
n=1
1
n.
Step 2: Form the ratio of the two series. Let bn=1
n. We will find the
limit of the ratio lim
n→∞
an
bn
.
Step 3: Calculate the limit and make a conclusion.
lim
n→∞
an
bn
= lim
n→∞
n3+5n2−3
4n4+n3−2
1
n
= lim
n→∞
n4+ 5n3−3n
4n4+n3−2
= lim
n→∞
1+5/n −3/n3
4+1/n2−2/n4
=1+0−0
4+0−0
=1
4.
Since lim
n→∞
an
bn
=1
4>0, the series ∞
X
n=1
n3+ 5n2−3
4n4+n3−2converges or diverges
with the same behavior as ∞
X
n=1
1
n. Therefore, the series ∞
X
n=1
n3+ 5n2−3
4n4+n3−2di-
verges (by divergence of the harmonic series).
Question 3
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
2
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
Step 1: Calculate the limit of the ratio limn→∞
an+1
an.
lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1
n)n
=1
e
Step 2: Determine the convergence based on the limit in Step 1. Since
the limit of the ratio is 1
e, according to the Ratio Test: - If 1
e<1, the series
P∞
n=1 n!
nnconverges. - If 1
e≥1, the series P∞
n=1 n!
nndiverges.
In this case, 1
e<1, so the series P∞
n=1 n!
nnconverges.
Question 4
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n3
n4+ 1
Solution
To determine the convergence or divergence of the series P∞
n=1 n3
n4+1 , we will
use the Limit Comparison Test.
Step 1: Find a series to compare with.
Consider the series P∞
n=1 1
n, which is a well-known divergent p-series.
Step 2: Form the limit of the ratio of the two series.
Let an=n3
n4+1 and bn=1
n. We will consider the limit:
lim
n→∞
an
bn
= lim
n→∞
n3
n4+1
1
n
= lim
n→∞
n4
n4+ 1 = 1
Step 3: Conclude based on the Limit Comparison Test.
Since limn→∞
an
bn= 1 and P∞
n=1 1
nis a divergent series, by the limit com-
parison test, the series P∞
n=1 n3
n4+1 also diverges.
3
Question 5
Question
Determine whether the series P∞
n=1 n2+1
2n3+n+1 converges or diverges.
Solution
To analyze the convergence of the series, we will use the Limit Comparison Test.
Step 1: Let’s consider the series bn=1
n, which we know diverges (it is a
p-series with p= 1). We will compare the given series to bn.
Step 2: Calculate the limit:
lim
n→∞
n2+1
2n3+n+1
1
n
= lim
n→∞
n3+n
2n3+n+ 1 = lim
n→∞
n3(1 + 1
n2)
n3(2 + 1
n2+1
n3)=1
2
Step 3: Since the limit is a finite positive number, by the Limit Compar-
ison Test, the given series P∞
n=1 n2+1
2n3+n+1 must converge if the series P∞
n=1 1
n
converges.
Step 4: Therefore, using the comparison with the divergent series P∞
n=1 1
n,
we can conclude that the given series P∞
n=1 n2+1
2n3+n+1 also diverges.
Question 6
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series, we will use the Ratio Test.
Step 1: Compute the ratio R.
R= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Calculate R.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n+ 1
1 + 1
nn+1
4
=e
where the above limit is a known result in calculus.
Step 3: Analyze the value of R. Since R=e > 1, the series P∞
n=1 n!
nn
diverges by the Ratio Test.
Therefore, the series P∞
n=1 n!
nndiverges.
Question 7
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ra-
tio test. Step 1: Let’s apply the ratio test. The ratio test states that if
limn→∞
an+1
an=L, then the series converges if L < 1 and diverges if L > 1.
Consider an=n!
nn. Then,
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!= lim
n→∞
n+ 1
(1 + 1
n)n=1
e<1
Since the limit is less than 1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 8
Question
Determine whether the series P∞
n=1 n2+3n
n3+2 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n
n3+2 , we will use the limit
comparison test.
Step 1: Let’s find a series that we can easily determine the convergence of.
We choose the series P∞
n=1 1
n.
