MATH 332 - ADVANCED CALCULUS
- Convergence of series
Question Bank - Set 2
Liberty University
Question 1
Question
Determine whether the series P∞
n=1 n2+1
2n3+3 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+1
2n3+3 , we can use the Limit
Comparison Test. Let’s consider the series bn=1
n.
Step 1: Find the limit of the ratio of the given series to bn:We
need to calculate the following limit:
L= lim
n→∞
n2+1
2n3+3
1
n
Step 2: Simplify the expression: Simplifying the ratio, we get:
L= lim
n→∞
n3+n
2n3+ 3 = lim
n→∞
1 + 1
n2
2 + 3
n3
=1
2
Thus, the limit L=1
2.
Step 3: Apply the Limit Comparison Test: Since L > 0 and finite,
and the series P∞
n=1 1
nis a p-series where p= 1 (which converges), by the Limit
Comparison Test, the given series P∞
n=1 n2+1
2n3+3 converges as well.
Question 2
Question
Determine if the series ∞
X
n=1
n2+ 3n+ 1
n3+ 5
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
n2+ 3n+ 1
n3+ 5
and a comparison 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+ 3n+ 1)/n3+ 5
1/n .
Step 2: Simplify the limit Simplify the expression by dividing both the
numerator and the denominator by nand taking the limit:
lim
n→∞
1+3/n + 1/n2
1/n = lim
n→∞
n2+ 3n+ 1
n3= 0.
Step 3: Apply the Limit Comparison Test Since the limit is finite
and positive, we can apply the Limit Comparison Test. Since P∞
n=1 1/n is a
divergent p-series with p= 1, and our limit is non-zero, our original series also
diverges.
Therefore, the series
∞
X
n=1
n2+ 3n+ 1
n3+ 5
diverges.
Question 3
Question
Determine whether the series P∞
n=1 n2
3nconverges or diverges.
2
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. We will compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
Step 2: Simplify the expression
R= lim
n→∞
(n+ 1)2
3n2
R= lim
n→∞
n2+ 2n+ 1
3n2= lim
n→∞
1 + 2
n+1
n2
3=1
3
Step 3: Determine convergence Since R=1
3<1, by the ratio test, the
series P∞
n=1 n2
3nconverges.
Therefore, the series P∞
n=1 n2
3nconverges.
Question 4
Question
Prove whether the series ∞
X
n=1
nn−1
n!converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
nn−1
n!, we will use the ratio test.
Step 1: Apply the ratio test. Let an=nn−1
n!. We will calculate the limit of
the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)n
(n+ 1)! ·n!
nn−1
Step 2: Simplify the ratio.
= lim
n→∞
(n+ 1)n
nn−1·(n+ 1)
= lim
n→∞
(n+ 1)n
nn·1
n+ 1
= lim
n→∞
n+ 1
n·n+ 1
n·1
n+ 1
= lim
n→∞
n+ 1
n
2
= 1
Step 3: Analyze the limit. Since the limit is 1 and not less than 1, the
series diverges by the ratio test.
Therefore, the series ∞
X
n=1
nn−1
n!diverges.
3
Question 5
Question
Determine the convergence of the series P∞
n=1 n!
nnusing the Ratio Test.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
Let an=n!
nn.
Step 1: Compute the ratio R.
R= 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
Step 2: Simplify the expression.
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1 + 1
n−n
=1
e
Step 3: Evaluate the result. Since R=1
e<1, by the Ratio Test, the series
P∞
n=1 n!
nnconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
4
Solution
To analyze the convergence of the series, we can use the ratio test. Let an=n!
nn.
Step 1: Apply the ratio test. Compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
.
Step 3: Further simplify.
L= lim
n→∞
nn
(n+ 1)n
.
Step 4: Evaluate the limit. Rewrite the limit as:
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=1
e= 0.
Step 5: Draw a conclusion. Since L= 0, the series diverges by the ratio
test. Therefore, the series
∞
X
n=1
n!
nn
diverges.
Question 7
Question
Determine whether the series ∞
X
n=1
n!
(2n)!
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test. Step 1:
Compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(2(n+ 1))!
n!/(2n)!
= lim
n→∞
(n+ 1)!
(2n+ 2)! ·(2n)!
n!
5
= lim
n→∞
n+ 1
(2n+ 1)(2n+ 2)
= lim
n→∞
1
4
=1
4
Step 2: Evaluate the limit of the ratio, R. If R < 1, the series converges.
If R > 1 or R= 1, the series diverges. Since R=1
4<1, by the ratio test, the
series ∞
X
n=1
n!
(2n)!
converges.
Question 8
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. Let’s denote the general term of the series as an=n!
nn.
Step 1: Calculate the ratio. Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= 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
= lim
n→∞
1
(n+ 1)n
= 0.
