MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 10
Liberty University
Question 1
Question
Determine whether the following series converges or diverges:
∞
X
n=1
n!
nn
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, then 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
(n+ 1)n+1 =1
n+ 1
Step 2: Apply the ratio test: We have limn→∞
an+1
an= limn→∞
1
n+1 =
0<1.
Step 3: Conclusion: Since the limit is less than 1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
Question 2
Question
Evaluate the convergence of the series P∞
n=1 1
n3+n2.
Solution
To determine the convergence of the series P∞
n=1 1
n3+n2, we will use the com-
parison test.
Step 1: Find a Series to Compare Let’s find a convergent series which
bounds the given series from above.
Consider the series P∞
n=1 1
n3. This is a p-series with p= 3, which converges
by the p-series test.
Step 2: Show the Comparison We will show that 1
n3+n2≤1
n3for all
n≥1. 1
n3+n2≤1
n3⇐⇒ n3≤n3+n2⇐⇒ 0≤n2
which is true for all n≥1.
Step 3: Conclude Convergence Since 1
n3+n2≤1
n3for all n≥1 and
P∞
n=1 1
n3converges, by the comparison test, the series P∞
n=1 1
n3+n2also con-
verges.
Question 3
Question
Determine whether the series P∞
n=1 n2
3nconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2
3n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n2
3n. Compute the ratio R:
R= lim
n→∞
an+1
an
R= lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
R= lim
n→∞
(n+ 1)2
3n2
R= lim
n→∞
n2+ 2n+ 1
3n2
R= lim
n→∞
1
3+2
3n+1
3n2
=1
3
Step 2: Analyze the ratio R. Since R=1
3<1, the series P∞
n=1 n2
3nconverges
by the ratio test.
Therefore, the series P∞
n=1 n2
3nconverges.
2
Question 4
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. Calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 2: Simplify and evaluate the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 3: Determine convergence based on the ratio test: - If the limit L < 1,
then the series converges. - If the limit L > 1 or is infinite, then the series
diverges. - If the limit L= 1, the test is inconclusive.
Since L= 1/e < 1, by the ratio test, the series ∞
X
n=1
n!
nnconverges.
Question 5
Question
Determine whether the series P∞
n=1 n2+3n−1
n4+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series P∞
n=1 n2+3n−1
n4+5 and a new series P∞
n=1 1
n2.
Step 1: Find the limit of the ratio of the two series. Let an=n2+3n−1
n4+5 and
bn=1
n2. We want to find the limit of the ratio limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n−1
n4+ 5 ·n2
1= lim
n→∞
n4+ 3n3−n2
n4+ 5 = 1
3
Step 2: Apply the Limit Comparison Test. Since the limit is a finite positive
number, then either both series P∞
n=1 anand P∞
n=1 bnconverge or both diverge.
We know that the series P∞
n=1 bn=P∞
n=1 1
n2is a p-series with p= 2, which
converges.
Step 3: Conclusion. By the Limit Comparison Test, since the series of
n2+3n−1
n4+5 and 1
n2have the same convergence behavior, we can conclude that the
series P∞
n=1 n2+3n−1
n4+5 also converges.
Question 6
Question
Consider the series P∞
n=1 n2+2
n3+3 . Determine whether the series converges or
diverges.
Solution
To analyze the convergence of the series P∞
n=1 n2+2
n3+3 , we will use the Limit
Comparison Test. Let’s denote the given series as an=n2+2
n3+3 .
Step 1: Find a comparison series Let’s consider the series bn=1
n, which
is a divergent p-series with p= 1.
Step 2: Calculate the limit We will compute the limit limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 2)/(n3+ 3)
1/n = lim
n→∞
n2+ 2
n2(n+ 3) = lim
n→∞
1+2/n2
n+ 3
Step 3: Evaluate the limit Looking at the expression inside the limit, as
napproaches infinity, the term 2/n2becomes negligible compared to 1/n, so we
focus on the dominant terms:
lim
n→∞
1
n= 0
Step 4: Check the conditions of the Limit Comparison Test Since
limn→∞
an
bn= 0 and P∞
n=1 bndiverges, we can conclude by the Limit Compari-
son Test that P∞
n=1 analso diverges.
Therefore, the series P∞
n=1 n2+2
n3+3 diverges.
Question 7
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
2n3+ 1.
4
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. Let’s consider the series
∞
X
n=1
n2
2n3+ 1.
and compare it with the series
∞
X
n=1
1
n.
Step 1: Find the limit of the ratio Compute the limit of the ratio of
the two series:
lim
n→∞
n2
2n3+1
1
n
= lim
n→∞
n3
2n3+ 1.
Step 2: Simplify the expression Simplify the above limit:
= lim
n→∞
1
2 + 1
n2
=1
2.
Step 3: Conclusion Since the limit is a positive finite number, by the limit
comparison test, the series
∞
X
n=1
n2
2n3+ 1
converges if the series
∞
X
n=1
1
n
converges. The harmonic series
∞
X
n=1
1
n
is a divergent series.
Conclusion: By the limit comparison test, the series
∞
X
n=1
n2
2n3+ 1
diverges since it can be compared with the harmonic series which is divergent.
Question 8
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
5
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn, then 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!
nn·nn
n!·n+ 1
n+ 1 =n+ 1
n= 1+ 1
n
Step 2: Apply the ratio test. The ratio test states that if limn→∞
an+1
an
<
1, then the series converges; if limn→∞
an+1
an
>1 or it does not exist, then the
series diverges; and if limn→∞
an+1
an
= 1, the ratio test is inconclusive.
Step 3: Find the limit. Taking the limit of the ratio as napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
1 + 1
n
= 1
Step 4: Conclusion. Since limn→∞
an+1
an
= 1, the ratio test is inconclusive.
We may need to try other tests to determine the convergence of the series
∞
X
n=1
n!
nn.
Question 9
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,
we will use the ratio test.
Step 1: Compute the nth term of the series. The general term of the series
is given by an=n!
nn.
6
Step 2: Apply the ratio test. Consider the limit of the ratio of the (n+ 1)th
term to the nth term:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 3: Simplify the ratio.
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n
(1 + 1
n)n
Step 4: Evaluate the limit. Take the limit of the last expression:
= lim
n→∞
n
e=∞
Step 5: Conclusion. Since the limit of the ratio is greater than 1, by the
ratio test, the series diverges. Thus, the series
∞
X
n=1
n!
nn
diverges.
Question 10
Question
Determine if the series ∞
X
n=1
n2+n+ 1
2n3+ 3 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+n+ 1
2n3+ 3 and the series ∞
X
n=1
1
n.
Step 1: Find the limit of an
bn: Let an=n2+n+1
2n3+3 and bn=1
n. We will find
the limit of an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+n+ 1
2n3+ 3 ·n
1
= lim
n→∞
n3+n2+n
2n3+ 3 = lim
n→∞
1 + 1
n+1
n2
2 + 3
n3
=1+0+0
2+0 =1
2
7
Step 2: Compare the limit with a known series: Since ∞
X
n=1
1
nis a harmonic
series that diverges, and an
bn=1
2(which is a finite positive value), by the Limit
Comparison Test, the given series ∞
X
n=1
n2+n+ 1
2n3+ 3 also diverges.
Therefore, the series ∞
X
n=1
n2+n+ 1
2n3+ 3 diverges.
Question 11
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We will
consider the ratio Rof consecutive terms:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify further.
R= lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n= lim
n→∞ 1 + 1
n−n
Step 4: Evaluate the limit.
R= lim
n→∞ 1 + 1
n−n
=1
e
Step 5: Determine convergence. Since R < 1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 12
Question
Determine whether the series
∞
X
n=1
n+ cos n
n2+n+ 1
8
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=n+cos n
n2+n+1 .
Step 1: Find a suitable series for comparison. Consider the series
P∞
n=1 1
n. This is a well-known divergent series.
Step 2: Calculate the limit. Let bn=1
n. We will calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n+cos n
n2+n+1
1
n
= lim
n→∞
n2+ncos n
n2+n+ 1
Since both the numerator and denominator have the same leading term, we can
apply L’Hˆopital’s Rule:
= lim
n→∞
2n+ cos n
2n+ 1
= 1
Thus, limn→∞
an
bn= 1.