Step 2: Compute the limit:
lim
n→∞
n2+ 3n
n3+ 2 ·n
1= lim
n→∞
n3+ 3n2
n3+ 2 = lim
n→∞
1 + 3
n
1 + 2
n3
= 1
Step 3: Since the limit is a finite positive number (not zero or infinity), by
the limit comparison test, the series P∞
n=1 n2+3n
n3+2 behaves like the series P∞
n=1 1
n.
5
Step 4: Since the harmonic series P∞
n=1 1
ndiverges, our original series
P∞
n=1 n2+3n
n3+2 also diverges by comparison.
Therefore, the series P∞
n=1 n2+3n
n3+2 diverges.
Question 9
Question
Determine if the series ∞
X
n=1
n2
2nconverges or diverges.
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Let an=n2
2n.
Step 2: Compute the limit of the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 3: Simplify the expression:
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
Step 4: Taking the limit, we get:
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 5: Since the limit is less than 1, by the ratio test, the series ∞
X
n=1
n2
2n
converges.
Therefore, the series converges.
Question 10
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
6
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. We want to
find limn→∞
an+1
an.
Step 2: Compute 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
= lim
n→∞
nn
(n+ 1)n
Step 3: Simplify the expression. Using the properties of limits, we get:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Analyze the result. 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 11
Question
Determine whether the series ∞
X
n=1
n2
2n
converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
2n, we will use the ratio test.
Step 1: Apply the ratio test by considering the limit
L= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Find an+1 and
an+1
an.
an+1 =(n+ 1)2
2n+1 and
an+1
an
=(n+ 1)2
2n+1 ·2n
n2
7
Step 3: Simplify the expression
an+1
an.
(n+ 1)2
2n+1 ·2n
n2=(n+ 1)2·2n
2n+1 ·n2=(n+ 1)2
2n2
Step 4: Evaluate the limit L.
L= lim
n→∞
(n+ 1)2
2n2=1
2
Step 5: Analyze the value of L. If L < 1, then the series converges. If
L > 1 or L=∞, then the series diverges. If L= 1, the test is inconclusive.
Since L=1
2<1, by the ratio test, the series P∞
n=1 n2
2nconverges.
Question 12
Question
Determine whether the series P∞
n=1(−1)n√n+1−√n
nconverges or diverges.
Solution
To determine the convergence of the given series, we will analyze it using the
alternating series test.
Step 1: Identify the terms Let an= (−1)n√n+1−√n
nbe the nth term of
the series.
Step 2: Check for absolute convergence First, we will consider the
absolute convergence of the series by finding the absolute value of the terms:
|an|=
√n+ 1 −√n
n
=√n+ 1 −√n
n
Step 3: Simplify the expression Simplify the absolute value of the terms:
|an|=√n+ 1 −√n
n=(√n+ 1 −√n)(√n+1+√n)
n(√n+1+√n)=√n+ 12−√n2
n(√n+1+√n)=1
n(√n+1+√n)
Step 4: Determine the limit Find the limit as napproaches infinity of
|an|to check for absolute convergence:
lim
n→∞ |an|= lim
n→∞
1
n(√n+1+√n)= lim
n→∞
1
n√n(1 + 1
n)= lim
n→∞
1
n3/2(1 + 1
n)= 0
Step 5: Apply the alternating series test Since the series P∞
n=1(−1)n√n+1−√n
n
satisfies the conditions of the alternating series test (terms are decreasing in ab-
solute value and approach 0), the series converges.
Therefore, the series P∞
n=1(−1)n√n+1−√n
nconverges.
8
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n2
2n.
Solution
To determine the convergence or divergence of the series P∞
n=1 n2
2n, we will use
the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the limit expression.
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: Apply the ratio test. - If the limit is less than 1, the series converges
absolutely. - If the limit is greater than 1 or does not exist, the series diverges.
- If the limit is equal to 1, the test is inconclusive.
Since the limit is 1
2which is less than 1, by the ratio test, the series P∞
n=1 n2
2n
converges absolutely.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
9
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n(n+ 1)
= lim
n→∞
nn
(n+ 1)n+1
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
n
= 1
Since the limit is equal to 1, the ratio test is inconclusive. Let’s try another
test, the root test.
lim
n→∞
n
p|an|= lim
n→∞
n
s
n!
nn
= lim
n→∞
n
√n!
n
=limn→∞
n
√n!
limn→∞ n
=1
∞
= 0
Since the limit is less than 1, the root test tells us that the series P∞
n=1 n!
nn
converges.
