MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 3
Liberty University
Question 1
Question
Let (an) be a sequence defined by an=n2
2n. Determine the convergence or
divergence of the series P∞
n=1 an.
Solution
To determine the convergence or divergence of the series P∞
n=1 an, we will use
the Ratio Test.
Step 1: Compute the limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1
2+1
n+1
2n2
=1
2
Step 2: Determine the convergence of the series by the Ratio Test. Since
limn→∞
an+1
an
=1
2<1, the series P∞
n=1 anconverges by the Ratio Test.
Therefore, the series P∞
n=1
n2
2nconverges.
Question 2
Question
Consider the series P∞
n=1
(−1)n
npwhere p > 0. Determine for which values of p
the series converges absolutely, converges conditionally, or diverges.
Solution
To analyze the convergence of the series P∞
n=1
(−1)n
np, we will apply the alter-
nating series test and the p-series test.
Step 1: Absolute Convergence To determine the values of pfor which
the series converges absolutely, we consider the series P∞
n=1
(−1)n
np
=P∞
n=1 1
np.
This series is a p-series, which converges if p > 1 and diverges if p≤1.
Therefore, the series P∞
n=1
(−1)n
npconverges absolutely for p > 1.
Step 2: Conditional Convergence To determine the values of pfor which
the series converges conditionally, we consider the series P∞
n=1 1
np.
This series converges by the p-series test if p > 1.
For 0 < p ≤1, we can consider the alternating series P∞
n=1
(−1)n
np.
This series satisfies the conditions of the alternating series test since 1
npis
monotonically decreasing and converges to 0.
Therefore, the series P∞
n=1
(−1)n
npconverges conditionally for 0 < p ≤1.
Step 3: Divergence The series P∞
n=1
(−1)n
npdiverges for p≤0 as the terms
do not approach 0 as napproaches infinity.
In conclusion: - The series converges absolutely for p > 1. - The series
converges conditionally for 0 < p ≤1. - The series diverges for p≤0.
Question 3
Question
Determine the convergence or divergence of the series: P∞
n=1
n3
2n.
Solution
To determine the convergence of the series P∞
n=1
n3
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n3
2n. We compute the ratio Ras
follows:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3·2n
2n+1 ·n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞ (n+ 1)3
2n3= lim
n→∞ n3+ 3n2+ 3n+ 1
2n3=1
2
Step 2: Evaluate the limit of the ratio. Since the limit of the ratio R=1
2
is less than 1, by the ratio test, the series converges.
Therefore, the series P∞
n=1
n3
2nconverges.
2
Question 4
Question
Determine whether the series P∞
n=1
n2+3n+2
4n3+5n+1 converges or diverges.
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s compare the given series with a known convergent series.
Step 1: Find a known convergent series Consider the series P∞
n=1 1
n,
which is a known convergent series.
Step 2: Find the limit and compare Let an=n2+3n+2
4n3+5n+1 and bn=1
n.
We will compare the two series by finding the limit of the ratio an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 2
4n3+ 5n+ 1 ·n
1= lim
n→∞
n3+ 3n2+ 2n
4n3+ 5n+ 1 =1
4
Step 3: Conclusion Since limn→∞
an
bn=1
4>0, and the series P∞
n=1 1
n
converges (by the p-series test), by the limit comparison test, the given series
P∞
n=1
n2+3n+2
4n3+5n+1 also converges.
Question 5
Question
Determine whether the series P∞
n=1
n2+3n+1
2n4+n3+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s choose a series Pbnthat is known to converge/diverge and compare
it to the given series.
Step 1: Choose a test series. Let’s consider the series P1
n2which is a
p-series with p= 2 and is known to converge.
Step 2: Compute the limit. Let an=n2+3n+1
2n4+n3+1 and bn=1
n2. We need to
find
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
2n4+n3+ 1 ·n2
1
Step 3: Simplify the limit.
lim
n→∞
n2+ 3n+ 1
2n4+n3+ 1 ·n2
1= lim
n→∞
n4+ 3n3+n2
2n4+n3+ 1 = lim
n→∞
1 + 3
n+1
n2
2 + 1
n+1
n2
=1
2
Step 4: Determine the convergence. Since the limit is a positive finite num-
ber, by the Limit Comparison Test, the given series P∞
n=1
n2+3n+1
2n4+n3+1 converges
3
or diverges in the same way as P1
n2. Since P1
n2converges, the given series
also converges.
Question 6
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 can use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
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
Step 3: Simplify the limit.
lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 4: Apply the ratio test. Since the limit is 0, the series converges
absolutely by the ratio test.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 7
Question
Determine whether the series P∞
n=1
n2+2n
n4+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+2n
n4+3 , we can use the limit
comparison test by comparing it to a known convergent or divergent series.
Step 1: Determine a comparison series
Let an=n2+2n
n4+3 . As napproaches infinity, the highest order terms dominate,
so we can simplify anto n2
n4=1
n2.
Thus, we will compare our series to the series P∞
n=1 1
n2.
4
Step 2: Use the limit comparison test
We will compute the limit:
L= lim
n→∞
n2+2n
n4+3
1
n2
= lim
n→∞
n4+ 2n3
n4+ 3 = lim
n→∞ 1 + 2n3−3
n4+ 3
Since the numerator has the same degree as the denominator, we can divide
each term by n4:
L= lim
n→∞ 1 + 2/n −3/n4
1+3/n4= 1
Step 3: Draw a conclusion
Since Lis a finite positive number, by the limit comparison test, P∞
n=1
n2+2n
n4+3
converges if and only if P∞
n=1 1
n2converges.
Since P∞
n=1 1
n2is a convergent p-series with p= 2 >1, then by the compar-
ison test, P∞
n=1
n2+2n
n4+3 also converges.
Question 8
Question
Determine whether the series P∞
n=1 2n+n
3n+n2converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series bn=1
3n.
Step 1: Find the limit of anbn:
lim
n→∞
2n+n
3n+n2·3n
1= lim
n→∞
2n+n
3n+n2·3n
Now, simplify the expression inside the limit:
= lim
n→∞
1 + n
2n
1 + n2
3n·3n
We know that limn→∞
n
an= 0 for any constant a > 1, so limn→∞
n
2n= 0 and
limn→∞
n2
3n= 0.
Therefore, the limit simplifies to:
= lim
n→∞
1+0
1+0· ∞ =∞
Step 2: Conclude based on the limit: Since limn→∞
2n+n
3n+n2·3n
1=∞,
and P∞
n=1 1
3nis a convergent geometric series, by the limit comparison test, the
given series P∞
n=1 2n+n
3n+n2also converges.
5
Question 9
Question
Determine whether the series ∞
X
n=1
3n2+ 1
5n3+ 2 converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series ∞
X
n=1
3n2+1
5n3+2
1
n
.
Step 1: Find the limit
Compute the limit of the ratio as napproaches infinity:
lim
n→∞
3n2+1
5n3+2
1
n
= lim
n→∞
(3n2+ 1)n
5n3+ 2 = lim
n→∞
3n3+n
5n3+ 2
Step 2: Simplify the limit
Divide all terms by the highest power of nin the denominator:
lim
n→∞
3n3+n
5n3+ 2 = lim
n→∞
n3(3 + 1
n2)
n3(5 + 2
n3)= lim
n→∞
3 + 1
n2
5 + 2
n3
Step 3: Evaluate the limit
As napproaches infinity, the terms with 1
n2and 1
n3will approach zero, so
the limit simplifies to:
lim
n→∞
3
5=3
5
Step 4: Determine convergence
Since the limit is a finite positive value, by the Limit Comparison Test, the
series ∞
X
n=1
3n2+ 1
5n3+ 2 has the same convergence behavior as P1
n.
Since ∞
X
n=1
1
ndiverges (harmonic series), the given series ∞
X
n=1
3n2+ 1
5n3+ 2 also
diverges.
Question 10
Question
Determine the convergence or divergence of the series P∞
n=1
n2+3n
n3+5 .
6
Solution
To analyze the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will find a series which is easier to work with and compare it
to the given series.
Consider the series P∞
n=1
n2
n3=P∞
n=1 1
n, which is a p-series with p= 1.
Step 2: Find the limit of the ratio of the terms.
Let an=n2+3n
n3+5 and bn=1
n. We will find L= limn→∞
an
bn.
L= limn→∞
n2+3n
n3+5
1
n
= limn→∞
n3+3n2
n3+5 = 1.
Step 3: Apply the limit comparison test.
Since L= 1 and P∞
n=1 1
ndiverges, by the limit comparison test, the given
series P∞
n=1
n2+3n
n3+5 also diverges.
Question 11
Question
Determine if the series ∞
X
n=1
n3+ 2
3n4−5converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Let’s consider the series ∞
X
n=1
n3+ 2
3n4−5and the series ∞
X
n=1
1
n.
We will calculate the following limit:
lim
n→∞
n3+2
3n4−5
1
n
Step 2: Simplifying the limit:
lim
n→∞
n3+2
3n4−5
1
n
= lim
n→∞
n4+ 2n
3n4−5= lim
n→∞
1 + 2
n3
3−5
n4
=1+0
3−0=1
3
Step 3: Since the limit is a positive finite number, by the limit comparison
test, ∞
X
n=1
n3+ 2
3n4−5converges if and only if ∞
X
n=1
1
nconverges.
Step 4: The series ∞
X
n=1
1
nis known as the harmonic series, which diverges.
7
Step 5: Therefore, by the limit comparison test, the series ∞
X
n=1
n3+ 2
3n4−5also
diverges.
Question 12
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series, we will apply the ratio test.