Step 3: Conclusion Since the limit is a positive finite number, the Limit
Comparison Test is inconclusive. Therefore, we cannot determine the conver-
gence of the series using this test. Further investigation or a different test may
be needed to determine the convergence or divergence of the series.
Question 13
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, we can use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
9
Step 2: Simplify the ratio.
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
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 4: Interpret the result. Since limn→∞
an+1
an=1
e<1, by the ratio
test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnis convergent.
Question 14
Question
Determine whether the series P∞
n=1 n3
2nconverges or diverges.
Solution
To determine the convergence 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. The ratio
of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1 + 3
n+3
n2+1
n3
2
=1
2
Step 3: Analyze the limit. 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.
10
Question 15
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test Let an=n!
nn. Compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/(nn)
Step 2: Simplify the ratio Simplify the ratio (n+ 1)!/(n+ 1)n+1
n!/(nn):
R= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n+ 1
(n+ 1)n
Step 3: Find the limit Take the limit of n+ 1
(n+ 1)n:
R= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 4: Make a conclusion Since R=1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Question 16
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+ 1
n3+ 2.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s consider the series ∞
X
n=1
n2+ 1
n3+ 2 and a new series ∞
X
n=1
1
n.
11
Step 1: Compute the limit:
lim
n→∞
n2+1
n3+2
1
n
= lim
n→∞
n3+n
n3+ 2 = 1.
Step 2: Since the limit is a positive finite value, we can conclude that both
series either converge or diverge together.
Step 3: ∞
X
n=1
1
nis known as the Harmonic Series, which diverges (by the
p-series test with p= 1).
Step 4: Therefore, by the Limit Comparison Test, as the Harmonic Series
diverges and our series has the same behavior as the Harmonic Series, the given
series ∞
X
n=1
n2+ 1
n3+ 2 also diverges.
Question 17
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n3+ 1
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find the limit of the term inside the series:
lim
n→∞
n2+ 1
n3+ 1
Step 2: Divide numerator and denominator by n3to simplify the expression:
lim
n→∞
n2
n3+1
n3
1 + 1
n3
Step 3: Simplify the expression to find the limit:
lim
n→∞
1/n + 0
1+0 = lim
n→∞
1
1= 1
Step 4: Since the limit is a finite positive value, we can apply the Limit
Comparison Test with the series P∞
n=1 1
n, which is a p-series with p= 1.
Step 5: Let’s define an=n2+1
n3+1 and bn=1
nfor the comparison test.
12
Step 6: Now, compute the limit:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 1 ·n
1= 1
Step 7: Since P∞
n=1 1
nis a divergent series, by the Limit Comparison Test,
we can conclude that P∞
n=1 n2+1
n3+1 also diverges.
Question 18
Question
Let P∞
n=1 anbe a convergent series with positive terms. If limn→∞
nan+1
an=
L > 1, determine whether P∞
n=1 anmust also converge.
Solution
Given that limn→∞
nan+1
an=L > 1, we know that for sufficiently large n, the
ratio nan+1
anis greater than 1. This gives us the general idea that the terms of
the series must grow rapidly.
Step 1: Use the Ratio Test with the given ratio expression. Let’s consider
the series P∞
n=1 an. According to the Ratio Test, if limn→∞
an+1
an=L > 1,
then the series diverges. Here, we are given limn→∞
nan+1
an=L > 1. Let
r=an+1
an. Hence, r= limn→∞
an+1
an>1.
Step 2: Understand the implications of r > 1. Since r > 1, we know that
the terms of the series {an}are increasing. This suggests that the terms of the
series do not approach 0 as quickly, which can make it difficult for the series to
converge.
Step 3: Conclude the convergence of the series. Since the terms of {an}are
increasing (meaning an>0 for all n) and limn→∞
nan+1
an=L > 1, we conclude
that the series diverges by the Ratio Test.
Therefore, for the given series to converge, it is necessary that the limit is
less than or equal to 1.
Question 19
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1 √anan+1 also converges.
Solution
To show that the series P∞
n=1 √anan+1 converges, we will use the Comparison
Test.
13
Step 1: Choose appropriate inequality. Since √xis a monotonically
increasing function, we have √anan+1 ≤1
2(an+an+1) for all n∈N.
Step 2: Prove the inequality. Note that (√an−√an+1)2≥0. Expanding
this inequality gives us an+an+1 −2√anan+1 ≥0. Rearranging terms, we find
that √anan+1 ≤1
2(an+an+1).
Step 3: Apply the Comparison Test. Since P∞
n=1 anconverges, there
exists some N∈Nsuch that P∞
n=Nan<1. For all n≥N, we have √anan+1 ≤
1
2(an+an+1). Thus, P∞
n=N√anan+1 ≤1
2P∞
n=N(an+an+1). Simplifying the in-
equality gives P∞
n=N√anan+1 ≤1
2P∞
n=Nan+1
2P∞
n=Nan+1. Since both series
P∞
n=Nanand P∞
n=Nan+1 converge, the series P∞
n=N√anan+1 also converges
by the Comparison Test. Therefore, the series P∞
n=1 √anan+1 converges.
Question 20
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. Then, compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
[(n+ 1)!/(n+ 1)n+1]·[nn/n!]
nn/n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Calculate the limit.
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)n=1
e<1
Since the limit is less than 1, by the Ratio Test, the series ∞
X
n=1
n!
nnconverges.
14
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Recall that for a series Pan, the ratio test states that if
L= lim
n→∞
an+1
an
,
then the series converges if L < 1 and diverges if L > 1.
Step 2: Let’s apply the ratio test to our series. We have
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 3: Simplifying the expression inside the absolute value, we get
L= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
.
Step 4: Simplifying further, we get
L= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
.
Step 5: After canceling out terms, we have
L= lim
n→∞
nn
(n+ 1)n
.
Step 6: Simplifying the limit, we get
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
.
Step 7: Since limn→∞ 1
1+ 1
nn=1
e, where eis the base of the natural
logarithm, we have L=1
e.
Step 8: Since 1
e<1, by the ratio test, the series Pn!
nnconverges.
Therefore, the given series converges.
15
Question 22
Question
Determine whether the series P∞
n=1 n!
nn+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn+1 , we can use the ratio
test.
Step 1: Apply the ratio test. Let’s calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+2
n!/nn+1
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+2
n!/nn+1
= lim
n→∞
(n+ 1)!nn+1
(n+ 1)n+2n!
= lim
n→∞
(n+ 1)nnn!
(n+ 1)n+1n!
= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
1
(1 + 1
n)n+1
=1
e<1
Step 3: Since the limit is less than 1, by the ratio test, the series P∞
n=1 n!
nn+1
converges.
Question 23
Question
Determine if 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: Calculate the ratio Let’s consider the ratio R:
R= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Find the expression for an+1 and anWe have:
an+1 =(n+ 1)2
2n+1 =n2+ 2n+ 1
2·2n
16
and
an=n2
2n.
Step 3: Calculate the limit Now, let’s compute the limit:
R= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1
2+1
n+1
n2
=1
2.
Step 4: Conclusion Since the ratio R= 1/2<1, by the ratio test, the
series P∞
n=1 n2
2nconverges.
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n4+ 3
Solution
To determine the convergence or divergence of the series, we will use the Com-
parison Test.
Step 1: Consider the series P∞
n=1 n2+1
n4+3 .
Step 2: Simplify the terms of the series: n2+1
n4+3 =n2+1
n4·1
1+ 3
n4
=1
n2·1
1+ 3
n4
.
Step 3: Note that for n≥1, we have 3
n4≤3 and therefore 1
1+ 3
n4≥1
4.
Step 4: Now, consider the series P∞
n=1 1
n2, which is a p-series with p= 2.
Step 5: Since 1
n2≤1
n2·1
1+ 3
n4
for all n≥1 and P∞
n=1 1
n2converges, by the
Comparison Test, we conclude that P∞
n=1 n2+1
n4+3 converges.
Question 25
Question
Determine whether the series ∞
X
n=1
n3
2n4+ 3 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n3
2n4+ 3, we will use the limit
comparison test.