Question 15
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
10
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. The ratio test states that if
L= lim
n→∞
an+1
an
,
then: - If L < 1, the series converges absolutely. - If L > 1, the series diverges.
- If L= 1, the test is inconclusive.
Step 1: Compute the limit involved in the ratio test:
L= 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
= lim
n→∞ 1
1+1/nn
=1
e,
where the last step follows from the limit limn→∞ 1 + 1
nn=e.
Step 2: Analyze the value of L: Since L=1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges absolutely.
Therefore, the series converges.
Question 16
Question
Determine whether the series ∞
X
n=1
n2+ 3
3n3+ 4 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2+ 3
3n3+ 4, we can use the limit
comparison test.
11
Step 1: Find the limit for comparison Let an=n2+3
3n3+4 be the general
term of the series. We will find limn→∞
an
1
n
to compare.
lim
n→∞
an
1
n
= lim
n→∞
n2+ 3
3n3+ 4 ·n= lim
n→∞
n3+ 3n
3n3+ 4 =1
3
Step 2: Interpret the limit Since limn→∞
an
1
n
=1
3>0, and ∞
X
n=1
1
nis
ap-series with p= 1 (which diverges), we can conclude that ∞
X
n=1
n2+ 3
3n3+ 4 also
diverges by the limit comparison test.
Question 17
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n2+ 1
n3+ 1
Solution
To analyze the convergence of the given series, we can use the limit comparison
test. We will compare it to the series P∞
n=1 1
n.
Step 1: Find the limit of the ratio: Let an=n2+1
n3+1 and bn=1
n. We
want to find:
lim
n→∞
an
bn
lim
n→∞
n2+1
n3+1
1
n
= lim
n→∞
n3+n
n3+ 1 = 1
Step 2: Conclude with the limit comparison test: Since the limit is
a finite nonzero number, we can conclude by the limit comparison test that the
series P∞
n=1 n2+1
n3+1 behaves the same as the divergent harmonic series P∞
n=1 1
n.
Therefore, by the limit comparison test, the series P∞
n=1 n2+1
n3+1 diverges.
Question 18
Question
Determine the convergence or divergence of the series P∞
n=1 n2
2n.
12
Solution
To determine the convergence or divergence of the series P∞
n=1 n2
2n, we will use
the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n2
2n. We will consider the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
Step 2: Simplify the ratio.
L= lim
n→∞
(n+ 1)2
2n2
Step 3: Calculate the limit.
L= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1
2+1
n+1
2n2
=1
2
Step 4: Determine convergence. Since L=1
2<1, by the Ratio Test, the
series P∞
n=1 n2
2nconverges.
Therefore, the series P∞
n=1 n2
2nconverges.
Question 19
Question
Determine whether the series P∞
n=1 n+1
2nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n+1
2n, we will use the ratio test.
Step 1: Compute the limit of the ratio of consecutive terms: Let an=n+1
2n.
We compute:
lim
n→∞
an+1
an
= lim
n→∞
n+2
2n+1
n+1
2n
= lim
n→∞
n+ 2
2(n+ 1)
= lim
n→∞
n
2n
= lim
n→∞
1
2
=1
2
Step 2: Apply the ratio test: - If limn→∞
an+1
an<1, the series converges
absolutely. - If limn→∞
an+1
an>1 or the limit is infinite, the series diverges. -
If limn→∞
an+1
an= 1, the test is inconclusive.
Since the limit we found is 1
2<1, by the ratio test, the series P∞
n=1 n+1
2n
converges absolutely.
13
Question 20
Question
Let P∞
n=1 anbe a series such that an=1
n2for n= 2kand an= 0 otherwise.
Determine whether the series converges or diverges.
Solution
Step 1: We will first rewrite the series using only the terms where anis nonzero.
Step 2: Notice that the terms where anis nonzero are when nis a power
of 2, i.e., n= 2,4,8,16, . . .. Step 3: Therefore, we can rewrite the series as
P∞
k=1 1
(2k)2=P∞
k=1 1
4k. Step 4: We recognize this series as a geometric series
with first term a= 1/4 and common ratio r= 1/4. Step 5: The geometric
series converges if |r|<1, which is true in this case since |1/4|<1. Step 6:
Therefore, the series P∞
n=1 anconverges.