Step 1: Compute 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)!nn
n!(n+ 1)n+1 =n+ 1
(1 + 1
n)n+1
Step 2: Apply the limit in the ratio test. Taking the limit of the ratio as n
approaches infinity, we have
lim
n→∞
an+1
an
= lim
n→∞
n+ 1
(1 + 1
n)n+1
We can rewrite the limit as
lim
n→∞
n+ 1
1 + 1
nn·(1 + 1
n)
Step 3: Evaluate the limit using the fact that limn→∞(1 + 1
n)n=e. This
simplifies the expression to 1
e>0
Step 4: Determine the convergence of the series. Since the limit of the ratio
is a positive finite number, by the ratio test, the series P∞
n=1
n!
nnconverges.
Question 13
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2−1
n3−2.
8
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series ∞
X
n=1
bnsuch that lim
n→∞
an
bn
exists and is a positive
finite number.
Let an=n2−1
n3−2and bn=1
n.
Step 2: Find lim
n→∞
an
bn
.
lim
n→∞
an
bn
= lim
n→∞ n(n2−1)(n3−2) = 1
Step 3: Since lim
n→∞
an
bn
is a positive finite number, by the Limit Comparison
Test, the series ∞
X
n=1
anand ∞
X
n=1
bneither both converge or both diverge.
Step 4: Since ∞
X
n=1
bn=∞
X
n=1
1
nis a harmonic series which diverges, the given
series ∞
X
n=1
n2−1
n3−2also diverges.
Therefore, the series ∞
X
n=1
n2−1
n3−2diverges.
Question 14
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. Then, the
9
ratio of consecutive terms is
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n
n+ 1
= lim
n→∞ 1 + 1
nn
=e.
Step 2: Apply the ratio test: Since the limit is e > 1, the series diverges by
the ratio test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 15
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: Compute the limit of the ratio test:
L= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Determine an+1:
an+1 =(n+ 1)!
(n+ 1)n+1 =(n+ 1)n!
(n+ 1)(n+ 1)n=n!
(n+ 1)n
10
Step 3: Calculate the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
n!/(n+ 1)n
n!/nn
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e(since lim
n→∞ 1 + 1
nn
=e)
Step 4: Analyze L: Since L=1
e<1, by the ratio test, the series P∞
n=1
n!
nn
converges.
Question 16
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n2+ 1
2n3−1.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s denote the general term of the given series as an:
an=3n2+ 1
2n3−1.
Step 1: Find a suitable series to compare it to. We will look for a series
of the form Pbnwhere bnis simpler to work with and easier to determine the
convergence of.
Let’s consider the series P1
n. This series is a p-series with p= 1, which we
know diverges. We will compare our series to this one.
Step 2: Compute the limit:
lim
n→∞
an
bn
= lim
n→∞
3n2+1
2n3−1
1
n
= lim
n→∞
3n3+n
2n3−1= lim
n→∞
3 + 1
n2
2−1
n3
=3
2.
Step 3: Interpret the limit. Since the limit is a finite positive number, and
P1
nis a divergent series, then by the Limit Comparison Test, we conclude that
the series P∞
n=1 3n2+1
2n3−1also diverges.
11
Question 17
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. Compute
the ratio ras follows:
r= lim
n→∞
an+1
an
Step 2: Calculate the ratio r.
r= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1 ·nn
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1n
= lim
n→∞
1
1 + 1
nn
= lim
n→∞
1
(1 + 1/n)n
=1
e
where eis the base of the natural logarithm.
Step 3: Apply the ratio test. If r < 1, the series converges. If r > 1 or
r= 1, the series diverges. Here, since r=1
e<1, the series P∞
n=1
n!
nnconverges
by the ratio test.
Question 18
Question
Determine whether the series P∞
n=1
n2+5
n3+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+5
n3+3 , we can use the Limit
Comparison Test.
Step 1: Find a suitable series to compare. Let’s consider the series P∞
n=1 1
n.
This is a p-series with p= 1, which we know diverges.
Step 2: Calculate the limit of the ratio. We will find the limit:
lim
n→∞
(n2+ 5)/(n3+ 3)
1/n = lim
n→∞
n2+ 5
n3+ 3 ·n
1= lim
n→∞
n3+ 5n
n3+ 3 = 1
Step 3: Conclusion. Since the limit is a positive finite number and the
series P∞
n=1 1
ndiverges, by the Limit Comparison Test, the series P∞
n=1
n2+5
n3+3
also diverges.
12
Question 19
Question
Determine the convergence or divergence of the series P∞
n=1
n
2n+1 using the
Limit Comparison Test.
Solution
To determine the convergence or divergence of the series P∞
n=1
n
2n+1 , we will
use the Limit Comparison Test. We will compare this series with the convergent
series P∞
n=1 1
n.
Step 1: Find the limit to compare the series. Let an=n
2n+1 and
bn=1
n. We consider the limit:
lim
n→∞
an
bn
= lim
n→∞
n
2n+1
1
n
= lim
n→∞
n2
2n2+n= lim
n→∞
1
2 + 1
n
=1
2
Step 2: Make a conclusion based on the limit. Since the limit is a
finite positive value (1
2), by the Limit Comparison Test, the series P∞
n=1
n
2n+1
has the same convergence behavior as the series P∞
n=1 1
n, which is a divergent
p-series with p= 1. Therefore, the series P∞
n=1
n
2n+1 also diverges.
Question 20
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
.
13
Step 2: Simplify the limit.
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
1 + 1
nn=1
e.
Step 3: Determine the convergence or divergence of the series. Since L=
1
e<1, by the ratio test, the series converges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Calculate the limit:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Determine the convergence or divergence based on the limit: Since
the limit is less than 1, by the ratio test, the series
∞
X
n=1
n!
nn
converges.
14
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1
n2
en.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms. Let an=n2
en. The ratio of
consecutive terms is given by
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
en+1 ·en
n2
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)2
en+1 ·en
n2
= lim
n→∞
(n+ 1)2
n2·e= lim
n→∞ 1
e+2
n+1
n2=1
e
Step 3: Apply the ratio test. Since the limit of the ratio is 1
e<1, by the
ratio test, the series P∞
n=1
n2
enconverges.
Therefore, the series P∞
n=1
n2
enconverges.
Question 23
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Compute the limit of the ratio test. Let an=n!
nn. We will
consider the limit
L= lim
n→∞
an+1
an
.
15
Step 2: Simplify the expression in the limit.
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
n+ 1 ·nn
(n+ 1)n
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1 + 1
n−n
.
Step 3: Compute the limit.
L= lim
n→∞
1 + 1
n−n
= lim
n→∞
1
1 + 1
nn
=1
e.
Step 4: Determine the convergence of the series. Since L=1
e<1, by the ratio
test, the series P∞
n=1
n!
nnconverges. Therefore, the series converges.
Question 24
Question
Determine whether the series P∞
n=1
n2+1
n4+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n4+1 , we can use the Limit
Comparison Test. Let’s consider the series an=n2+1
n4+1 .
Step 1: Find the limit of an
1
n2
as n→ ∞.
lim
n→∞
an
1
n2
= lim
n→∞
n2+ 1
n4+ 1 ·n2= lim
n→∞
n2+ 1
n2(n2+1
n2)= lim
n→∞
1 + 1
n2
1 + 1
n2
= 1.
Step 2: Conclusion based on the Limit Comparison Test Since
limn→∞
an
1
n2
= 1 >0, and P∞
n=1 1
n2is a convergent p-series with p= 2 >1, by
the Limit Comparison Test, we conclude that the series P∞
n=1
n2+1
n4+1 converges.
16
Question 25
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is given by:
R= lim
n→∞
an+1
an
Step 2: Find an+1 and an. We have:
an+1 =(n+ 1)!
(n+ 1)n+1 =(n+ 1)!
(n+ 1)n(n+ 1) =(n+ 1)n!
(n+ 1)n=n!
(n+ 1)n
Therefore,
R= lim
n→∞
n!/(n+ 1)n
n!/nn
Step 3: Simplify the expression.
R= lim
n→∞
nn
(n+ 1)n
Step 4: Calculate the limit.
R= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
where eis the base of the natural logarithm.
Step 5: Analyze the result. Since R < 1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
17
Solution
We will use the ratio test to determine the convergence of the series.
Step 1: Consider the ratio Rn:
Rn=an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
n!·nn
(n+ 1)n+1 .
Step 2: Simplify Rn:
Rn= (n+ 1) n
n+ 1n
= (n+ 1) 1
1 + 1
nn
.
Step 3: Consider the limit of Rnas napproaches infinity:
lim
n→∞ Rn= lim
n→∞(n+ 1) 1
1 + 1
nn
= lim
n→∞(n+ 1) n
n+ 1n
= lim
n→∞
n+ 1
1 + 1
nn.
Step 4: Use the limit definition of e:
lim
n→∞
n+ 1
1 + 1
nn=e
e= 1.
Step 5: Analyze the limit of Rn: Since the limit of Rnis equal to 1, the
ratio test is inconclusive. Therefore, we cannot determine the convergence or
divergence of the series using the ratio test.
Step 6: Conclusion: The ratio test is inconclusive for the given series, so
we cannot determine its convergence or divergence using this test.
Question 27
Question
Determine the convergence or divergence of the series P∞
n=1
n
2n.
Solution
To determine the convergence of the series P∞
n=1
n
2n, we will use the ratio test.
Step 1: Calculate the ratio Compute the ratio of consecutive terms:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2n
n2n+1
Step 2: Simplify the ratio Simplify the ratio:
R= lim
n→∞
n+ 1
2n
= lim
n→∞
1
21 + 1
n=1
2
Step 3: Apply the ratio test Since R=1
2<1, the series P∞
n=1
n
2n
converges by the ratio test.