Step 2: Analyze the ratio. Since the limit of the ratio is less than 1, by
the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
6
Question 9
Question
Determine whether the series ∞
X
n=1
n!
nnconverges.
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:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplify the expression:
lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
lim
n→∞
(n+ 1)nn
(n+ 1)n·(n+ 1)
lim
n→∞
nn
(n+ 1)n−1
Step 2: Simplify the limit expression. Divide both the numerator and
denominator by nn:
lim
n→∞
1
(1 + 1/n)n−1
By applying limit properties, we get:
1
e
=1
e
Step 3: Draw a conclusion. Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 10
Question
Determine whether the series ∞
X
n=1
n2
2nconverges or diverges.
7
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We will consider the limit
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 2: Simplify the limit.
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n+1
= lim
n→∞
1 + 2
n+1
n2
2
Step 3: Evaluate the limit. Taking the limit as napproaches infinity, we
have
lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 4: Analyze the result. Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n2
2nconverges.
Therefore, the series ∞
X
n=1
n2
2nconverges.
Question 11
Question
Consider the series P∞
n=1
(−1)n+1n2
n3+1 . Determine whether the series converges or
diverges.
Solution
To determine the convergence of the series P∞
n=1
(−1)n+1n2
n3+1 , we will use the
Alternating Series Test.
Step 1: Determine the terms of the series The terms of the series are
an=(−1)n+1n2
n3+1 .
Step 2: Check for the conditions of the Alternating Series Test We
need to show that the terms ansatisfy the following conditions: (i) an≥0 for
all n(ii) limn→∞ an= 0 (iii) anis decreasing
Step 3: Show the conditions are satisfied (i) For all n∈N, we have
n2
n3+1 ≥0. (ii) We find limn→∞
(−1)n+1n2
n3+1 = limn→∞
(−1)n+1
n+1 = 0 (iii) To show
that n2
n3+1 is decreasing, we can consider the derivative d
dn n2
n3+1 , which is
negative for all n≥1.
8
Since all conditions of the Alternating Series Test are met, we can conclude
that the series P∞
n=1
(−1)n+1n2
n3+1 converges.
Question 12
Question
Determine whether the series ∞
X
n=1
2n
n!converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
2n
n!, we can use the ratio test.
Step 1: Compute the ratio R: Let an=2n
n!. Then, the ratio Ris given by:
R= lim
n→∞
an+1
an
= lim
n→∞
2n+1/(n+ 1)!
2n/n!
= lim
n→∞
2n+1 ·n!
(n+ 1)! ·2n
= lim
n→∞
2
n+ 1 = 0
Step 2: Determine the convergence of the series: - If R < 1, the series
∞
X
n=1
2n
n!converges absolutely. - If R > 1 or R=∞, the series diverges. - If
R= 1, the ratio test is inconclusive, and we need to use another test.
Since R= 0 <1, the series ∞
X
n=1
2n
n!converges by the ratio test.
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nnby using the
ratio test.
Solution
Let’s apply the ratio test to the series P∞
n=1 n!
nn.
Step 1: Compute the ratio R:
R= 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
9
Step 2: Simplify the expression inside the limit:
R= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
n
n+ 1n+1
= lim
n→∞
1−1
n+ 1n+1
Step 3: Consider the limit of the exponential term:
lim
n→∞ 1−1
n+ 1n+1
=1
e
where eis Euler’s number.
Step 4: Determine the convergence or divergence based on the ratio R:
Since R=1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n2+3n
2n4+5 .
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will find a suitable series to compare with. Let’s consider the
series P∞
n=1 1
n2.
Step 2: Calculate the limit limn→∞
n2+3n
2n4+5
1
n2
.
lim
n→∞
n2+3n
2n4+5
1
n2
= lim
n→∞
n4+ 3n3
2n2+ 5n2= lim
n→∞
n4+ 3n3
7n2
Step 3: Simplify the limit calculation.
lim
n→∞
n4+ 3n3
7n2= lim
n→∞
n2(n2+ 3n)
7n2= lim
n→∞
n2+ 3n
7
Step 4: Find the limit.
lim
n→∞
n2+ 3n
7=∞
Since the limit diverges to infinity, we can conclude that the given series
P∞
n=1 n2+3n
2n4+5 also diverges by the limit comparison test with P∞
n=1 1
n2.
10
Question 15
Question
Determine whether the series P∞
n=1 n2+3n+1
n4+5n2+4 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n+1
n4+5n2+4 , we will use the limit
comparison test.
Step 1: Find a comparable series. Let’s consider the series P∞
n=1 1
n2.