Question 21
Question
Let P∞
n=1 anbe a series such that limn→∞ nan=c= 0. Determine whether
the series converges, diverges, or cannot be determined. Justify your answer.
Solution
Given: limn→∞ nan=c= 0
It is known that if limn→∞ nanis a non-zero finite value, then the series
P∞
n=1 andiverges.
Step 1: Let’s prove this result using the Limit Comparison Test.
Since limn→∞ nan=c= 0, we can consider the series bn=1
n.
Step 2: Now, we calculate the limit of the ratio limn→∞
an
bn:
lim
n→∞
an
bn
= lim
n→∞
nan
n= lim
n→∞ an=c
Step 3: According to the Limit Comparison Test, if limn→∞
an
bnexists and is
a positive finite number, then either both series P∞
n=1 anand P∞
n=1 bnconverge
or both diverge.
Since the Harmonic Series P∞
n=1 1
nis divergent, and we have shown that
limn→∞
an
bn=c= 0, it follows that the series P∞
n=1 analso diverges.
Hence, the given series P∞
n=1 andiverges.
14
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test.
Step 1: Apply the ratio test.
Let an=n!
nn. We will calculate the following limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the limit.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
= lim
n→∞
1
e
=1
e
Step 3: Analyze the limit.
Since the limit of the absolute value of the ratio is less than 1 (specifically,
it is 1
e<1), by the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 23
Question
Let P∞
n=1 anbe a convergent series such that an>0 for all n≥1. Prove or
disprove the following statement: If limn→∞ an= 0, then the series P∞
n=1 √an
converges.
Solution
To prove or disprove the statement, we will provide a counterexample.
Counterexample: Consider the series where an=1
n2. We know that
P∞
n=1 1
n2is a convergent series by the p-series test.
Now, let’s calculate bn=√an=q1
n2=1
n. We know that the harmonic
series P∞
n=1 1
nis a divergent series.
15
Since the series P∞
n=1 1
ndiverges, this provides a counterexample to the
statement. Therefore, the statement is disproved.
In conclusion, the statement ”If limn→∞ an= 0, then the series P∞
n=1 √an
converges” is false, as shown by the counterexample provided.
Question 24
Question
Determine whether the series ∞
X
k=1
3k2−1
4k3+ 2 converges or diverges.
Solution
To determine the convergence of the series ∞
X
k=1
3k2−1
4k3+ 2, we will first simplify
the terms and then apply the limit comparison test.
Step 1: Simplifying the terms Consider the term 3k2−1
4k3+2 . We notice that
the degree of the numerator and denominator is the same, so we can simplify
by dividing each term by the highest power of kpresent in the denominator:
3k2−1
4k3+ 2 =k2(3 −1
k2)
k3(4 + 2
k3)=3−1
k2
4 + 2
k3
Step 2: Applying the limit comparison test Let’s find the limit of
the absolute value of the ratio of the given series term and a known convergent
series term. We will compare it to the alternating harmonic series ∞
X
k=1
(−1)k
k.
Consider the limit:
lim
k→∞
3−1
k2
4+ 2
k3
(−1)k
k
Simplifying the expression inside the limit further, we get:
lim
k→∞
3k+ 2
4k3(−1)k+k(−1)k+ 2k4(−1)k
Since we have both positive and negative terms in the denominator, the
absolute values will alternate.
For the limit comparison test, we consider the positive terms only and find
that the above series is comparable to ∞
X
k=1
1
k2.
16
Step 3: Conclusion Since ∞
X
k=1
1
k2is a convergent p-series with p= 2 >1,
by the limit comparison test, the original series ∞
X
k=1
3k2−1
4k3+ 2 also converges.
Question 25
Question
Determine whether the series
∞
X
n=1
3n+ (−2)n
5n
converges or diverges.
Solution
To determine the convergence of the series, we will split it into two separate
series and analyze each one individually.
Let’s denote the two series as an=3n
5nand bn=(−2)n
5n.
Step 1: Analyzing the series anFor the series an:
an=3n
5n=3
5n
This is a geometric series with r=3
5. The series converges if |r|<1, so:
|r|=
3
5
=3
5<1
Therefore, the series P∞
n=1 3n
5nconverges.
Step 2: Analyzing the series bnFor the series bn:
bn=(−2)n
5n=−2
5n
This is also a geometric series with r=−2
5. The series converges if |r|<1, so:
|r|=
−2
5
=2
5<1
Therefore, the series P∞
n=1
(−2)n
5nconverges.