18
Question 28
Question
Determine whether the series P∞
n=1 3n2+5n+1
5n3−n+4 converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test.
Step 1: Consider the series P∞
n=1 3n2+5n+1
5n3−n+4 and another series P∞
n=1 bn
such that bn=1
n.
Step 2: Calculate the limit:
lim
n→∞
(3n2+ 5n+ 1)/5n3−n+ 4
1/n
= lim
n→∞
(3n2+ 5n+ 1)/n
5n3−n+ 4
= lim
n→∞
3/n + 5/n2+ 1/n3
5−1/n2+ 4/n3
= lim
n→∞
0+0+0
5
= 0
Step 3: Since the limit above is a finite positive value, we can use the Limit
Comparison Test. Therefore, the convergence of the series P∞
n=1 3n2+5n+1
5n3−n+4 is
the same as the convergence of the series P∞
n=1 1
n.
Step 4: The series P∞
n=1 1
nis the harmonic series which diverges.
Thus, by the Limit Comparison Test, the series P∞
n=1 3n2+5n+1
5n3−n+4 also diverges.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1
n!
nn.
Solution
To analyze the convergence or divergence of the series P∞
n=1
n!
nn, we will use the
Ratio Test.
Step 1: Compute the ratio R:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
19
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(n+ 1)n
Step 2: Simplify the expression and evaluate the limit:
lim
n→∞
n+ 1
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 3: Determine the convergence or divergence of the series: If R < 1, the
series converges. If R > 1, the series diverges. If R= 1, the test is inconclusive.
Since 1
e<1, by the Ratio Test, the series P∞
n=1
n!
nnconverges.
Question 30
Question
Prove or disprove the convergence of the series:
∞
X
n=1
n2+ 1
n3+ 2n+ 1.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
an=∞
X
n=1
n2+ 1
n3+ 2n+ 1
and the series ∞
X
n=1
bn=∞
X
n=1
1
n.
Step 1: Find the limit of the ratio: Let’s find the limit of the ratio an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+1
n3+2n+1
1
n
.
Step 2: Simplify the expression:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 2n+ 1 ·n= lim
n→∞
n3+n
n3+ 2n+ 1.
Step 3: Divide by the highest power of n:
lim
n→∞
an
bn
= lim
n→∞
1 + 1
n2
1 + 2
n+1
n3
=1+0
1+0+0 = 1.
Step 4: Apply the Limit Comparison Test: Since the limit is a positive
finite number, and P∞
n=1 bn=P∞
n=1 1
nis a harmonic series which diverges, we
conclude by the Limit Comparison Test that P∞
n=1 analso diverges.
Therefore, the series P∞
n=1
n2+1
n3+2n+1 diverges.
20
Question 31
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: Compute the limit of the ratio. Let an=n3
2n. We will consider the
limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3=1
2
Step 2: Analyze the limit obtained. Since the limit of the ratio is less than
1, by the Ratio Test, the series P∞
n=1
n3
2nconverges.
Therefore, the series P∞
n=1
n3
2nconverges.
Question 32
Question
Determine whether the series P∞
n=1
√n+1−√n
n3
2
converges or diverges.
Solution
To determine whether the given series converges or diverges, we will analyze the
convergence of the general term using the limit comparison test.
Step 1: Find the general term anof the series.
The general term anof the series is √n+1−√n
n3
2
.
Step 2: Simplify the general term an.
We can simplify the general term anas follows:
an=√n+ 1 −√n
n3
2
=√n+ 1 −√n
n3
2·√n+1+√n
√n+1+√n
an=(√n+ 1)2−(√n)2
n3
2(√n+1+√n)=n+ 1 −n
n3
2(√n+1+√n)=1
n3
2(√n+1+√n)
21
Step 3: Find the limit of the general term as napproaches infinity.
Taking the limit as napproaches infinity, we get:
lim
n→∞ an= lim
n→∞
1
n3
2(√n+1+√n)= lim
n→∞
1
n3
2(2√n)= lim
n→∞
1
2n2= 0
Step 4: Apply the limit comparison test.
Since limn→∞ an= 0 and the series P∞
n=1 1
n2converges (by the p-series test
with p= 2 >1), we can conclude by the limit comparison test that the series
P∞
n=1
√n+1−√n
n3
2
converges.
Question 33
Question
Determine the convergence or divergence of the series P∞
n=1
n3+2
n4+n+1 .
Solution
To determine the convergence or divergence of the series, we can use the limit
comparison test. Let’s compare the given series with a simpler series whose
convergence is known.
Step 1: Find the simpler series
Consider the series P∞
n=1 1
n, which is a p-series with p= 1 and known to be
divergent.
Step 2: Compute the limit
We will compute the limit of the ratio of the two series:
L= lim
n→∞
n3+2
n4+n+1
1
n
Step 3: Simplify the expression
Simplify the expression within the limit by dividing both the numerator and
denominator by n4:
L= lim
n→∞
1+ 2
n3
1+ 1
n3+1
n4
1
n
L= lim
n→∞
1
n4+2
n3
1
n+1
n2+1
n3
L= lim
n→∞
0+0
0+0+0 = 0
Step 4: Conclusion
Since L= 0 and the series P∞
n=1 1
nis divergent, by the limit comparison test,
the given series P∞
n=1
n3+2
n4+n+1 also diverges.
22
Question 34
Question
Determine the convergence of the series ∞
X
n=1
n!
nn.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
L= lim
n→∞
an+1
an
Step 2: Calculate an+1 and an+1
an
.
an+1 =(n+ 1)!
(n+ 1)n+1
=(n+ 1)n!
(n+ 1)n·(n+ 1)
=n!
(n+ 1)n
an+1
an
=
n!
(n+1)n
n!
nn
=nn
(n+ 1)n
=n
n+ 1n
Step 3: Evaluate the limit L.
L= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Determine the convergence of the series. Since L=1
e<1, by the
ratio test, the series ∞
X
n=1
n!
nnconverges.
23
Question 35
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: Apply the Ratio Test. Let an=n!
nn. We consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + 1
n−n
=1
e
Step 3: Analyze the limit. Since 1
e<1, by the Ratio Test, the series ∞
X
n=1
n!
nn
converges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
24
Solution
To analyze the convergence of the series P∞
n=1
(−1)n
np, we will apply the alter-
nating series test and the p-series test.
Step 1: Absolute Convergence To determine the values of pfor which
the series converges absolutely, we consider the series P∞
n=1
(−1)n
np
=P∞
n=1 1
np.
This series is a p-series, which converges if p > 1 and diverges if p≤1.
Therefore, the series P∞
n=1
(−1)n
npconverges absolutely for p > 1.
Step 2: Conditional Convergence To determine the values of pfor which
the series converges conditionally, we consider the series P∞
n=1 1
np.
This series converges by the p-series test if p > 1.
For 0 < p ≤1, we can consider the alternating series P∞
n=1
(−1)n
np.
This series satisfies the conditions of the alternating series test since 1
npis
monotonically decreasing and converges to 0.
Therefore, the series P∞
n=1
(−1)n
npconverges conditionally for 0 < p ≤1.
Step 3: Divergence The series P∞
n=1
(−1)n
npdiverges for p≤0 as the terms
do not approach 0 as napproaches infinity.
In conclusion: - The series converges absolutely for p > 1. - The series
converges conditionally for 0 < p ≤1. - The series diverges for p≤0.
Question 3
Question
Determine the convergence or divergence of the series: P∞
n=1
n3
2n.
Solution
To determine the convergence of the series P∞
n=1
n3
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n3
2n. We compute the ratio Ras
follows:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3·2n
2n+1 ·n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞ (n+ 1)3
2n3= lim
n→∞ n3+ 3n2+ 3n+ 1
2n3=1
2
Step 2: Evaluate the limit of the ratio. Since the limit of the ratio R=1
2
is less than 1, by the ratio test, the series converges.
Therefore, the series P∞
n=1
n3
2nconverges.
2
Question 4
Question
Determine whether the series P∞
n=1
n2+3n+2
4n3+5n+1 converges or diverges.
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s compare the given series with a known convergent series.
Step 1: Find a known convergent series Consider the series P∞
n=1 1
n,
which is a known convergent series.
Step 2: Find the limit and compare Let an=n2+3n+2
4n3+5n+1 and bn=1
n.
We will compare the two series by finding the limit of the ratio an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 2
4n3+ 5n+ 1 ·n
1= lim
n→∞
n3+ 3n2+ 2n
4n3+ 5n+ 1 =1
4
Step 3: Conclusion Since limn→∞
an
bn=1
4>0, and the series P∞
n=1 1
n
converges (by the p-series test), by the limit comparison test, the given series
P∞
n=1
n2+3n+2
4n3+5n+1 also converges.
Question 5
Question
Determine whether the series P∞
n=1
n2+3n+1
2n4+n3+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s choose a series Pbnthat is known to converge/diverge and compare
it to the given series.
Step 1: Choose a test series. Let’s consider the series P1
n2which is a
p-series with p= 2 and is known to converge.
Step 2: Compute the limit. Let an=n2+3n+1
2n4+n3+1 and bn=1
n2. We need to
find
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
2n4+n3+ 1 ·n2
1
Step 3: Simplify the limit.
lim
n→∞
n2+ 3n+ 1
2n4+n3+ 1 ·n2
1= lim
n→∞
n4+ 3n3+n2
2n4+n3+ 1 = lim
n→∞
1 + 3
n+1
n2
2 + 1
n+1
n2
=1
2
Step 4: Determine the convergence. Since the limit is a positive finite num-
ber, by the Limit Comparison Test, the given series P∞
n=1
n2+3n+1
2n4+n3+1 converges
3
or diverges in the same way as P1
n2. Since P1
n2converges, the given series
also converges.