17
Step 1: Find the limit Consider the series ∞
X
n=1
n3
2n4+ 3 and let an=
n3
2n4+ 3. We will find the limit of an
1/n as napproaches infinity:
lim
n→∞
an
1
n
= lim
n→∞
n4
2n4+ 3 = lim
n→∞
1
2 + 3
n4
=1
2
Step 2: Conclusion Since the limit is a finite positive number ( 1
2), by
the limit comparison test, the series Pn3
2n4+3 converges if and only if the series
P1
nconverges. And since the harmonic series P1
ndiverges, our original series
Pn3
2n4+3 also diverges.
Question 26
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Apply the ratio test. Consider the ratio of consecutive terms:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
r= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
Step 3: Simplify further and evaluate the limit.
r= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
Step 4: Use the limit definition of e.
r= lim
n→∞
n
(1 + 1/n)n
= lim
n→∞
n
e=1
e
Step 5: Determine the series convergence. If r < 1, then the series con-
verges. Since 1
e<1, the series P∞
n=1 n!
nnconverges by the ratio test.
18
Question 27
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: Apply the Ratio Test. Let an=n!
nn.
We calculate the limit:
L= lim
n→∞
an+1
an
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
L= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
L= lim
n→∞
nn
(n+ 1)n
L= lim
n→∞
1
(1 + 1/n)n
Step 2: Simplify the limit. Applying L’Hopital’s Rule leads to:
L= lim
n→∞
1
e=1
e
Step 3: Analyze the limit value. Since L=1
e<1, by the Ratio Test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 28
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
19
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Let’s consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplifying the expression, we get:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Further simplifying gives:
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 4: Taking the limit, we have:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 5: Since the limit is equal to 1, the ratio test is inconclusive. We need
to use another test to determine the convergence of the series.
Step 6: Let’s apply the root test to the series. Consider the limit:
lim
n→∞
n
s
n!
nn
= lim
n→∞
n
√n!
n
Step 7: By applying Stirling’s approximation, we find that limn→∞
n
√n!
n=1
e.
Step 8: Since 1
e<1, the series P∞
n=1 n!
nnconverges by the root test.
Question 29
Question
Determine whether the series P∞
n=1 nn
n!converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 nn
n!, we will use the ratio test.
Step 1: Compute an: Let an=nn
n!.
Step 2: Apply the ratio test: Consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)n+1/(n+ 1)!
nn/n!
20
Simplify the expression:
= lim
n→∞
(n+ 1) n+ 1
nn
= lim
n→∞(n+ 1) 1 + 1
nn
= lim
n→∞(n+ 1)e
=∞
Step 3: Analyze the limit: Since the limit from the ratio test is greater than
1, the series P∞
n=1 nn
n!diverges.
Therefore, the given series diverges.
Question 30
Question
Determine the convergence or divergence of the series
∞
X
n=1
(−1)n·n
2n+n2.
Solution
To determine the convergence or divergence of the given series, we can use the
alternating series test and compare it to a simpler series.
Step 1: Apply the alternating series test. The series P∞
n=1
(−1)n·n
2n+n2is an
alternating series because each term has a sign that alternates. Let’s verify the
conditions of the alternating series test:
1. The terms an=n
2n+n2are positive for all n≥1. 2. The terms an+1 =
n+1
2n+1+(n+1)2are decreasing for all n≥1. 3. limn→∞ an= 0.
Since all conditions are met, we can conclude that the series converges by
the alternating series test.
Step 2: Compare the series to a simpler series. We can compare the series
P∞
n=1 n
2n+n2to P∞
n=1 n
2nsince 2n+n2>2nfor all n≥1.
Consider the series P∞
n=1 n
2n. This is a convergent series since it is a geo-
metric series with common ratio r=1
2where −1< r < 1.
Therefore, by the comparison test, the series P∞
n=1 n
2n+n2also converges.
21
Question 31
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
Let’s use the ratio test to determine the convergence or divergence of the series.
Step 1: Calculate the limit of the ratio using the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the limit:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
.
Step 3: Further simplify the limit:
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn+1(n+ 1)
(n+ 1)n+1
.
Step 4: Evaluate the limit:
lim
n→∞
nn+1(n+ 1)
(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
.
Step 5: Since limn→∞
n
(1+1/n)ntends to ∞, the series P∞
n=1 n!
nndiverges by
the ratio test.
Question 32
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Determine whether the series P∞
n=1
nan
1+n2converges or diverges.
Solution
We will use the Limit Comparison Test to determine the convergence of the
series P∞
n=1
nan
1+n2.
Step 1: Find the limit of the ratio. Let bn=nan
1+n2. We want to find
limn→∞
bn
an.
22
lim
n→∞
bn
an
= lim
n→∞
nan
1+n2
an
= lim
n→∞
n
1 + n2= 0
Step 2: Apply the Limit Comparison Test. Since limn→∞
bn
an= 0 and
P∞
n=1 anconverges, by the Limit Comparison Test, P∞
n=1 bnconverges as well.
Step 3: Conclusion. Therefore, the series P∞
n=1
nan
1+n2converges.
Question 33
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms using 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
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
Step 3: Take the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 4: Apply the ratio test: Since the limit 1
eis less than 1, by the ratio
test, the series P∞
n=1 n!
nnconverges.
Question 34
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
23
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Let’s denote an=n!
nn.
Step 1: Compute the ratio test. The ratio test states that if limn→∞
an+1
an<
1, then the series converges; if limn→∞
an+1
an>1 or diverges to ∞, then the
series diverges; and if the limit equals 1, the test is inconclusive.
Compute the ratio R= lim
n→∞
an+1
an
.
R= lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 2: Analyze the value of R. Since the limit R=1
e<1, by the ratio
test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 35
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1
√an
nis convergent.
Solution
To prove that the series P∞
n=1
√an
nis convergent, we will make use of the Com-
parison Test.
Step 1: Since P∞
n=1 anis a convergent series with positive terms, we know
that the terms anmust converge to zero as napproaches infinity. This implies
that limn→∞ an= 0.
24
Question 4
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. Calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 2: Simplify and evaluate the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 3: Determine convergence based on the ratio test: - If the limit L < 1,
then the series converges. - If the limit L > 1 or is infinite, then the series
diverges. - If the limit L= 1, the test is inconclusive.
Since L= 1/e < 1, by the ratio test, the series ∞
X
n=1
n!
nnconverges.
Question 5
Question
Determine whether the series P∞
n=1 n2+3n−1
n4+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series P∞
n=1 n2+3n−1
n4+5 and a new series P∞
n=1 1
n2.
Step 1: Find the limit of the ratio of the two series. Let an=n2+3n−1
n4+5 and
bn=1
n2. We want to find the limit of the ratio limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n−1
n4+ 5 ·n2
1= lim
n→∞
n4+ 3n3−n2
n4+ 5 = 1
3
Step 2: Apply the Limit Comparison Test. Since the limit is a finite positive
number, then either both series P∞
n=1 anand P∞
n=1 bnconverge or both diverge.
We know that the series P∞
n=1 bn=P∞
n=1 1
n2is a p-series with p= 2, which
converges.
Step 3: Conclusion. By the Limit Comparison Test, since the series of
n2+3n−1
n4+5 and 1
n2have the same convergence behavior, we can conclude that the
series P∞
n=1 n2+3n−1
n4+5 also converges.
Question 6
Question
Consider the series P∞
n=1 n2+2
n3+3 . Determine whether the series converges or
diverges.
Solution
To analyze the convergence of the series P∞
n=1 n2+2
n3+3 , we will use the Limit
Comparison Test. Let’s denote the given series as an=n2+2
n3+3 .
Step 1: Find a comparison series Let’s consider the series bn=1
n, which
is a divergent p-series with p= 1.
Step 2: Calculate the limit We will compute the limit limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 2)/(n3+ 3)
1/n = lim
n→∞
n2+ 2
n2(n+ 3) = lim
n→∞
1+2/n2
n+ 3
Step 3: Evaluate the limit Looking at the expression inside the limit, as
napproaches infinity, the term 2/n2becomes negligible compared to 1/n, so we
focus on the dominant terms:
lim
n→∞
1
n= 0
Step 4: Check the conditions of the Limit Comparison Test Since
limn→∞
an
bn= 0 and P∞
n=1 bndiverges, we can conclude by the Limit Compari-
son Test that P∞
n=1 analso diverges.