Step 3: Combining the series Since both series anand bnconverge, and
the original series is the sum of these two convergent series, the series
∞
X
n=1
3n+ (−2)n
5n
also converges by the Comparison Test.
17
Question 26
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: Calculate the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 2: Simplify the expression and calculate the limit.
= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 3: Analyze the limit. Since 1
e<1, the series converges by the ratio
test.
Therefore, the series
∞
X
n=1
n!
nn
converges.
Question 27
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series
∞
X
n=1
n!
nn,
18
we can use the ratio test. Let’s compute the limit of the ratio test.
Step 1: Compute the ratio test. Consider the ratio rn=an+1
an, where
an=n!
nn. Then,
rn=(n+ 1)!/(n+ 1)n+1
n!/nn
=(n+ 1)!
(n+ 1)n+1 ·nn
n!
=(n+ 1)n!
(n+ 1)n+1 ·nn
n!
=nn
(n+ 1)n
=n
n+ 1n
=1−1
n+ 1n
.
Step 2: Find the limit of the ratio. As napproaches infinity, we have
lim
n→∞ 1−1
n+ 1n
=1
e,
where eis the base of the natural logarithm.
Step 3: Determine the convergence of the series. Since the limit of the ratio
test is 1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 28
Question
Determine whether the series
∞
X
n=1
2n2+ 1
3n3−1
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 an=P∞
n=1 2n2+1
3n3−1and the series P∞
n=1 bn=
P∞
n=1 1
n.
19
Step 1: Find limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
2n2+1
3n3−1
1
n
= lim
n→∞
(2n2+ 1)n
3n3−1
= lim
n→∞
2n3+n
3n3−1
= lim
n→∞
2 + 1
n2
3−1
n3
=2
3.
Step 2: Analyze the limit. Since limn→∞
an
bn=2
3<∞, by the Limit
Comparison Test, either both series P∞
n=1 anand P∞
n=1 bnconverge or both
diverge.
Step 3: Determine the convergence of P∞
n=1 bn. The series P∞
n=1 1
nis the
harmonic series which is known to diverge.
Step 4: Conclusion. Since P∞
n=1 1
ndiverges and limn→∞
an
bn=2
3<∞, by
the Limit Comparison Test, the series P∞
n=1 2n2+1
3n3−1also diverges.
Question 29
Question
Let (an)∞
n=1 be a sequence such that an=n2+5
n3+3 . Determine whether the series
P∞
n=1 anconverges or diverges.
Solution
To investigate the convergence of the series P∞
n=1 an, we will apply the Limit
Comparison Test.
Step 1: Find a reference series. Let’s consider the series P∞
n=1 1
n, which
is known to be a divergent p-series with p= 1.
Step 2: Compute the limit. We will calculate the limit of the ratio
limn→∞
an
1/n :
lim
n→∞
an
1/n = lim
n→∞
n2+ 5
n3+ 3 ×n
1= lim
n→∞
n3+ 5n
n3+ 3 = 1
Step 3: Apply the Limit Comparison Test. Since the limit from Step 2
is a positive finite value, the series P∞
n=1 anhas the same convergence behavior
as the reference series P∞
n=1 1
n. Thus, P∞
n=1 andiverges.
20
Question 30
Question
Determine the convergence or divergence of the series P∞
n=1 3n+n2
n!.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. The ratio test states that if limn→∞
an+1
an<1, then the series Pan
converges. If limn→∞
an+1
an>1 or the limit does not exist, then the series
diverges.
Step 1: Compute the ratio an+1
an.
an+1
an
=
3n+1+(n+1)2
(n+1)!
3n+n2
n!
an+1
an
=3n+1 + (n+ 1)2
(n+ 1)! ·n!
3n+n2
an+1
an
=3·3n+ (n+ 1)2
(n+ 1)! ·n!
3n+n2
an+1
an
=3n+1 + 3(n+ 1)2
(n+ 1)! ·(3n+n2)
Step 2: Simplify the expression and find the limit.
lim
n→∞
an+1
an
= lim
n→∞
3n+1 + 3(n+ 1)2
(n+ 1)! ·(3n+n2)
lim
n→∞
3n+1 + 3n2+ 6n+ 3
(n+ 1)! ·(3n+n2)
As napproaches infinity, the terms with highest power dominate, so the limit
simplifies to:
lim
n→∞
3n
nn
= 0
Step 3: Determine convergence or divergence. Since the limit is less than
1, by the ratio test, the series P∞
n=1 3n+n2
n!converges.