Question 6
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 can use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
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
Step 3: Simplify the limit.
lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 4: Apply the ratio test. Since the limit is 0, the series converges
absolutely by the ratio test.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 7
Question
Determine whether the series P∞
n=1
n2+2n
n4+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+2n
n4+3 , we can use the limit
comparison test by comparing it to a known convergent or divergent series.
Step 1: Determine a comparison series
Let an=n2+2n
n4+3 . As napproaches infinity, the highest order terms dominate,
so we can simplify anto n2
n4=1
n2.
Thus, we will compare our series to the series P∞
n=1 1
n2.
4
Step 2: Use the limit comparison test
We will compute the limit:
L= lim
n→∞
n2+2n
n4+3
1
n2
= lim
n→∞
n4+ 2n3
n4+ 3 = lim
n→∞ 1 + 2n3−3
n4+ 3
Since the numerator has the same degree as the denominator, we can divide
each term by n4:
L= lim
n→∞ 1 + 2/n −3/n4
1+3/n4= 1
Step 3: Draw a conclusion
Since Lis a finite positive number, by the limit comparison test, P∞
n=1
n2+2n
n4+3
converges if and only if P∞
n=1 1
n2converges.
Since P∞
n=1 1
n2is a convergent p-series with p= 2 >1, then by the compar-
ison test, P∞
n=1
n2+2n
n4+3 also converges.
Question 8
Question
Determine whether the series P∞
n=1 2n+n
3n+n2converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series bn=1
3n.
Step 1: Find the limit of anbn:
lim
n→∞
2n+n
3n+n2·3n
1= lim
n→∞
2n+n
3n+n2·3n
Now, simplify the expression inside the limit:
= lim
n→∞
1 + n
2n
1 + n2
3n·3n
We know that limn→∞
n
an= 0 for any constant a > 1, so limn→∞
n
2n= 0 and
limn→∞
n2
3n= 0.
Therefore, the limit simplifies to:
= lim
n→∞
1+0
1+0· ∞ =∞
Step 2: Conclude based on the limit: Since limn→∞
2n+n
3n+n2·3n
1=∞,
and P∞
n=1 1
3nis a convergent geometric series, by the limit comparison test, the
given series P∞
n=1 2n+n
3n+n2also converges.
5
Question 9
Question
Determine whether the series ∞
X
n=1
3n2+ 1
5n3+ 2 converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series ∞
X
n=1
3n2+1
5n3+2
1
n
.
Step 1: Find the limit
Compute the limit of the ratio as napproaches infinity:
lim
n→∞
3n2+1
5n3+2
1
n
= lim
n→∞
(3n2+ 1)n
5n3+ 2 = lim
n→∞
3n3+n
5n3+ 2
Step 2: Simplify the limit
Divide all terms by the highest power of nin the denominator:
lim
n→∞
3n3+n
5n3+ 2 = lim
n→∞
n3(3 + 1
n2)
n3(5 + 2
n3)= lim
n→∞
3 + 1
n2
5 + 2
n3
Step 3: Evaluate the limit
As napproaches infinity, the terms with 1
n2and 1
n3will approach zero, so
the limit simplifies to:
lim
n→∞
3
5=3
5
Step 4: Determine convergence
Since the limit is a finite positive value, by the Limit Comparison Test, the
series ∞
X
n=1
3n2+ 1
5n3+ 2 has the same convergence behavior as P1
n.
Since ∞
X
n=1
1
ndiverges (harmonic series), the given series ∞
X
n=1
3n2+ 1
5n3+ 2 also
diverges.
Question 10
Question
Determine the convergence or divergence of the series P∞
n=1
n2+3n
n3+5 .
6
Solution
To analyze the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will find a series which is easier to work with and compare it
to the given series.
Consider the series P∞
n=1
n2
n3=P∞
n=1 1
n, which is a p-series with p= 1.
Step 2: Find the limit of the ratio of the terms.
Let an=n2+3n
n3+5 and bn=1
n. We will find L= limn→∞
an
bn.
L= limn→∞
n2+3n
n3+5
1
n
= limn→∞
n3+3n2
n3+5 = 1.
Step 3: Apply the limit comparison test.
Since L= 1 and P∞
n=1 1
ndiverges, by the limit comparison test, the given
series P∞
n=1
n2+3n
n3+5 also diverges.
Question 11
Question
Determine if the series ∞
X
n=1
n3+ 2
3n4−5converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Let’s consider the series ∞
X
n=1
n3+ 2
3n4−5and the series ∞
X
n=1
1
n.
We will calculate the following limit:
lim
n→∞
n3+2
3n4−5
1
n
Step 2: Simplifying the limit:
lim
n→∞
n3+2
3n4−5
1
n
= lim
n→∞
n4+ 2n
3n4−5= lim
n→∞
1 + 2
n3
3−5
n4
=1+0
3−0=1
3
Step 3: Since the limit is a positive finite number, by the limit comparison
test, ∞
X
n=1
n3+ 2
3n4−5converges if and only if ∞
X
n=1
1
nconverges.
Step 4: The series ∞
X
n=1
1
nis known as the harmonic series, which diverges.
7
Step 5: Therefore, by the limit comparison test, the series ∞
X
n=1
n3+ 2
3n4−5also
diverges.
Question 12
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series, we will apply the ratio test.
Step 1: Compute 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)!nn
n!(n+ 1)n+1 =n+ 1
(1 + 1
n)n+1
Step 2: Apply the limit in the ratio test. Taking the limit of the ratio as n
approaches infinity, we have
lim
n→∞
an+1
an
= lim
n→∞
n+ 1
(1 + 1
n)n+1
We can rewrite the limit as
lim
n→∞
n+ 1
1 + 1
nn·(1 + 1
n)
Step 3: Evaluate the limit using the fact that limn→∞(1 + 1
n)n=e. This
simplifies the expression to 1
e>0
Step 4: Determine the convergence of the series. Since the limit of the ratio
is a positive finite number, by the ratio test, the series P∞
n=1
n!
nnconverges.
Question 13
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2−1
n3−2.
8
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series ∞
X
n=1
bnsuch that lim
n→∞
an
bn
exists and is a positive
finite number.
Let an=n2−1
n3−2and bn=1
n.
Step 2: Find lim
n→∞
an
bn
.
lim
n→∞
an
bn
= lim
n→∞ n(n2−1)(n3−2) = 1
Step 3: Since lim
n→∞
an
bn
is a positive finite number, by the Limit Comparison
Test, the series ∞
X
n=1
anand ∞
X
n=1
bneither both converge or both diverge.
Step 4: Since ∞
X
n=1
bn=∞
X
n=1
1
nis a harmonic series which diverges, the given
series ∞
X
n=1
n2−1
n3−2also diverges.
Therefore, the series ∞
X
n=1
n2−1
n3−2diverges.
Question 14
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. Then, the
9
ratio of consecutive terms is
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n
n+ 1
= lim
n→∞ 1 + 1
nn
=e.
Step 2: Apply the ratio test: Since the limit is e > 1, the series diverges by
the ratio test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 15
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: Compute the limit of the ratio test:
L= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Determine an+1:
an+1 =(n+ 1)!
(n+ 1)n+1 =(n+ 1)n!
(n+ 1)(n+ 1)n=n!
(n+ 1)n
10
Step 3: Calculate the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
n!/(n+ 1)n
n!/nn
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e(since lim
n→∞ 1 + 1
nn
=e)
Step 4: Analyze L: Since L=1
e<1, by the ratio test, the series P∞
n=1
n!
nn
converges.
Question 16
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n2+ 1
2n3−1.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s denote the general term of the given series as an:
an=3n2+ 1
2n3−1.
Step 1: Find a suitable series to compare it to. We will look for a series
of the form Pbnwhere bnis simpler to work with and easier to determine the
convergence of.
Let’s consider the series P1
n. This series is a p-series with p= 1, which we
know diverges. We will compare our series to this one.
Step 2: Compute the limit:
lim
n→∞
an
bn
= lim
n→∞
3n2+1
2n3−1
1
n
= lim
n→∞
3n3+n
2n3−1= lim
n→∞
3 + 1
n2
2−1
n3
=3
2.
Step 3: Interpret the limit. Since the limit is a finite positive number, and
P1
nis a divergent series, then by the Limit Comparison Test, we conclude that
the series P∞
n=1 3n2+1
2n3−1also diverges.
11
Question 17
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. Compute
the ratio ras follows:
r= lim
n→∞
an+1
an
Step 2: Calculate the ratio r.
r= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1 ·nn
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1n
= lim
n→∞
1
1 + 1
nn
= lim
n→∞
1
(1 + 1/n)n
=1
e
where eis the base of the natural logarithm.
Step 3: Apply the ratio test. If r < 1, the series converges. If r > 1 or
r= 1, the series diverges. Here, since r=1
e<1, the series P∞
n=1
n!
nnconverges
by the ratio test.
Question 18
Question
Determine whether the series P∞
n=1
n2+5
n3+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+5
n3+3 , we can use the Limit
Comparison Test.
Step 1: Find a suitable series to compare. Let’s consider the series P∞
n=1 1
n.
This is a p-series with p= 1, which we know diverges.
Step 2: Calculate the limit of the ratio. We will find the limit:
lim
n→∞
(n2+ 5)/(n3+ 3)
1/n = lim
n→∞
n2+ 5
n3+ 3 ·n
1= lim
n→∞
n3+ 5n
n3+ 3 = 1
Step 3: Conclusion. Since the limit is a positive finite number and the
series P∞
n=1 1
ndiverges, by the Limit Comparison Test, the series P∞
n=1
n2+5
n3+3
also diverges.