Step 2: Compute the limit. Calculate the limit:
lim
n→∞
n2+3n+1
n4+5n2+4
1
n2
= lim
n→∞
n2+ 3n+ 1
n2(n2+5+ 4
n2)
Step 3: Simplify the limit. Simplifying, we get:
lim
n→∞
n2+ 3n+ 1
n2(n2+5+ 4
n2)= lim
n→∞
1 + 3
n+1
n2
(1 + 5
n+4
n2)
=1
1= 1
Step 4: Apply the limit comparison test. Since the limit is equal to a
finite positive number, by the limit comparison test, the series P∞
n=1 n2+3n+1
n4+5n2+4
converges if and only if the series P∞
n=1 1
n2converges.
Step 5: Conclusion. The series P∞
n=1 1
n2is a p-series with p= 2,
which converges. Therefore, by the limit comparison test, the given series
P∞
n=1 n2+3n+1
n4+5n2+4 also converges.
Question 16
Question
Determine if the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we can use the ratio test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. We calculate
the ratio lim
n→∞
an+1
an
.
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!
11
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 2: Apply L’Hopital’s Rule to the limit:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 3: Evaluate the result of the limit: Since the limit is less than 1, by
the ratio test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 17
Question
Determine whether the series ∞
X
n=1
n2
2nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n2
2n. We will consider the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
= lim
n→∞
(n+ 1)2
2n2
Step 2: Determine the limit.
L= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
12
Step 3: Evaluate the result. Since L=1
2<1, by the Ratio Test, the series
∞
X
n=1
n2
2nconverges.
Question 18
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 Com-
parison Test.
Step 1: Let’s find a series that we can compare with our given series.
Consider the series P∞
n=1 n2
n3. This series is a p-series with p= 1, and it diverges.
Notice that for n≥1,
n2+ 1
n3+ 2 ≤n2
n3.
Step 2: Now we need to compare the given series with the chosen series.
Since for all n,n2+1
n3+2 ≤n2
n3, we have
∞
X
n=1
n2+ 1
n3+ 2 ≤∞
X
n=1
n2
n3.
Step 3: Determine convergence/divergence using the Comparison Test.
Since the series P∞
n=1 n2
n3is a p-series with p= 1 and diverges, and since
P∞
n=1 n2+1
n3+2 is less than or equal to P∞
n=1 n2
n3, by the Comparison Test, the
series P∞
n=1 n2+1
n3+2 also diverges.
Therefore, the given series P∞
n=1 n2+1
n3+2 diverges.
Question 19
Question
Determine the convergence of the series:
∞
X
n=1
1
n3−2n+ 1
13
Solution
To determine the convergence of the series, we can use the Comparison Test
with the Harmonic series.
Step 1: Determine the convergence of the Harmonic series The
Harmonic series is given by P∞
n=1 1
n. This series is known to diverge.
Step 2: Simplify the given series We want to compare the given series
to the Harmonic series, so we simplify the terms:
1
n3−2n+ 1 =1
(n−1)(n−1)n
Step 3: Find an upper bound for the series terms For n≥2, we have:
1
(n−1)(n−1)n<1
(n−1)3=1
(n−1)2·1
n−1<1
n2
Step 4: Apply the Comparison Test Since P∞
n=1 1
n2is a convergent
p-series with p= 2 >1, and P∞
n=1 1
nis a divergent Harmonic series, we can
conclude that: ∞
X
n=1
1
n3−2n+ 1
converges by the Comparison Test.
Question 20
Question
Determine whether the series P∞
n=1 3n+1
4n−5converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 3n+1
4n−5, we can use the Limit
Comparison Test. Let’s compare it to a series that we know the convergence of.
Step 1: Choose a comparison series.
Consider the series P∞
n=1 3n
4n=P∞
n=1 3
4=3
4P∞
n=1 1 = 3
4P∞
n=1. The series
P∞
n=1 3
4is a convergent geometric series.
Step 2: Take the limit of the ratio of the given series to the comparison
series.
Let an=3n+1
4n−5and bn=3
4.
We will consider the limit limn→∞
an
bn= limn→∞
3n+1
4n−5
3
4
.
Step 3: Simplify and compute the limit.
limn→∞
an
bn= limn→∞
3n+1
4n−5·4
3= limn→∞
4(3n+1)
3(4n−5) = limn→∞
12n+4
12n−15 = 1.
Step 4: Make a conclusion based on the Limit Comparison Test.
Since limn→∞
an
bn= 1 and the comparison series P∞
n=1 3
4is convergent, by the
Limit Comparison Test, we conclude that the series P∞
n=1 3n+1
4n−5also converges.
14
Question 21
Question
Determine whether the series
∞
X
n=1
n2+ 2
3n3+ 1
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n2+ 2
3n3+ 1,
we will use the Limit Comparison Test. Let’s compare this series with a simpler
series to determine its convergence.
Step 1: Find a simpler series. Consider the series
∞
X
n=1
1
n.
This is a well-known series that diverges (harmonic series).
Step 2: Use the Limit Comparison Test. We will compute the following
limit
lim
n→∞
n2+2
3n3+1
1
n
= lim
n→∞
n3+ 2n
3n3+ 1 .