Question 31
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n3+ 2.
21
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test.
Step 1: Let’s consider the series P∞
n=1 n2+1
n3+2 and the series P∞
n=1 1
n.
Step 2: We will find the limit of their ratio:
lim
n→∞
n2+1
n3+2
1
n
= lim
n→∞
n(n−1+n−3)
1+2n−3= lim
n→∞
1 + n−2
1= 1
Step 3: Since the limit is a finite positive value, we can conclude that both
series either converge or diverge together.
Step 4: Since P∞
n=1 1
nis a divergent p-series with p= 1, by the Limit
Comparison Test, we can assert that P∞
n=1 n2+1
n3+2 also diverges. Hence, the
series diverges.
Question 32
Question
Determine the convergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the limit limn→∞
an+1
an.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 2: Analyze the limit. Since the limit limn→∞
an+1
an= 0 <1, by the
ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
22
Question 33
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 test. Let an=n!
nn. We will calculate the
limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio. Simplifying the above expression, we get:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
n
n+ 1
n
= lim
n→∞
1
1+1/n
= 1
Step 3: Analyze the result. Since the limit of the absolute value of the
ratio is equal to 1, the ratio test is inconclusive.
Step 4: Apply another test. Let’s try to use the root test. Consider the
auxiliary series bn=n
p|an|=n
qn!
nn.
Step 5: Calculate the root test. We will calculate the limit:
lim
n→∞ bn= lim
n→∞
n
rn!
nn
Step 6: Simplify the root. Simplifying the above expression, we get:
lim
n→∞
n
rn!
nn= lim
n→∞
n
√n!
n
Step 7: Apply the root test. As n→ ∞, we have n
√n!≈n/e, where e
is the base of the natural logarithm. Therefore,
lim
n→∞
n
√n!
n= lim
n→∞
n/e
n=1
e<1
Step 8: Conclusion. Since the limit of n
p|an|is less than 1, by the root
test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
23
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 3n+ 1
n4+ 1 .
Solution
To analyze the convergence of the series, we will use the limit comparison test.
Step 1: Find a suitable series for comparison. Consider the series
P∞
n=1 1
n2. This series is a p-series with p= 2, which converges.
Step 2: Establish a comparison criterion. We will compare the given
series to P∞
n=1 1
n2.
Step 3: Calculate the limit. Let an=n2+3n+1
n4+1 and bn=1
n2. We need
to evaluate the following limit:
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
n4+ 1 ·n2
1= lim
n→∞
n4+ 3n3+n2
n4+ 1 = 1.
Step 4: Apply the comparison test. Since the limit in Step 3 is a positive
finite number, the series P∞
n=1 n2+3n+1
n4+1 and P∞
n=1 1
n2either both converge or
both diverge.
Step 5: Final conclusion. Since P∞
n=1 1
n2converges, by the limit com-
parison test, the series P∞
n=1 n2+3n+1
n4+1 also converges.
Question 35
Question
Determine whether the series P∞
n=1
log(n+1)
n2converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series an=log(n+1)
n2.
Step 1: Find the limit of the ratio of the terms We will find the limit
of the ratio limn→∞
an
1
n2
.
lim
n→∞
an
1
n2
= lim
n→∞
log(n+1)
n2
1
n2
= lim
n→∞ log(n+ 1) = ∞
Step 2: Interpret the limit Since the limit is ∞, we cannot conclude
anything yet. Therefore, let’s use the Limit Comparison Test.
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Step 3: Apply the Limit Comparison Test We will compare the series
P∞
n=1
log(n+1)
n2with the series P∞
n=1 1
n.
Let bn=1
n. We want to show that anand bnhave the same behavior.
Step 4: Take the limit of the ratio of anand bnCalculate limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
log(n+1)
n2
1
n
= lim
n→∞
log(n+ 1)
n= 0
Step 5: Conclusion Since P∞
n=1 1
nis a divergent harmonic series, and the
series P∞
n=1
log(n+1)
n2has the same behavior, by the Limit Comparison Test, the
given series diverges.
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