12
Question 19
Question
Determine the convergence or divergence of the series P∞
n=1
n
2n+1 using the
Limit Comparison Test.
Solution
To determine the convergence or divergence of the series P∞
n=1
n
2n+1 , we will
use the Limit Comparison Test. We will compare this series with the convergent
series P∞
n=1 1
n.
Step 1: Find the limit to compare the series. Let an=n
2n+1 and
bn=1
n. We consider the limit:
lim
n→∞
an
bn
= lim
n→∞
n
2n+1
1
n
= lim
n→∞
n2
2n2+n= lim
n→∞
1
2 + 1
n
=1
2
Step 2: Make a conclusion based on the limit. Since the limit is a
finite positive value (1
2), by the Limit Comparison Test, the series P∞
n=1
n
2n+1
has the same convergence behavior as the series P∞
n=1 1
n, which is a divergent
p-series with p= 1. Therefore, the series P∞
n=1
n
2n+1 also diverges.
Question 20
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
.
13
Step 2: Simplify the limit.
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
1 + 1
nn=1
e.
Step 3: Determine the convergence or divergence of the series. Since L=
1
e<1, by the ratio test, the series converges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Calculate the limit:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Determine the convergence or divergence based on the limit: Since
the limit is less than 1, by the ratio test, the series
∞
X
n=1
n!
nn
converges.
14
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1
n2
en.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms. Let an=n2
en. The ratio of
consecutive terms is given by
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
en+1 ·en
n2
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)2
en+1 ·en
n2
= lim
n→∞
(n+ 1)2
n2·e= lim
n→∞ 1
e+2
n+1
n2=1
e
Step 3: Apply the ratio test. Since the limit of the ratio is 1
e<1, by the
ratio test, the series P∞
n=1
n2
enconverges.
Therefore, the series P∞
n=1
n2
enconverges.
Question 23
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Compute the limit of the ratio test. Let an=n!
nn. We will
consider the limit
L= lim
n→∞
an+1
an
.
15
Step 2: Simplify the expression in the limit.
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
n+ 1 ·nn
(n+ 1)n
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1 + 1
n−n
.
Step 3: Compute the limit.
L= lim
n→∞
1 + 1
n−n
= lim
n→∞
1
1 + 1
nn
=1
e.
Step 4: Determine the convergence of the series. Since L=1
e<1, by the ratio
test, the series P∞
n=1
n!
nnconverges. Therefore, the series converges.
Question 24
Question
Determine whether the series P∞
n=1
n2+1
n4+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n4+1 , we can use the Limit
Comparison Test. Let’s consider the series an=n2+1
n4+1 .
Step 1: Find the limit of an
1
n2
as n→ ∞.
lim
n→∞
an
1
n2
= lim
n→∞
n2+ 1
n4+ 1 ·n2= lim
n→∞
n2+ 1
n2(n2+1
n2)= lim
n→∞
1 + 1
n2
1 + 1
n2
= 1.
Step 2: Conclusion based on the Limit Comparison Test Since
limn→∞
an
1
n2
= 1 >0, and P∞
n=1 1
n2is a convergent p-series with p= 2 >1, by
the Limit Comparison Test, we conclude that the series P∞
n=1
n2+1
n4+1 converges.
16
Question 25
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is given by:
R= lim
n→∞
an+1
an
Step 2: Find an+1 and an. We have:
an+1 =(n+ 1)!
(n+ 1)n+1 =(n+ 1)!
(n+ 1)n(n+ 1) =(n+ 1)n!
(n+ 1)n=n!
(n+ 1)n
Therefore,
R= lim
n→∞
n!/(n+ 1)n
n!/nn
Step 3: Simplify the expression.
R= lim
n→∞
nn
(n+ 1)n
Step 4: Calculate the limit.
R= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
where eis the base of the natural logarithm.
Step 5: Analyze the result. Since R < 1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
17
Solution
We will use the ratio test to determine the convergence of the series.
Step 1: Consider the ratio Rn:
Rn=an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
n!·nn
(n+ 1)n+1 .
Step 2: Simplify Rn:
Rn= (n+ 1) n
n+ 1n
= (n+ 1) 1
1 + 1
nn
.
Step 3: Consider the limit of Rnas napproaches infinity:
lim
n→∞ Rn= lim
n→∞(n+ 1) 1
1 + 1
nn
= lim
n→∞(n+ 1) n
n+ 1n
= lim
n→∞
n+ 1
1 + 1
nn.
Step 4: Use the limit definition of e:
lim
n→∞
n+ 1
1 + 1
nn=e
e= 1.
Step 5: Analyze the limit of Rn: Since the limit of Rnis equal to 1, the
ratio test is inconclusive. Therefore, we cannot determine the convergence or
divergence of the series using the ratio test.
Step 6: Conclusion: The ratio test is inconclusive for the given series, so
we cannot determine its convergence or divergence using this test.
Question 27
Question
Determine the convergence or divergence of the series P∞
n=1
n
2n.
Solution
To determine the convergence of the series P∞
n=1
n
2n, we will use the ratio test.
Step 1: Calculate the ratio Compute the ratio of consecutive terms:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2n
n2n+1
Step 2: Simplify the ratio Simplify the ratio:
R= lim
n→∞
n+ 1
2n
= lim
n→∞
1
21 + 1
n=1
2
Step 3: Apply the ratio test Since R=1
2<1, the series P∞
n=1
n
2n
converges by the ratio test.
18
Question 28
Question
Determine whether the series P∞
n=1 3n2+5n+1
5n3−n+4 converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test.
Step 1: Consider the series P∞
n=1 3n2+5n+1
5n3−n+4 and another series P∞
n=1 bn
such that bn=1
n.
Step 2: Calculate the limit:
lim
n→∞
(3n2+ 5n+ 1)/5n3−n+ 4
1/n
= lim
n→∞
(3n2+ 5n+ 1)/n
5n3−n+ 4
= lim
n→∞
3/n + 5/n2+ 1/n3
5−1/n2+ 4/n3
= lim
n→∞
0+0+0
5
= 0
Step 3: Since the limit above is a finite positive value, we can use the Limit
Comparison Test. Therefore, the convergence of the series P∞
n=1 3n2+5n+1
5n3−n+4 is
the same as the convergence of the series P∞
n=1 1
n.
Step 4: The series P∞
n=1 1
nis the harmonic series which diverges.
Thus, by the Limit Comparison Test, the series P∞
n=1 3n2+5n+1
5n3−n+4 also diverges.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1
n!
nn.
Solution
To analyze the convergence or divergence of the series P∞
n=1
n!
nn, we will use the
Ratio Test.
Step 1: Compute the ratio R:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
19
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(n+ 1)n
Step 2: Simplify the expression and evaluate the limit:
lim
n→∞
n+ 1
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 3: Determine the convergence or divergence of the series: If R < 1, the
series converges. If R > 1, the series diverges. If R= 1, the test is inconclusive.
Since 1
e<1, by the Ratio Test, the series P∞
n=1
n!
nnconverges.
Question 30
Question
Prove or disprove the convergence of the series:
∞
X
n=1
n2+ 1
n3+ 2n+ 1.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
an=∞
X
n=1
n2+ 1
n3+ 2n+ 1
and the series ∞
X
n=1
bn=∞
X
n=1
1
n.
Step 1: Find the limit of the ratio: Let’s find the limit of the ratio an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+1
n3+2n+1
1
n
.
Step 2: Simplify the expression:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 2n+ 1 ·n= lim
n→∞
n3+n
n3+ 2n+ 1.
Step 3: Divide by the highest power of n:
lim
n→∞
an
bn
= lim
n→∞
1 + 1
n2
1 + 2
n+1
n3
=1+0
1+0+0 = 1.
Step 4: Apply the Limit Comparison Test: Since the limit is a positive
finite number, and P∞
n=1 bn=P∞
n=1 1
nis a harmonic series which diverges, we
conclude by the Limit Comparison Test that P∞
n=1 analso diverges.
Therefore, the series P∞
n=1
n2+1
n3+2n+1 diverges.
20
Question 31
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: Compute the limit of the ratio. Let an=n3
2n. We will consider the
limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3=1
2
Step 2: Analyze the limit obtained. Since the limit of the ratio is less than
1, by the Ratio Test, the series P∞
n=1
n3
2nconverges.
Therefore, the series P∞
n=1
n3
2nconverges.
Question 32
Question
Determine whether the series P∞
n=1
√n+1−√n
n3
2
converges or diverges.
Solution
To determine whether the given series converges or diverges, we will analyze the
convergence of the general term using the limit comparison test.
Step 1: Find the general term anof the series.
The general term anof the series is √n+1−√n
n3
2
.
Step 2: Simplify the general term an.
We can simplify the general term anas follows:
an=√n+ 1 −√n
n3
2
=√n+ 1 −√n
n3
2·√n+1+√n
√n+1+√n
an=(√n+ 1)2−(√n)2
n3
2(√n+1+√n)=n+ 1 −n
n3
2(√n+1+√n)=1
n3
2(√n+1+√n)
21
Step 3: Find the limit of the general term as napproaches infinity.
Taking the limit as napproaches infinity, we get:
lim
n→∞ an= lim
n→∞
1
n3
2(√n+1+√n)= lim
n→∞
1
n3
2(2√n)= lim
n→∞
1
2n2= 0
Step 4: Apply the limit comparison test.
Since limn→∞ an= 0 and the series P∞
n=1 1
n2converges (by the p-series test
with p= 2 >1), we can conclude by the limit comparison test that the series
P∞
n=1
√n+1−√n
n3
2
converges.
Question 33
Question
Determine the convergence or divergence of the series P∞
n=1
n3+2
n4+n+1 .