Therefore, the series P∞
n=1 n2+2
n3+3 diverges.
Question 7
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
2n3+ 1.
4
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. Let’s consider the series
∞
X
n=1
n2
2n3+ 1.
and compare it with the series
∞
X
n=1
1
n.
Step 1: Find the limit of the ratio Compute the limit of the ratio of
the two series:
lim
n→∞
n2
2n3+1
1
n
= lim
n→∞
n3
2n3+ 1.
Step 2: Simplify the expression Simplify the above limit:
= lim
n→∞
1
2 + 1
n2
=1
2.
Step 3: Conclusion Since the limit is a positive finite number, by the limit
comparison test, the series
∞
X
n=1
n2
2n3+ 1
converges if the series
∞
X
n=1
1
n
converges. The harmonic series
∞
X
n=1
1
n
is a divergent series.
Conclusion: By the limit comparison test, the series
∞
X
n=1
n2
2n3+ 1
diverges since it can be compared with the harmonic series which is divergent.
Question 8
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
5
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn, then 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!
nn·nn
n!·n+ 1
n+ 1 =n+ 1
n= 1+ 1
n
Step 2: Apply the ratio test. The ratio test states that if limn→∞
an+1
an
<
1, then the series converges; if limn→∞
an+1
an
>1 or it does not exist, then the
series diverges; and if limn→∞
an+1
an
= 1, the ratio test is inconclusive.
Step 3: Find the limit. Taking the limit of the ratio as napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
1 + 1
n
= 1
Step 4: Conclusion. Since limn→∞
an+1
an
= 1, the ratio test is inconclusive.
We may need to try other tests to determine the convergence of the series
∞
X
n=1
n!
nn.
Question 9
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,
we will use the ratio test.
Step 1: Compute the nth term of the series. The general term of the series
is given by an=n!
nn.
6
Step 2: Apply the ratio test. Consider the limit of the ratio of the (n+ 1)th
term to the nth term:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 3: Simplify the ratio.
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n
(1 + 1
n)n
Step 4: Evaluate the limit. Take the limit of the last expression:
= lim
n→∞
n
e=∞
Step 5: Conclusion. Since the limit of the ratio is greater than 1, by the
ratio test, the series diverges. Thus, the series
∞
X
n=1
n!
nn
diverges.
Question 10
Question
Determine if the series ∞
X
n=1
n2+n+ 1
2n3+ 3 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+n+ 1
2n3+ 3 and the series ∞
X
n=1
1
n.
Step 1: Find the limit of an
bn: Let an=n2+n+1
2n3+3 and bn=1
n. We will find
the limit of an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+n+ 1
2n3+ 3 ·n
1
= lim
n→∞
n3+n2+n
2n3+ 3 = lim
n→∞
1 + 1
n+1
n2
2 + 3
n3
=1+0+0
2+0 =1
2
7
Step 2: Compare the limit with a known series: Since ∞
X
n=1
1
nis a harmonic
series that diverges, and an
bn=1
2(which is a finite positive value), by the Limit
Comparison Test, the given series ∞
X
n=1
n2+n+ 1
2n3+ 3 also diverges.
Therefore, the series ∞
X
n=1
n2+n+ 1
2n3+ 3 diverges.
Question 11
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We will
consider the ratio Rof consecutive terms:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify further.
R= lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n= lim
n→∞ 1 + 1
n−n
Step 4: Evaluate the limit.
R= lim
n→∞ 1 + 1
n−n
=1
e
Step 5: Determine convergence. Since R < 1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 12
Question
Determine whether the series
∞
X
n=1
n+ cos n
n2+n+ 1
8
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=n+cos n
n2+n+1 .
Step 1: Find a suitable series for comparison. Consider the series
P∞
n=1 1
n. This is a well-known divergent series.
Step 2: Calculate the limit. Let bn=1
n. We will calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n+cos n
n2+n+1
1
n
= lim
n→∞
n2+ncos n
n2+n+ 1
Since both the numerator and denominator have the same leading term, we can
apply L’Hˆopital’s Rule:
= lim
n→∞
2n+ cos n
2n+ 1
= 1
Thus, limn→∞
an
bn= 1.
Step 3: Conclusion Since the limit is a positive finite number, the Limit
Comparison Test is inconclusive. Therefore, we cannot determine the conver-
gence of the series using this test. Further investigation or a different test may
be needed to determine the convergence or divergence of the series.
Question 13
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, we can use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
9
Step 2: Simplify the ratio.
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
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 4: Interpret the result. Since limn→∞
an+1
an=1
e<1, by the ratio
test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnis convergent.
Question 14
Question
Determine whether the series P∞
n=1 n3
2nconverges or diverges.
Solution
To determine the convergence 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. The ratio
of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1 + 3
n+3
n2+1
n3
2
=1
2
Step 3: Analyze the limit. 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.
10
Question 15
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test Let an=n!
nn. Compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/(nn)
Step 2: Simplify the ratio Simplify the ratio (n+ 1)!/(n+ 1)n+1
n!/(nn):
R= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n+ 1
(n+ 1)n
Step 3: Find the limit Take the limit of n+ 1
(n+ 1)n:
R= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 4: Make a conclusion Since R=1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Question 16
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+ 1
n3+ 2.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s consider the series ∞
X
n=1
n2+ 1
n3+ 2 and a new series ∞
X
n=1
1
n.
11
Step 1: Compute the limit:
lim
n→∞
n2+1
n3+2
1
n
= lim
n→∞
n3+n
n3+ 2 = 1.
Step 2: Since the limit is a positive finite value, we can conclude that both
series either converge or diverge together.
Step 3: ∞
X
n=1
1
nis known as the Harmonic Series, which diverges (by the
p-series test with p= 1).
Step 4: Therefore, by the Limit Comparison Test, as the Harmonic Series
diverges and our series has the same behavior as the Harmonic Series, the given
series ∞
X
n=1
n2+ 1
n3+ 2 also diverges.
Question 17
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n3+ 1
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find the limit of the term inside the series:
lim
n→∞
n2+ 1
n3+ 1
Step 2: Divide numerator and denominator by n3to simplify the expression:
lim
n→∞
n2
n3+1
n3
1 + 1
n3
Step 3: Simplify the expression to find the limit:
lim
n→∞
1/n + 0
1+0 = lim
n→∞
1
1= 1
Step 4: Since the limit is a finite positive value, we can apply the Limit
Comparison Test with the series P∞
n=1 1
n, which is a p-series with p= 1.
Step 5: Let’s define an=n2+1
n3+1 and bn=1
nfor the comparison test.
12
Step 6: Now, compute the limit:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 1 ·n
1= 1
Step 7: Since P∞
n=1 1
nis a divergent series, by the Limit Comparison Test,
we can conclude that P∞
n=1 n2+1
n3+1 also diverges.
Question 18
Question
Let P∞
n=1 anbe a convergent series with positive terms. If limn→∞
nan+1
an=
L > 1, determine whether P∞
n=1 anmust also converge.
Solution
Given that limn→∞
nan+1
an=L > 1, we know that for sufficiently large n, the
ratio nan+1
anis greater than 1. This gives us the general idea that the terms of
the series must grow rapidly.
Step 1: Use the Ratio Test with the given ratio expression. Let’s consider
the series P∞
n=1 an. According to the Ratio Test, if limn→∞
an+1
an=L > 1,
then the series diverges. Here, we are given limn→∞
nan+1
an=L > 1. Let
r=an+1
an. Hence, r= limn→∞
an+1
an>1.
Step 2: Understand the implications of r > 1. Since r > 1, we know that
the terms of the series {an}are increasing. This suggests that the terms of the
series do not approach 0 as quickly, which can make it difficult for the series to
converge.
Step 3: Conclude the convergence of the series. Since the terms of {an}are
increasing (meaning an>0 for all n) and limn→∞
nan+1
an=L > 1, we conclude
that the series diverges by the Ratio Test.
Therefore, for the given series to converge, it is necessary that the limit is
less than or equal to 1.
Question 19
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1 √anan+1 also converges.
Solution
To show that the series P∞
n=1 √anan+1 converges, we will use the Comparison
Test.