Solution
To determine the convergence or divergence of the series, we can use the limit
comparison test. Let’s compare the given series with a simpler series whose
convergence is known.
Step 1: Find the simpler series
Consider the series P∞
n=1 1
n, which is a p-series with p= 1 and known to be
divergent.
Step 2: Compute the limit
We will compute the limit of the ratio of the two series:
L= lim
n→∞
n3+2
n4+n+1
1
n
Step 3: Simplify the expression
Simplify the expression within the limit by dividing both the numerator and
denominator by n4:
L= lim
n→∞
1+ 2
n3
1+ 1
n3+1
n4
1
n
L= lim
n→∞
1
n4+2
n3
1
n+1
n2+1
n3
L= lim
n→∞
0+0
0+0+0 = 0
Step 4: Conclusion
Since L= 0 and the series P∞
n=1 1
nis divergent, by the limit comparison test,
the given series P∞
n=1
n3+2
n4+n+1 also diverges.
22
Question 34
Question
Determine the convergence of the series ∞
X
n=1
n!
nn.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
L= lim
n→∞
an+1
an
Step 2: Calculate an+1 and an+1
an
.
an+1 =(n+ 1)!
(n+ 1)n+1
=(n+ 1)n!
(n+ 1)n·(n+ 1)
=n!
(n+ 1)n
an+1
an
=
n!
(n+1)n
n!
nn
=nn
(n+ 1)n
=n
n+ 1n
Step 3: Evaluate the limit L.
L= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Determine the convergence of the series. Since L=1
e<1, by the
ratio test, the series ∞
X
n=1
n!
nnconverges.
23
Question 35
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: Apply the Ratio Test. Let an=n!
nn. We consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + 1
n−n
=1
e
Step 3: Analyze the limit. Since 1
e<1, by the Ratio Test, the series ∞
X
n=1
n!
nn
converges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
24
Solution
To analyze the convergence of the series P∞
n=1
(−1)n
np, we will apply the alter-
nating series test and the p-series test.
Step 1: Absolute Convergence To determine the values of pfor which
the series converges absolutely, we consider the series P∞
n=1
(−1)n
np
=P∞
n=1 1
np.
This series is a p-series, which converges if p > 1 and diverges if p≤1.
Therefore, the series P∞
n=1
(−1)n
npconverges absolutely for p > 1.
Step 2: Conditional Convergence To determine the values of pfor which
the series converges conditionally, we consider the series P∞
n=1 1
np.
This series converges by the p-series test if p > 1.
For 0 < p ≤1, we can consider the alternating series P∞
n=1
(−1)n
np.
This series satisfies the conditions of the alternating series test since 1
npis
monotonically decreasing and converges to 0.
Therefore, the series P∞
n=1
(−1)n
npconverges conditionally for 0 < p ≤1.
Step 3: Divergence The series P∞
n=1
(−1)n
npdiverges for p≤0 as the terms
do not approach 0 as napproaches infinity.
In conclusion: - The series converges absolutely for p > 1. - The series
converges conditionally for 0 < p ≤1. - The series diverges for p≤0.
Question 3
Question
Determine the convergence or divergence of the series: P∞
n=1
n3
2n.
Solution
To determine the convergence of the series P∞
n=1
n3
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n3
2n. We compute the ratio Ras
follows:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3/2n+1
n3/2n
= lim
n→∞
(n+ 1)3·2n
2n+1 ·n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞ (n+ 1)3
2n3= lim
n→∞ n3+ 3n2+ 3n+ 1
2n3=1
2
Step 2: Evaluate the limit of the ratio. Since the limit of the ratio R=1
2
is less than 1, by the ratio test, the series converges.
Therefore, the series P∞
n=1
n3
2nconverges.
2
Question 4
Question
Determine whether the series P∞
n=1
n2+3n+2
4n3+5n+1 converges or diverges.
Solution
To determine the convergence of the series, we can use the limit comparison
test. Let’s compare the given series with a known convergent series.
Step 1: Find a known convergent series Consider the series P∞
n=1 1
n,
which is a known convergent series.
Step 2: Find the limit and compare Let an=n2+3n+2
4n3+5n+1 and bn=1
n.
We will compare the two series by finding the limit of the ratio an
bn.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 2
4n3+ 5n+ 1 ·n
1= lim
n→∞
n3+ 3n2+ 2n
4n3+ 5n+ 1 =1
4
Step 3: Conclusion Since limn→∞
an
bn=1
4>0, and the series P∞
n=1 1
n
converges (by the p-series test), by the limit comparison test, the given series
P∞
n=1
n2+3n+2
4n3+5n+1 also converges.
Question 5
Question
Determine whether the series P∞
n=1
n2+3n+1
2n4+n3+1 converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s choose a series Pbnthat is known to converge/diverge and compare
it to the given series.
Step 1: Choose a test series. Let’s consider the series P1
n2which is a
p-series with p= 2 and is known to converge.
Step 2: Compute the limit. Let an=n2+3n+1
2n4+n3+1 and bn=1
n2. We need to
find
lim
n→∞
an
bn
= lim
n→∞
n2+ 3n+ 1
2n4+n3+ 1 ·n2
1
Step 3: Simplify the limit.
lim
n→∞
n2+ 3n+ 1
2n4+n3+ 1 ·n2
1= lim
n→∞
n4+ 3n3+n2
2n4+n3+ 1 = lim
n→∞
1 + 3
n+1
n2
2 + 1
n+1
n2
=1
2
Step 4: Determine the convergence. Since the limit is a positive finite num-
ber, by the Limit Comparison Test, the given series P∞
n=1
n2+3n+1
2n4+n3+1 converges
3
or diverges in the same way as P1
n2. Since P1
n2converges, the given series
also converges.
Question 6
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 can use the ratio test.
Step 1: Calculate the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is given by:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio.
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
Step 3: Simplify the limit.
lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 4: Apply the ratio test. Since the limit is 0, the series converges
absolutely by the ratio test.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 7
Question
Determine whether the series P∞
n=1
n2+2n
n4+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+2n
n4+3 , we can use the limit
comparison test by comparing it to a known convergent or divergent series.
Step 1: Determine a comparison series
Let an=n2+2n
n4+3 . As napproaches infinity, the highest order terms dominate,
so we can simplify anto n2
n4=1
n2.
Thus, we will compare our series to the series P∞
n=1 1
n2.
4
Step 2: Use the limit comparison test
We will compute the limit:
L= lim
n→∞
n2+2n
n4+3
1
n2
= lim
n→∞
n4+ 2n3
n4+ 3 = lim
n→∞ 1 + 2n3−3
n4+ 3
Since the numerator has the same degree as the denominator, we can divide
each term by n4:
L= lim
n→∞ 1 + 2/n −3/n4
1+3/n4= 1
Step 3: Draw a conclusion
Since Lis a finite positive number, by the limit comparison test, P∞
n=1
n2+2n
n4+3
converges if and only if P∞
n=1 1
n2converges.
Since P∞
n=1 1
n2is a convergent p-series with p= 2 >1, then by the compar-
ison test, P∞
n=1
n2+2n
n4+3 also converges.
Question 8
Question
Determine whether the series P∞
n=1 2n+n
3n+n2converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test. Let’s consider the series bn=1
3n.
Step 1: Find the limit of anbn:
lim
n→∞
2n+n
3n+n2·3n
1= lim
n→∞
2n+n
3n+n2·3n
Now, simplify the expression inside the limit:
= lim
n→∞
1 + n
2n
1 + n2
3n·3n
We know that limn→∞
n
an= 0 for any constant a > 1, so limn→∞
n
2n= 0 and
limn→∞
n2
3n= 0.
Therefore, the limit simplifies to:
= lim
n→∞
1+0
1+0· ∞ =∞
Step 2: Conclude based on the limit: Since limn→∞
2n+n
3n+n2·3n
1=∞,
and P∞
n=1 1
3nis a convergent geometric series, by the limit comparison test, the
given series P∞
n=1 2n+n
3n+n2also converges.
5
Question 9
Question
Determine whether the series ∞
X
n=1
3n2+ 1
5n3+ 2 converges or diverges.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series ∞
X
n=1
3n2+1
5n3+2
1
n
.
Step 1: Find the limit
Compute the limit of the ratio as napproaches infinity:
lim
n→∞
3n2+1
5n3+2
1
n
= lim
n→∞
(3n2+ 1)n
5n3+ 2 = lim
n→∞
3n3+n
5n3+ 2
Step 2: Simplify the limit
Divide all terms by the highest power of nin the denominator:
lim
n→∞
3n3+n
5n3+ 2 = lim
n→∞
n3(3 + 1
n2)
n3(5 + 2
n3)= lim
n→∞
3 + 1
n2
5 + 2
n3
Step 3: Evaluate the limit
As napproaches infinity, the terms with 1
n2and 1
n3will approach zero, so
the limit simplifies to:
lim
n→∞
3
5=3
5
Step 4: Determine convergence
Since the limit is a finite positive value, by the Limit Comparison Test, the
series ∞
X
n=1
3n2+ 1
5n3+ 2 has the same convergence behavior as P1
n.
Since ∞
X
n=1
1
ndiverges (harmonic series), the given series ∞
X
n=1
3n2+ 1
5n3+ 2 also
diverges.
Question 10
Question
Determine the convergence or divergence of the series P∞
n=1
n2+3n
n3+5 .
6
Solution
To analyze the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will find a series which is easier to work with and compare it
to the given series.
Consider the series P∞
n=1
n2
n3=P∞
n=1 1
n, which is a p-series with p= 1.
Step 2: Find the limit of the ratio of the terms.
Let an=n2+3n
n3+5 and bn=1
n. We will find L= limn→∞
an
bn.
L= limn→∞
n2+3n
n3+5
1
n
= limn→∞
n3+3n2
n3+5 = 1.
Step 3: Apply the limit comparison test.
Since L= 1 and P∞
n=1 1
ndiverges, by the limit comparison test, the given
series P∞
n=1
n2+3n
n3+5 also diverges.
Question 11
Question
Determine if the series ∞
X
n=1
n3+ 2
3n4−5converges or diverges.
Solution
To determine the convergence of the series, we will use the limit comparison
test.
Step 1: Let’s consider the series ∞
X
n=1
n3+ 2
3n4−5and the series ∞
X
n=1
1
n.
We will calculate the following limit:
lim
n→∞
n3+2
3n4−5
1
n
Step 2: Simplifying the limit:
lim
n→∞
n3+2
3n4−5
1
n
= lim
n→∞
n4+ 2n
3n4−5= lim
n→∞
1 + 2
n3
3−5
n4
=1+0
3−0=1
3
Step 3: Since the limit is a positive finite number, by the limit comparison
test, ∞
X
n=1
n3+ 2
3n4−5converges if and only if ∞
X
n=1
1
nconverges.
Step 4: The series ∞
X
n=1
1
nis known as the harmonic series, which diverges.
7
Step 5: Therefore, by the limit comparison test, the series ∞
X
n=1
n3+ 2
3n4−5also
diverges.
Question 12
Question
Determine whether the series ∞
X
n=1
n!
nn
converges or diverges.
Solution
To determine the convergence of the series, we will apply the ratio test.
Step 1: Compute 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)!nn
n!(n+ 1)n+1 =n+ 1
(1 + 1
n)n+1
Step 2: Apply the limit in the ratio test. Taking the limit of the ratio as n
approaches infinity, we have
lim
n→∞
an+1
an
= lim
n→∞
n+ 1
(1 + 1
n)n+1
We can rewrite the limit as
lim
n→∞
n+ 1
1 + 1
nn·(1 + 1
n)
Step 3: Evaluate the limit using the fact that limn→∞(1 + 1
n)n=e. This
simplifies the expression to 1
e>0
Step 4: Determine the convergence of the series. Since the limit of the ratio
is a positive finite number, by the ratio test, the series P∞
n=1
n!
nnconverges.
Question 13
Question
Determine the convergence or divergence of the series ∞
X
n=1
n2−1
n3−2.
8
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test.
Step 1: Let’s find a series ∞
X
n=1
bnsuch that lim
n→∞
an
bn
exists and is a positive
finite number.
Let an=n2−1
n3−2and bn=1
n.
Step 2: Find lim
n→∞
an
bn
.
lim
n→∞
an
bn
= lim
n→∞ n(n2−1)(n3−2) = 1
Step 3: Since lim
n→∞
an
bn
is a positive finite number, by the Limit Comparison
Test, the series ∞
X
n=1
anand ∞
X
n=1
bneither both converge or both diverge.
Step 4: Since ∞
X
n=1
bn=∞
X
n=1
1
nis a harmonic series which diverges, the given
series ∞
X
n=1
n2−1
n3−2also diverges.
Therefore, the series ∞
X
n=1
n2−1
n3−2diverges.
Question 14
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. Then, the
9
ratio of consecutive terms is
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n
n+ 1
= lim
n→∞ 1 + 1
nn
=e.
Step 2: Apply the ratio test: Since the limit is e > 1, the series diverges by
the ratio test.
Therefore, the series P∞
n=1
n!
nndiverges.
Question 15
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: Compute the limit of the ratio test:
L= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Determine an+1:
an+1 =(n+ 1)!
(n+ 1)n+1 =(n+ 1)n!
(n+ 1)(n+ 1)n=n!
(n+ 1)n
10
Step 3: Calculate the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
n!/(n+ 1)n
n!/nn
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e(since lim
n→∞ 1 + 1
nn
=e)
Step 4: Analyze L: Since L=1
e<1, by the ratio test, the series P∞
n=1
n!
nn
converges.
Question 16
Question
Determine the convergence or divergence of the series
∞
X
n=1
3n2+ 1
2n3−1.
Solution
To determine the convergence or divergence of the series, we will use the Limit
Comparison Test. Let’s denote the general term of the given series as an:
an=3n2+ 1
2n3−1.
Step 1: Find a suitable series to compare it to. We will look for a series
of the form Pbnwhere bnis simpler to work with and easier to determine the
convergence of.
Let’s consider the series P1
n. This series is a p-series with p= 1, which we
know diverges. We will compare our series to this one.
Step 2: Compute the limit:
lim
n→∞
an
bn
= lim
n→∞
3n2+1
2n3−1
1
n
= lim
n→∞
3n3+n
2n3−1= lim
n→∞
3 + 1
n2
2−1
n3
=3
2.
Step 3: Interpret the limit. Since the limit is a finite positive number, and
P1
nis a divergent series, then by the Limit Comparison Test, we conclude that
the series P∞
n=1 3n2+1
2n3−1also diverges.
11
Question 17
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. Compute
the ratio ras follows:
r= lim
n→∞
an+1
an
Step 2: Calculate the ratio r.
r= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1 ·nn
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1n
= lim
n→∞
1
1 + 1
nn
= lim
n→∞
1
(1 + 1/n)n
=1
e
where eis the base of the natural logarithm.
Step 3: Apply the ratio test. If r < 1, the series converges. If r > 1 or
r= 1, the series diverges. Here, since r=1
e<1, the series P∞
n=1
n!
nnconverges
by the ratio test.
Question 18
Question
Determine whether the series P∞
n=1
n2+5
n3+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+5
n3+3 , we can use the Limit
Comparison Test.
Step 1: Find a suitable series to compare. Let’s consider the series P∞
n=1 1
n.
This is a p-series with p= 1, which we know diverges.
Step 2: Calculate the limit of the ratio. We will find the limit:
lim
n→∞
(n2+ 5)/(n3+ 3)
1/n = lim
n→∞
n2+ 5
n3+ 3 ·n
1= lim
n→∞
n3+ 5n
n3+ 3 = 1
Step 3: Conclusion. Since the limit is a positive finite number and the
series P∞
n=1 1
ndiverges, by the Limit Comparison Test, the series P∞
n=1
n2+5
n3+3
also diverges.
12
Question 19
Question
Determine the convergence or divergence of the series P∞
n=1
n
2n+1 using the
Limit Comparison Test.
Solution
To determine the convergence or divergence of the series P∞
n=1
n
2n+1 , we will
use the Limit Comparison Test. We will compare this series with the convergent
series P∞
n=1 1
n.
Step 1: Find the limit to compare the series. Let an=n
2n+1 and
bn=1
n. We consider the limit:
lim
n→∞
an
bn
= lim
n→∞
n
2n+1
1
n
= lim
n→∞
n2
2n2+n= lim
n→∞
1
2 + 1
n
=1
2
Step 2: Make a conclusion based on the limit. Since the limit is a
finite positive value (1
2), by the Limit Comparison Test, the series P∞
n=1
n
2n+1
has the same convergence behavior as the series P∞
n=1 1
n, which is a divergent
p-series with p= 1. Therefore, the series P∞
n=1
n
2n+1 also diverges.
Question 20
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
.
13
Step 2: Simplify the limit.
L= lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)n!
(n+ 1)n+1
= lim
n→∞
n!
(n+ 1)n
= lim
n→∞
1
1 + 1
nn=1
e.
Step 3: Determine the convergence or divergence of the series. Since L=
1
e<1, by the ratio test, the series converges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 21
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the ratio:
= lim
n→∞
(n+ 1)!nn
n!(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 3: Calculate the limit:
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 4: Determine the convergence or divergence based on the limit: Since
the limit is less than 1, by the ratio test, the series
∞
X
n=1
n!
nn
converges.
14
Question 22
Question
Determine the convergence or divergence of the series P∞
n=1
n2
en.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test.
Step 1: Compute the ratio of consecutive terms. Let an=n2
en. The ratio of
consecutive terms is given by
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
en+1 ·en
n2
Step 2: Simplify the expression.
lim
n→∞
(n+ 1)2
en+1 ·en
n2
= lim
n→∞
(n+ 1)2
n2·e= lim
n→∞ 1
e+2
n+1
n2=1
e
Step 3: Apply the ratio test. Since the limit of the ratio is 1
e<1, by the
ratio test, the series P∞
n=1
n2
enconverges.
Therefore, the series P∞
n=1
n2
enconverges.
Question 23
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we will use the ratio
test. Step 1: Compute the limit of the ratio test. Let an=n!
nn. We will
consider the limit
L= lim
n→∞
an+1
an
.
15
Step 2: Simplify the expression in the limit.
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
n+ 1 ·nn
(n+ 1)n
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1 + 1
n−n
.
Step 3: Compute the limit.
L= lim
n→∞
1 + 1
n−n
= lim
n→∞
1
1 + 1
nn
=1
e.
Step 4: Determine the convergence of the series. Since L=1
e<1, by the ratio
test, the series P∞
n=1
n!
nnconverges. Therefore, the series converges.
Question 24
Question
Determine whether the series P∞
n=1
n2+1
n4+1 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1
n2+1
n4+1 , we can use the Limit
Comparison Test. Let’s consider the series an=n2+1
n4+1 .
Step 1: Find the limit of an
1
n2
as n→ ∞.
lim
n→∞
an
1
n2
= lim
n→∞
n2+ 1
n4+ 1 ·n2= lim
n→∞
n2+ 1
n2(n2+1
n2)= lim
n→∞
1 + 1
n2
1 + 1
n2
= 1.