13
Step 1: Choose appropriate inequality. Since √xis a monotonically
increasing function, we have √anan+1 ≤1
2(an+an+1) for all n∈N.
Step 2: Prove the inequality. Note that (√an−√an+1)2≥0. Expanding
this inequality gives us an+an+1 −2√anan+1 ≥0. Rearranging terms, we find
that √anan+1 ≤1
2(an+an+1).
Step 3: Apply the Comparison Test. Since P∞
n=1 anconverges, there
exists some N∈Nsuch that P∞
n=Nan<1. For all n≥N, we have √anan+1 ≤
1
2(an+an+1). Thus, P∞
n=N√anan+1 ≤1
2P∞
n=N(an+an+1). Simplifying the in-
equality gives P∞
n=N√anan+1 ≤1
2P∞
n=Nan+1
2P∞
n=Nan+1. Since both series
P∞
n=Nanand P∞
n=Nan+1 converge, the series P∞
n=N√anan+1 also converges
by the Comparison Test. Therefore, the series P∞
n=1 √anan+1 converges.
Question 20
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. Then, compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
[(n+ 1)!/(n+ 1)n+1]·[nn/n!]
nn/n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Calculate the limit.
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)n=1
e<1
Since the limit is less than 1, by the Ratio Test, the series ∞
X
n=1
n!
nnconverges.
14
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Recall that for a series Pan, the ratio test states that if
L= lim
n→∞
an+1
an
,
then the series converges if L < 1 and diverges if L > 1.
Step 2: Let’s apply the ratio test to our series. We have
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 3: Simplifying the expression inside the absolute value, we get
L= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
.
Step 4: Simplifying further, we get
L= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
.
Step 5: After canceling out terms, we have
L= lim
n→∞
nn
(n+ 1)n
.
Step 6: Simplifying the limit, we get
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
.
Step 7: Since limn→∞ 1
1+ 1
nn=1
e, where eis the base of the natural
logarithm, we have L=1
e.
Step 8: Since 1
e<1, by the ratio test, the series Pn!
nnconverges.
Therefore, the given series converges.
15
Question 22
Question
Determine whether the series P∞
n=1 n!
nn+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn+1 , we can use the ratio
test.
Step 1: Apply the ratio test. Let’s calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+2
n!/nn+1
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+2
n!/nn+1
= lim
n→∞
(n+ 1)!nn+1
(n+ 1)n+2n!
= lim
n→∞
(n+ 1)nnn!
(n+ 1)n+1n!
= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
1
(1 + 1
n)n+1
=1
e<1
Step 3: Since the limit is less than 1, by the ratio test, the series P∞
n=1 n!
nn+1
converges.
Question 23
Question
Determine if 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: Calculate the ratio Let’s consider the ratio R:
R= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Find the expression for an+1 and anWe have:
an+1 =(n+ 1)2
2n+1 =n2+ 2n+ 1
2·2n
16
and
an=n2
2n.
Step 3: Calculate the limit Now, let’s compute the limit:
R= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1
2+1
n+1
n2
=1
2.
Step 4: Conclusion Since the ratio R= 1/2<1, by the ratio test, the
series P∞
n=1 n2
2nconverges.
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n4+ 3
Solution
To determine the convergence or divergence of the series, we will use the Com-
parison Test.
Step 1: Consider the series P∞
n=1 n2+1
n4+3 .
Step 2: Simplify the terms of the series: n2+1
n4+3 =n2+1
n4·1
1+ 3
n4
=1
n2·1
1+ 3
n4
.
Step 3: Note that for n≥1, we have 3
n4≤3 and therefore 1
1+ 3
n4≥1
4.
Step 4: Now, consider the series P∞
n=1 1
n2, which is a p-series with p= 2.
Step 5: Since 1
n2≤1
n2·1
1+ 3
n4
for all n≥1 and P∞
n=1 1
n2converges, by the
Comparison Test, we conclude that P∞
n=1 n2+1
n4+3 converges.
Question 25
Question
Determine whether the series ∞
X
n=1
n3
2n4+ 3 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n3
2n4+ 3, we will use the limit
comparison test.
17
Step 1: Find the limit Consider the series ∞
X
n=1
n3
2n4+ 3 and let an=
n3
2n4+ 3. We will find the limit of an
1/n as napproaches infinity:
lim
n→∞
an
1
n
= lim
n→∞
n4
2n4+ 3 = lim
n→∞
1
2 + 3
n4
=1
2
Step 2: Conclusion Since the limit is a finite positive number ( 1
2), by
the limit comparison test, the series Pn3
2n4+3 converges if and only if the series
P1
nconverges. And since the harmonic series P1
ndiverges, our original series
Pn3
2n4+3 also diverges.
Question 26
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Apply the ratio test. Consider the ratio of consecutive terms:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
r= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
Step 3: Simplify further and evaluate the limit.
r= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
Step 4: Use the limit definition of e.
r= lim
n→∞
n
(1 + 1/n)n
= lim
n→∞
n
e=1
e
Step 5: Determine the series convergence. If r < 1, then the series con-
verges. Since 1
e<1, the series P∞
n=1 n!
nnconverges by the ratio test.
18
Question 27
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: Apply the Ratio Test. Let an=n!
nn.
We calculate the limit:
L= lim
n→∞
an+1
an
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
L= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
L= lim
n→∞
nn
(n+ 1)n
L= lim
n→∞
1
(1 + 1/n)n
Step 2: Simplify the limit. Applying L’Hopital’s Rule leads to:
L= lim
n→∞
1
e=1
e
Step 3: Analyze the limit value. Since L=1
e<1, by the Ratio Test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 28
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
19
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Let’s consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplifying the expression, we get:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Further simplifying gives:
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 4: Taking the limit, we have:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 5: Since the limit is equal to 1, the ratio test is inconclusive. We need
to use another test to determine the convergence of the series.
Step 6: Let’s apply the root test to the series. Consider the limit:
lim
n→∞
n
s
n!
nn
= lim
n→∞
n
√n!
n
Step 7: By applying Stirling’s approximation, we find that limn→∞
n
√n!
n=1
e.
Step 8: Since 1
e<1, the series P∞
n=1 n!
nnconverges by the root test.
Question 29
Question
Determine whether the series P∞
n=1 nn
n!converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 nn
n!, we will use the ratio test.
Step 1: Compute an: Let an=nn
n!.
Step 2: Apply the ratio test: Consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)n+1/(n+ 1)!
nn/n!
20
Simplify the expression:
= lim
n→∞
(n+ 1) n+ 1
nn
= lim
n→∞(n+ 1) 1 + 1
nn
= lim
n→∞(n+ 1)e
=∞
Step 3: Analyze the limit: Since the limit from the ratio test is greater than
1, the series P∞
n=1 nn
n!diverges.
Therefore, the given series diverges.
Question 30
Question
Determine the convergence or divergence of the series
∞
X
n=1
(−1)n·n
2n+n2.
Solution
To determine the convergence or divergence of the given series, we can use the
alternating series test and compare it to a simpler series.
Step 1: Apply the alternating series test. The series P∞
n=1
(−1)n·n
2n+n2is an
alternating series because each term has a sign that alternates. Let’s verify the
conditions of the alternating series test:
1. The terms an=n
2n+n2are positive for all n≥1. 2. The terms an+1 =
n+1
2n+1+(n+1)2are decreasing for all n≥1. 3. limn→∞ an= 0.
Since all conditions are met, we can conclude that the series converges by
the alternating series test.
Step 2: Compare the series to a simpler series. We can compare the series
P∞
n=1 n
2n+n2to P∞
n=1 n
2nsince 2n+n2>2nfor all n≥1.
Consider the series P∞
n=1 n
2n. This is a convergent series since it is a geo-
metric series with common ratio r=1
2where −1< r < 1.
Therefore, by the comparison test, the series P∞
n=1 n
2n+n2also converges.
21
Question 31
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
Let’s use the ratio test to determine the convergence or divergence of the series.
Step 1: Calculate the limit of the ratio using the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the limit:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
.
Step 3: Further simplify the limit:
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn+1(n+ 1)
(n+ 1)n+1
.
Step 4: Evaluate the limit:
lim
n→∞
nn+1(n+ 1)
(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
.
Step 5: Since limn→∞
n
(1+1/n)ntends to ∞, the series P∞
n=1 n!
nndiverges by
the ratio test.
Question 32
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Determine whether the series P∞
n=1
nan
1+n2converges or diverges.
Solution
We will use the Limit Comparison Test to determine the convergence of the
series P∞
n=1
nan
1+n2.
Step 1: Find the limit of the ratio. Let bn=nan
1+n2. We want to find
limn→∞
bn
an.
22
lim
n→∞
bn
an
= lim
n→∞
nan
1+n2
an
= lim
n→∞
n
1 + n2= 0
Step 2: Apply the Limit Comparison Test. Since limn→∞
bn
an= 0 and
P∞
n=1 anconverges, by the Limit Comparison Test, P∞
n=1 bnconverges as well.
Step 3: Conclusion. Therefore, the series P∞
n=1
nan
1+n2converges.
Question 33
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms using 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
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
Step 3: Take the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 4: Apply the ratio test: Since the limit 1
eis less than 1, by the ratio
test, the series P∞
n=1 n!
nnconverges.
Question 34
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
23
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Let’s denote an=n!
nn.
Step 1: Compute the ratio test. The ratio test states that if limn→∞
an+1
an<
1, then the series converges; if limn→∞
an+1
an>1 or diverges to ∞, then the
series diverges; and if the limit equals 1, the test is inconclusive.
Compute the ratio R= lim
n→∞
an+1
an
.
R= lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 2: Analyze the value of R. Since the limit R=1
e<1, by the ratio
test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 35
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1
√an
nis convergent.
Solution
To prove that the series P∞
n=1
√an
nis convergent, we will make use of the Com-
parison Test.
Step 1: Since P∞
n=1 anis a convergent series with positive terms, we know
that the terms anmust converge to zero as napproaches infinity. This implies
that limn→∞ an= 0.
24
Question 4
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. Calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
Step 2: Simplify and evaluate the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 3: Determine convergence based on the ratio test: - If the limit L < 1,
then the series converges. - If the limit L > 1 or is infinite, then the series
diverges. - If the limit L= 1, the test is inconclusive.
Since L= 1/e < 1, by the ratio test, the series ∞
X
n=1
n!
nnconverges.
Question 5
Question
Determine whether the series P∞
n=1 n2+3n−1
n4+5 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series P∞
n=1 n2+3n−1
n4+5 and a new series P∞
n=1 1
n2.
Step 1: Find the limit of the ratio of the two series. Let an=n2+3n−1
n4+5 and
bn=1
n2. We want to find the limit of the ratio limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n−1
n4+ 5 ·n2
1= lim
n→∞
n4+ 3n3−n2
n4+ 5 = 1
3
Step 2: Apply the Limit Comparison Test. Since the limit is a finite positive
number, then either both series P∞
n=1 anand P∞
n=1 bnconverge or both diverge.
We know that the series P∞
n=1 bn=P∞
n=1 1
n2is a p-series with p= 2, which
converges.
Step 3: Conclusion. By the Limit Comparison Test, since the series of
n2+3n−1
n4+5 and 1
n2have the same convergence behavior, we can conclude that the
series P∞
n=1 n2+3n−1
n4+5 also converges.
Question 6
Question
Consider the series P∞
n=1 n2+2
n3+3 . Determine whether the series converges or
diverges.
Solution
To analyze the convergence of the series P∞
n=1 n2+2
n3+3 , we will use the Limit
Comparison Test. Let’s denote the given series as an=n2+2
n3+3 .
Step 1: Find a comparison series Let’s consider the series bn=1
n, which
is a divergent p-series with p= 1.
Step 2: Calculate the limit We will compute the limit limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
(n2+ 2)/(n3+ 3)
1/n = lim
n→∞
n2+ 2
n2(n+ 3) = lim
n→∞
1+2/n2
n+ 3
Step 3: Evaluate the limit Looking at the expression inside the limit, as
napproaches infinity, the term 2/n2becomes negligible compared to 1/n, so we
focus on the dominant terms:
lim
n→∞
1
n= 0
Step 4: Check the conditions of the Limit Comparison Test Since
limn→∞
an
bn= 0 and P∞
n=1 bndiverges, we can conclude by the Limit Compari-
son Test that P∞
n=1 analso diverges.
Therefore, the series P∞
n=1 n2+2
n3+3 diverges.
Question 7
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
2n3+ 1.
4
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test. Let’s consider the series
∞
X
n=1
n2
2n3+ 1.
and compare it with the series
∞
X
n=1
1
n.
Step 1: Find the limit of the ratio Compute the limit of the ratio of
the two series:
lim
n→∞
n2
2n3+1
1
n
= lim
n→∞
n3
2n3+ 1.
Step 2: Simplify the expression Simplify the above limit:
= lim
n→∞
1
2 + 1
n2
=1
2.
Step 3: Conclusion Since the limit is a positive finite number, by the limit
comparison test, the series
∞
X
n=1
n2
2n3+ 1
converges if the series
∞
X
n=1
1
n
converges. The harmonic series
∞
X
n=1
1
n
is a divergent series.
Conclusion: By the limit comparison test, the series
∞
X
n=1
n2
2n3+ 1
diverges since it can be compared with the harmonic series which is divergent.
Question 8
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
5
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn, then 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!
nn·nn
n!·n+ 1
n+ 1 =n+ 1
n= 1+ 1
n
Step 2: Apply the ratio test. The ratio test states that if limn→∞
an+1
an
<
1, then the series converges; if limn→∞
an+1
an
>1 or it does not exist, then the
series diverges; and if limn→∞
an+1
an
= 1, the ratio test is inconclusive.
Step 3: Find the limit. Taking the limit of the ratio as napproaches infinity:
lim
n→∞
an+1
an
= lim
n→∞
1 + 1
n
= 1
Step 4: Conclusion. Since limn→∞
an+1
an
= 1, the ratio test is inconclusive.
We may need to try other tests to determine the convergence of the series
∞
X
n=1
n!
nn.
Question 9
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,
we will use the ratio test.
Step 1: Compute the nth term of the series. The general term of the series
is given by an=n!
nn.
6
Step 2: Apply the ratio test. Consider the limit of the ratio of the (n+ 1)th
term to the nth term:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 3: Simplify the ratio.
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n
(1 + 1
n)n
Step 4: Evaluate the limit. Take the limit of the last expression:
= lim
n→∞
n
e=∞
Step 5: Conclusion. Since the limit of the ratio is greater than 1, by the
ratio test, the series diverges. Thus, the series
∞
X
n=1
n!
nn
diverges.
Question 10
Question
Determine if the series ∞
X
n=1
n2+n+ 1
2n3+ 3 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+n+ 1
2n3+ 3 and the series ∞
X
n=1
1
n.
Step 1: Find the limit of an
bn: Let an=n2+n+1
2n3+3 and bn=1
n. We will find
the limit of an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+n+ 1
2n3+ 3 ·n
1
= lim
n→∞
n3+n2+n
2n3+ 3 = lim
n→∞
1 + 1
n+1
n2
2 + 3
n3
=1+0+0
2+0 =1
2
7
Step 2: Compare the limit with a known series: Since ∞
X
n=1
1
nis a harmonic
series that diverges, and an
bn=1
2(which is a finite positive value), by the Limit
Comparison Test, the given series ∞
X
n=1
n2+n+ 1
2n3+ 3 also diverges.
Therefore, the series ∞
X
n=1
n2+n+ 1
2n3+ 3 diverges.
Question 11
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We will
consider the ratio Rof consecutive terms:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
R= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
Step 3: Simplify further.
R= lim
n→∞
(n+ 1)nn
(n+ 1)n+1 = lim
n→∞
nn
(n+ 1)n= lim
n→∞ 1 + 1
n−n
Step 4: Evaluate the limit.
R= lim
n→∞ 1 + 1
n−n
=1
e
Step 5: Determine convergence. Since R < 1, by the ratio test, the series
P∞
n=1 n!
nnconverges.
Question 12
Question
Determine whether the series
∞
X
n=1
n+ cos n
n2+n+ 1
8
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let an=n+cos n
n2+n+1 .