Step 2: Conclusion based on the Limit Comparison Test Since
limn→∞
an
1
n2
= 1 >0, and P∞
n=1 1
n2is a convergent p-series with p= 2 >1, by
the Limit Comparison Test, we conclude that the series P∞
n=1
n2+1
n4+1 converges.
16
Question 25
Question
Determine whether the series P∞
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series, we can use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. The ratio
of consecutive terms is given by:
R= lim
n→∞
an+1
an
Step 2: Find an+1 and an. We have:
an+1 =(n+ 1)!
(n+ 1)n+1 =(n+ 1)!
(n+ 1)n(n+ 1) =(n+ 1)n!
(n+ 1)n=n!
(n+ 1)n
Therefore,
R= lim
n→∞
n!/(n+ 1)n
n!/nn
Step 3: Simplify the expression.
R= lim
n→∞
nn
(n+ 1)n
Step 4: Calculate the limit.
R= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
where eis the base of the natural logarithm.
Step 5: Analyze the result. Since R < 1, by the ratio test, the series
P∞
n=1
n!
nnconverges.
Therefore, the series P∞
n=1
n!
nnconverges.
Question 26
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
17
Solution
We will use the ratio test to determine the convergence of the series.
Step 1: Consider the ratio Rn:
Rn=an+1
an
=(n+ 1)!/(n+ 1)n+1
n!/nn=(n+ 1)!
n!·nn
(n+ 1)n+1 .
Step 2: Simplify Rn:
Rn= (n+ 1) n
n+ 1n
= (n+ 1) 1
1 + 1
nn
.
Step 3: Consider the limit of Rnas napproaches infinity:
lim
n→∞ Rn= lim
n→∞(n+ 1) 1
1 + 1
nn
= lim
n→∞(n+ 1) n
n+ 1n
= lim
n→∞
n+ 1
1 + 1
nn.
Step 4: Use the limit definition of e:
lim
n→∞
n+ 1
1 + 1
nn=e
e= 1.
Step 5: Analyze the limit of Rn: Since the limit of Rnis equal to 1, the
ratio test is inconclusive. Therefore, we cannot determine the convergence or
divergence of the series using the ratio test.
Step 6: Conclusion: The ratio test is inconclusive for the given series, so
we cannot determine its convergence or divergence using this test.
Question 27
Question
Determine the convergence or divergence of the series P∞
n=1
n
2n.
Solution
To determine the convergence of the series P∞
n=1
n
2n, we will use the ratio test.
Step 1: Calculate the ratio Compute the ratio of consecutive terms:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2n
n2n+1
Step 2: Simplify the ratio Simplify the ratio:
R= lim
n→∞
n+ 1
2n
= lim
n→∞
1
21 + 1
n=1
2
Step 3: Apply the ratio test Since R=1
2<1, the series P∞
n=1
n
2n
converges by the ratio test.
18
Question 28
Question
Determine whether the series P∞
n=1 3n2+5n+1
5n3−n+4 converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test.
Step 1: Consider the series P∞
n=1 3n2+5n+1
5n3−n+4 and another series P∞
n=1 bn
such that bn=1
n.
Step 2: Calculate the limit:
lim
n→∞
(3n2+ 5n+ 1)/5n3−n+ 4
1/n
= lim
n→∞
(3n2+ 5n+ 1)/n
5n3−n+ 4
= lim
n→∞
3/n + 5/n2+ 1/n3
5−1/n2+ 4/n3
= lim
n→∞
0+0+0
5
= 0
Step 3: Since the limit above is a finite positive value, we can use the Limit
Comparison Test. Therefore, the convergence of the series P∞
n=1 3n2+5n+1
5n3−n+4 is
the same as the convergence of the series P∞
n=1 1
n.
Step 4: The series P∞
n=1 1
nis the harmonic series which diverges.
Thus, by the Limit Comparison Test, the series P∞
n=1 3n2+5n+1
5n3−n+4 also diverges.
Question 29
Question
Determine the convergence or divergence of the series P∞
n=1
n!
nn.
Solution
To analyze the convergence or divergence of the series P∞
n=1
n!
nn, we will use the
Ratio Test.
Step 1: Compute the ratio R:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
19
= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!
= lim
n→∞
n+ 1
(n+ 1)n
Step 2: Simplify the expression and evaluate the limit:
lim
n→∞
n+ 1
(n+ 1)n
= lim
n→∞
1
(1 + 1/n)n
=1
e
Step 3: Determine the convergence or divergence of the series: If R < 1, the
series converges. If R > 1, the series diverges. If R= 1, the test is inconclusive.
Since 1
e<1, by the Ratio Test, the series P∞
n=1
n!
nnconverges.
Question 30
Question
Prove or disprove the convergence of the series:
∞
X
n=1
n2+ 1
n3+ 2n+ 1.
Solution
To determine the convergence of the series, we can use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
an=∞
X
n=1
n2+ 1
n3+ 2n+ 1
and the series ∞
X
n=1
bn=∞
X
n=1
1
n.
Step 1: Find the limit of the ratio: Let’s find the limit of the ratio an
bn:
lim
n→∞
an
bn
= lim
n→∞
n2+1
n3+2n+1
1
n
.
Step 2: Simplify the expression:
lim
n→∞
an
bn
= lim
n→∞
n2+ 1
n3+ 2n+ 1 ·n= lim
n→∞
n3+n
n3+ 2n+ 1.
Step 3: Divide by the highest power of n:
lim
n→∞
an
bn
= lim
n→∞
1 + 1
n2
1 + 2
n+1
n3
=1+0
1+0+0 = 1.
Step 4: Apply the Limit Comparison Test: Since the limit is a positive
finite number, and P∞
n=1 bn=P∞
n=1 1
nis a harmonic series which diverges, we
conclude by the Limit Comparison Test that P∞
n=1 analso diverges.
Therefore, the series P∞
n=1
n2+1
n3+2n+1 diverges.
20
Question 31
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: Compute the limit of the ratio. Let an=n3
2n. We will consider the
limit of the ratio limn→∞
an+1
an
.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)3
2n+1 ·2n
n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
(n+ 1)3
2n3
= lim
n→∞
n3+ 3n2+ 3n+ 1
2n3=1
2
Step 2: Analyze the limit obtained. Since the limit of the ratio is less than
1, by the Ratio Test, the series P∞
n=1
n3
2nconverges.
Therefore, the series P∞
n=1
n3
2nconverges.
Question 32
Question
Determine whether the series P∞
n=1
√n+1−√n
n3
2
converges or diverges.
Solution
To determine whether the given series converges or diverges, we will analyze the
convergence of the general term using the limit comparison test.
Step 1: Find the general term anof the series.
The general term anof the series is √n+1−√n
n3
2
.
Step 2: Simplify the general term an.
We can simplify the general term anas follows:
an=√n+ 1 −√n
n3
2
=√n+ 1 −√n
n3
2·√n+1+√n
√n+1+√n
an=(√n+ 1)2−(√n)2
n3
2(√n+1+√n)=n+ 1 −n
n3
2(√n+1+√n)=1
n3
2(√n+1+√n)
21
Step 3: Find the limit of the general term as napproaches infinity.
Taking the limit as napproaches infinity, we get:
lim
n→∞ an= lim
n→∞
1
n3
2(√n+1+√n)= lim
n→∞
1
n3
2(2√n)= lim
n→∞
1
2n2= 0
Step 4: Apply the limit comparison test.
Since limn→∞ an= 0 and the series P∞
n=1 1
n2converges (by the p-series test
with p= 2 >1), we can conclude by the limit comparison test that the series
P∞
n=1
√n+1−√n
n3
2
converges.
Question 33
Question
Determine the convergence or divergence of the series P∞
n=1
n3+2
n4+n+1 .
Solution
To determine the convergence or divergence of the series, we can use the limit
comparison test. Let’s compare the given series with a simpler series whose
convergence is known.
Step 1: Find the simpler series
Consider the series P∞
n=1 1
n, which is a p-series with p= 1 and known to be
divergent.
Step 2: Compute the limit
We will compute the limit of the ratio of the two series:
L= lim
n→∞
n3+2
n4+n+1
1
n
Step 3: Simplify the expression
Simplify the expression within the limit by dividing both the numerator and
denominator by n4:
L= lim
n→∞
1+ 2
n3
1+ 1
n3+1
n4
1
n
L= lim
n→∞
1
n4+2
n3
1
n+1
n2+1
n3
L= lim
n→∞
0+0
0+0+0 = 0
Step 4: Conclusion
Since L= 0 and the series P∞
n=1 1
nis divergent, by the limit comparison test,
the given series P∞
n=1
n3+2
n4+n+1 also diverges.
22
Question 34
Question
Determine the convergence of the series ∞
X
n=1
n!
nn.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn. We will compute the limit:
L= lim
n→∞
an+1
an
Step 2: Calculate an+1 and an+1
an
.
an+1 =(n+ 1)!
(n+ 1)n+1
=(n+ 1)n!
(n+ 1)n·(n+ 1)
=n!
(n+ 1)n
an+1
an
=
n!
(n+1)n
n!
nn
=nn
(n+ 1)n
=n
n+ 1n
Step 3: Evaluate the limit L.
L= lim
n→∞ n
n+ 1n
= lim
n→∞
1
1 + 1
nn
=1
e
Step 4: Determine the convergence of the series. Since L=1
e<1, by the
ratio test, the series ∞
X
n=1
n!
nnconverges.
23
Question 35
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: Apply the Ratio Test. Let an=n!
nn. We consider the limit:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Step 2: Simplify the expression.
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1) ·nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + 1
n−n
=1
e
Step 3: Analyze the limit. Since 1
e<1, by the Ratio Test, the series ∞
X
n=1
n!
nn
converges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
24