Step 1: Find a suitable series for comparison. Consider the series
P∞
n=1 1
n. This is a well-known divergent series.
Step 2: Calculate the limit. Let bn=1
n. We will calculate the limit:
lim
n→∞
an
bn
= lim
n→∞
n+cos n
n2+n+1
1
n
= lim
n→∞
n2+ncos n
n2+n+ 1
Since both the numerator and denominator have the same leading term, we can
apply L’Hˆopital’s Rule:
= lim
n→∞
2n+ cos n
2n+ 1
= 1
Thus, limn→∞
an
bn= 1.
Step 3: Conclusion Since the limit is a positive finite number, the Limit
Comparison Test is inconclusive. Therefore, we cannot determine the conver-
gence of the series using this test. Further investigation or a different test may
be needed to determine the convergence or divergence of the series.
Question 13
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, we can use the
ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
9
Step 2: Simplify the ratio.
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
Step 3: Evaluate the limit.
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 4: Interpret the result. Since limn→∞
an+1
an=1
e<1, by the ratio
test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnis convergent.
Question 14
Question
Determine whether the series P∞
n=1 n3
2nconverges or diverges.
Solution
To determine the convergence 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. The ratio
of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3
= lim
n→∞
1 + 3
n+3
n2+1
n3
2
=1
2
Step 3: Analyze the limit. 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.
10
Question 15
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test Let an=n!
nn. Compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/(nn)
Step 2: Simplify the ratio Simplify the ratio (n+ 1)!/(n+ 1)n+1
n!/(nn):
R= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
n+ 1
(n+ 1)n
Step 3: Find the limit Take the limit of n+ 1
(n+ 1)n:
R= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 4: Make a conclusion Since R=1
e<1, by the ratio test, the series
∞
X
n=1
n!
nnconverges.
Question 16
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2+ 1
n3+ 2.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s consider the series ∞
X
n=1
n2+ 1
n3+ 2 and a new series ∞
X
n=1
1
n.
11
Step 1: Compute the limit:
lim
n→∞
n2+1
n3+2
1
n
= lim
n→∞
n3+n
n3+ 2 = 1.
Step 2: Since the limit is a positive finite value, we can conclude that both
series either converge or diverge together.
Step 3: ∞
X
n=1
1
nis known as the Harmonic Series, which diverges (by the
p-series test with p= 1).
Step 4: Therefore, by the Limit Comparison Test, as the Harmonic Series
diverges and our series has the same behavior as the Harmonic Series, the given
series ∞
X
n=1
n2+ 1
n3+ 2 also diverges.
Question 17
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n3+ 1
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find the limit of the term inside the series:
lim
n→∞
n2+ 1
n3+ 1
Step 2: Divide numerator and denominator by n3to simplify the expression:
lim
n→∞
n2
n3+1
n3
1 + 1
n3
Step 3: Simplify the expression to find the limit:
lim
n→∞
1/n + 0
1+0 = lim
n→∞
1
1= 1
Step 4: Since the limit is a finite positive value, we can apply the Limit
Comparison Test with the series P∞
n=1 1
n, which is a p-series with p= 1.
Step 5: Let’s define an=n2+1
n3+1 and bn=1
nfor the comparison test.
12
Step 6: Now, compute the limit:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 1 ·n
1= 1
Step 7: Since P∞
n=1 1
nis a divergent series, by the Limit Comparison Test,
we can conclude that P∞
n=1 n2+1
n3+1 also diverges.
Question 18
Question
Let P∞
n=1 anbe a convergent series with positive terms. If limn→∞
nan+1
an=
L > 1, determine whether P∞
n=1 anmust also converge.
Solution
Given that limn→∞
nan+1
an=L > 1, we know that for sufficiently large n, the
ratio nan+1
anis greater than 1. This gives us the general idea that the terms of
the series must grow rapidly.
Step 1: Use the Ratio Test with the given ratio expression. Let’s consider
the series P∞
n=1 an. According to the Ratio Test, if limn→∞
an+1
an=L > 1,
then the series diverges. Here, we are given limn→∞
nan+1
an=L > 1. Let
r=an+1
an. Hence, r= limn→∞
an+1
an>1.
Step 2: Understand the implications of r > 1. Since r > 1, we know that
the terms of the series {an}are increasing. This suggests that the terms of the
series do not approach 0 as quickly, which can make it difficult for the series to
converge.
Step 3: Conclude the convergence of the series. Since the terms of {an}are
increasing (meaning an>0 for all n) and limn→∞
nan+1
an=L > 1, we conclude
that the series diverges by the Ratio Test.
Therefore, for the given series to converge, it is necessary that the limit is
less than or equal to 1.
Question 19
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1 √anan+1 also converges.
Solution
To show that the series P∞
n=1 √anan+1 converges, we will use the Comparison
Test.
13
Step 1: Choose appropriate inequality. Since √xis a monotonically
increasing function, we have √anan+1 ≤1
2(an+an+1) for all n∈N.
Step 2: Prove the inequality. Note that (√an−√an+1)2≥0. Expanding
this inequality gives us an+an+1 −2√anan+1 ≥0. Rearranging terms, we find
that √anan+1 ≤1
2(an+an+1).
Step 3: Apply the Comparison Test. Since P∞
n=1 anconverges, there
exists some N∈Nsuch that P∞
n=Nan<1. For all n≥N, we have √anan+1 ≤
1
2(an+an+1). Thus, P∞
n=N√anan+1 ≤1
2P∞
n=N(an+an+1). Simplifying the in-
equality gives P∞
n=N√anan+1 ≤1
2P∞
n=Nan+1
2P∞
n=Nan+1. Since both series
P∞
n=Nanand P∞
n=Nan+1 converge, the series P∞
n=N√anan+1 also converges
by the Comparison Test. Therefore, the series P∞
n=1 √anan+1 converges.
Question 20
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To analyze the convergence of the series ∞
X
n=1
n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn. Then, compute the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
lim
n→∞
[(n+ 1)!/(n+ 1)n+1]·[nn/n!]
nn/n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Calculate the limit.
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)n=1
e<1
Since the limit is less than 1, by the Ratio Test, the series ∞
X
n=1
n!
nnconverges.
14
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Recall that for a series Pan, the ratio test states that if
L= lim
n→∞
an+1
an
,
then the series converges if L < 1 and diverges if L > 1.
Step 2: Let’s apply the ratio test to our series. We have
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 3: Simplifying the expression inside the absolute value, we get
L= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
.
Step 4: Simplifying further, we get
L= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
.
Step 5: After canceling out terms, we have
L= lim
n→∞
nn
(n+ 1)n
.
Step 6: Simplifying the limit, we get
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
.
Step 7: Since limn→∞ 1
1+ 1
nn=1
e, where eis the base of the natural
logarithm, we have L=1
e.
Step 8: Since 1
e<1, by the ratio test, the series Pn!
nnconverges.
Therefore, the given series converges.
15
Question 22
Question
Determine whether the series P∞
n=1 n!
nn+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn+1 , we can use the ratio
test.
Step 1: Apply the ratio test. Let’s calculate the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+2
n!/nn+1
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!/(n+ 1)n+2
n!/nn+1
= lim
n→∞
(n+ 1)!nn+1
(n+ 1)n+2n!
= lim
n→∞
(n+ 1)nnn!
(n+ 1)n+1n!
= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
1
(1 + 1
n)n+1
=1
e<1
Step 3: Since the limit is less than 1, by the ratio test, the series P∞
n=1 n!
nn+1
converges.
Question 23
Question
Determine if 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: Calculate the ratio Let’s consider the ratio R:
R= lim
n→∞
an+1
an
,
where an=n2
2n.
Step 2: Find the expression for an+1 and anWe have:
an+1 =(n+ 1)2
2n+1 =n2+ 2n+ 1
2·2n
16
and
an=n2
2n.
Step 3: Calculate the limit Now, let’s compute the limit:
R= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1
2+1
n+1
n2
=1
2.
Step 4: Conclusion Since the ratio R= 1/2<1, by the ratio test, the
series P∞
n=1 n2
2nconverges.
Question 24
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2+ 1
n4+ 3
Solution
To determine the convergence or divergence of the series, we will use the Com-
parison Test.
Step 1: Consider the series P∞
n=1 n2+1
n4+3 .
Step 2: Simplify the terms of the series: n2+1
n4+3 =n2+1
n4·1
1+ 3
n4
=1
n2·1
1+ 3
n4
.
Step 3: Note that for n≥1, we have 3
n4≤3 and therefore 1
1+ 3
n4≥1
4.
Step 4: Now, consider the series P∞
n=1 1
n2, which is a p-series with p= 2.
Step 5: Since 1
n2≤1
n2·1
1+ 3
n4
for all n≥1 and P∞
n=1 1
n2converges, by the
Comparison Test, we conclude that P∞
n=1 n2+1
n4+3 converges.
Question 25
Question
Determine whether the series ∞
X
n=1
n3
2n4+ 3 converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n3
2n4+ 3, we will use the limit
comparison test.
17
Step 1: Find the limit Consider the series ∞
X
n=1
n3
2n4+ 3 and let an=
n3
2n4+ 3. We will find the limit of an
1/n as napproaches infinity:
lim
n→∞
an
1
n
= lim
n→∞
n4
2n4+ 3 = lim
n→∞
1
2 + 3
n4
=1
2
Step 2: Conclusion Since the limit is a finite positive number ( 1
2), by
the limit comparison test, the series Pn3
2n4+3 converges if and only if the series
P1
nconverges. And since the harmonic series P1
ndiverges, our original series
Pn3
2n4+3 also diverges.
Question 26
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nn.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Apply the ratio test. Consider the ratio of consecutive terms:
r= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
r= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
Step 3: Simplify further and evaluate the limit.
r= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
Step 4: Use the limit definition of e.
r= lim
n→∞
n
(1 + 1/n)n
= lim
n→∞
n
e=1
e
Step 5: Determine the series convergence. If r < 1, then the series con-
verges. Since 1
e<1, the series P∞
n=1 n!
nnconverges by the ratio test.
18
Question 27
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: Apply the Ratio Test. Let an=n!
nn.
We calculate the limit:
L= lim
n→∞
an+1
an
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
L= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
L= lim
n→∞
nn
(n+ 1)n
L= lim
n→∞
1
(1 + 1/n)n
Step 2: Simplify the limit. Applying L’Hopital’s Rule leads to:
L= lim
n→∞
1
e=1
e
Step 3: Analyze the limit value. Since L=1
e<1, by the Ratio Test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 28
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
19
Solution
Step 1: We will use the ratio test to determine the convergence of the series.
Let’s consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplifying the expression, we get:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
Step 3: Further simplifying gives:
lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 4: Taking the limit, we have:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 5: Since the limit is equal to 1, the ratio test is inconclusive. We need
to use another test to determine the convergence of the series.
Step 6: Let’s apply the root test to the series. Consider the limit:
lim
n→∞
n
s
n!
nn
= lim
n→∞
n
√n!
n
Step 7: By applying Stirling’s approximation, we find that limn→∞
n
√n!
n=1
e.
Step 8: Since 1
e<1, the series P∞
n=1 n!
nnconverges by the root test.
Question 29
Question
Determine whether the series P∞
n=1 nn
n!converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 nn
n!, we will use the ratio test.
Step 1: Compute an: Let an=nn
n!.
Step 2: Apply the ratio test: Consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)n+1/(n+ 1)!
nn/n!
20
Simplify the expression:
= lim
n→∞
(n+ 1) n+ 1
nn
= lim
n→∞(n+ 1) 1 + 1
nn
= lim
n→∞(n+ 1)e
=∞
Step 3: Analyze the limit: Since the limit from the ratio test is greater than
1, the series P∞
n=1 nn
n!diverges.
Therefore, the given series diverges.
Question 30
Question
Determine the convergence or divergence of the series
∞
X
n=1
(−1)n·n
2n+n2.
Solution
To determine the convergence or divergence of the given series, we can use the
alternating series test and compare it to a simpler series.
Step 1: Apply the alternating series test. The series P∞
n=1
(−1)n·n
2n+n2is an
alternating series because each term has a sign that alternates. Let’s verify the
conditions of the alternating series test:
1. The terms an=n
2n+n2are positive for all n≥1. 2. The terms an+1 =
n+1
2n+1+(n+1)2are decreasing for all n≥1. 3. limn→∞ an= 0.
Since all conditions are met, we can conclude that the series converges by
the alternating series test.
Step 2: Compare the series to a simpler series. We can compare the series
P∞
n=1 n
2n+n2to P∞
n=1 n
2nsince 2n+n2>2nfor all n≥1.
Consider the series P∞
n=1 n
2n. This is a convergent series since it is a geo-
metric series with common ratio r=1
2where −1< r < 1.
Therefore, by the comparison test, the series P∞
n=1 n
2n+n2also converges.
21
Question 31
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
Let’s use the ratio test to determine the convergence or divergence of the series.
Step 1: Calculate the limit of the ratio using the ratio test:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the limit:
lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
.
Step 3: Further simplify the limit:
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
nn+1(n+ 1)
(n+ 1)n+1
.
Step 4: Evaluate the limit:
lim
n→∞
nn+1(n+ 1)
(n+ 1)n+1
= lim
n→∞
nn+1
(n+ 1)n
= lim
n→∞
n
(1 + 1/n)n
.
Step 5: Since limn→∞
n
(1+1/n)ntends to ∞, the series P∞
n=1 n!
nndiverges by
the ratio test.
Question 32
Question
Let {an}be a sequence of positive real numbers such that P∞
n=1 anconverges.
Determine whether the series P∞
n=1
nan
1+n2converges or diverges.
Solution
We will use the Limit Comparison Test to determine the convergence of the
series P∞
n=1
nan
1+n2.
Step 1: Find the limit of the ratio. Let bn=nan
1+n2. We want to find
limn→∞
bn
an.
22
lim
n→∞
bn
an
= lim
n→∞
nan
1+n2
an
= lim
n→∞
n
1 + n2= 0
Step 2: Apply the Limit Comparison Test. Since limn→∞
bn
an= 0 and
P∞
n=1 anconverges, by the Limit Comparison Test, P∞
n=1 bnconverges as well.
Step 3: Conclusion. Therefore, the series P∞
n=1
nan
1+n2converges.
Question 33
Question
Determine whether the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence or divergence of the series P∞
n=1 n!
nn, we will use
the ratio test.
Step 1: Compute the ratio of consecutive terms using 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
Step 2: Simplify the expression:
lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
Step 3: Take the limit:
lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1 + 1
nn
=1
e
Step 4: Apply the ratio test: Since the limit 1
eis less than 1, by the ratio
test, the series P∞
n=1 n!
nnconverges.
Question 34
Question
Determine whether the series ∞
X
n=1
n!
nnconverges or diverges.
23
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Let’s denote an=n!
nn.
Step 1: Compute the ratio test. The ratio test states that if limn→∞
an+1
an<
1, then the series converges; if limn→∞
an+1
an>1 or diverges to ∞, then the
series diverges; and if the limit equals 1, the test is inconclusive.
Compute the ratio R= lim
n→∞
an+1
an
.
R= lim
n→∞
(n+1)!
(n+1)n+1
n!
nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 2: Analyze the value of R. Since the limit R=1
e<1, by the ratio
test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 35
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1
√an
nis convergent.
Solution
To prove that the series P∞
n=1
√an
nis convergent, we will make use of the Com-
parison Test.
Step 1: Since P∞
n=1 anis a convergent series with positive terms, we know
that the terms anmust converge to zero as napproaches infinity. This implies
that limn→∞ an= 0.
24
Step 2: Consider the sequence √an
n. We will show that it is bounded
above by a convergent series.
Since limn→∞ an= 0, there exists an N∈Nsuch that for all n>N,an<1
(say). Therefore, √an<1 for all n>N.
Step 3: Now, we compare the series P∞
n=N+1
√an
nwith the convergent series
P∞
n=1 1
n2.
For all n>N,
0<√an
n<1
n2.
Step 4: Therefore, by the Comparison Test, since P∞
n=1 1
n2is a convergent
series (p-series with p= 2 >1), the series P∞
n=1
√an
nis also convergent.
Thus, the series P∞
n=1
√an
nconverges.
25