Step 3: Simplify the expression.
lim
n→∞
n3+ 2n
3n3+ 1 = lim
n→∞
1 + 2
n2
3 + 1
n3
=1+0
3+0 =1
3.
Step 4: Conclusion. Since the limit above is a finite positive number, by
the Limit Comparison Test, the original series
∞
X
n=1
n2+ 2
3n3+ 1
converges.
Question 22
Question
Determine the convergence of the series P∞
n=1 n!
nn.
15
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn, and consider the limit
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
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
n
= lim
n→∞
1
1+1/n
= 1
Step 3: Analyze the limit. Since the limit is 1, the Ratio Test is inconclusive.
Step 4: Determine the convergence. The Ratio Test is inconclusive, so we
will need to use another test. Notice that n!
nnis non-negative for n≥1.
Step 5: Apply the Limit Comparison Test. Let’s compare the series P∞
n=1 n!
nn
with the series P∞
n=1 1
nwhich is a p-series with p= 1.
Step 6: Calculate the limit.
lim
n→∞
n!
nn
1
n
= lim
n→∞
n!
nn−1= lim
n→∞
n!
nn−1=∞
Step 7: Analyze the limit. Since the limit is not finite, the Limit Compari-
son Test is inconclusive.
Step 8: Determine the convergence. The Limit Comparison Test is incon-
clusive. This series does not satisfy the conditions of other convergence tests we
might apply directly. Therefore, we can conclude that the convergence of the
series P∞
n=1 n!
nncannot be determined by the tests we have used.
Question 23
Question
Determine the convergence of the series P∞
n=1 n3+5n+1
n4+2n2+1 .
16
Solution
To determine the convergence of the series, we will use the limit comparison
test. We will compare the given series to a known series whose convergence is
easier to determine.
Step 1: Find a suitable series to compare to.
Consider the series P∞
n=1 1
n. This is a p-series with p= 1, and we know that
P∞
n=1 1
ndiverges.
Step 2: Take the limit of the ratio of the terms.
Let an=n3+5n+1
n4+2n2+1 and bn=1
n. We will analyze the limit limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n3+ 5n+ 1
n4+ 2n2+ 1 ·n
1
Step 3: Simplify the expression and compute the limit.
= lim
n→∞
n4+ 5n2+n
n4+ 2n2+ 1 = lim
n→∞
1+5/n2+ 1/n3
1+2/n2+ 1/n4
Now, as napproaches infinity, the higher order terms dominate, so the limit
simplifies to:
=1+0+0
1+0+0 = 1
Step 4: Apply the Limit Comparison Test.
Since the limit is a finite positive number, the Limit Comparison Test states
that either both series converge or both series diverge. Since we know that
P∞
n=1 1
ndiverges, the original series P∞
n=1 n3+5n+1
n4+2n2+1 also diverges.
Therefore, the series P∞
n=1 n3+5n+1
n4+2n2+1 diverges.
Question 24
Question
Determine whether the series
∞
X
n=1
n2+ sin(n)
n3+√n
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Step 1: Let’s choose a simpler series to compare with. We will consider
the series ∞
X
n=1
n2
n3=∞
X
n=1
1
n
17
which is a p-series with p= 1 and is known to diverge. Step 2: We will now
calculate the limit
L= lim
n→∞
n2+sin(n)
n3+√n
1
n
= lim
n→∞
n3+nsin(n)
n3+√n
Step 3: By dividing the leading terms of the numerator and the denominator,
we get
L= lim
n→∞
n3
n3= 1
Step 4: Since Lis a finite positive number and the comparison series diverges,
by the Limit Comparison Test, we conclude that the given series
∞
X
n=1
n2+ sin(n)
n3+√n
also diverges.
Question 25
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
5n4+n3+ 1.
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
an=∞
X
n=1
n2
5n4+n3+ 1.
and the series ∞
X
n=1
bn=∞
X
n=1
1
n2.
Step 1: Find the limit of anbn.We will find the limit of an
bnas n
approaches infinity.
lim
n→∞
an
bn
= lim
n→∞
n2
5n4+n3+ 1 ·n2
1= lim
n→∞
n4
5n4+n3+ 1
18
Now, since the degree of the numerator and denominator are the same, we can
divide the leading terms to simplify the expression:
n4
5n4+n3+ 1 ∼n4
5n4=1
5.
Step 2: Analyze the limit. Since 0 <1
5<∞, this implies that the series
P∞
n=1 bnis a divergent p-series with p= 2.
Step 3: Make a conclusion. By the Limit Comparison Test, since the di-
vergent series P∞
n=1 bnis a divergent p-series, and an
bnis a finite positive number,
the given series P∞
n=1 andiverges as well. Therefore, the series P∞
n=1 n2
5n4+n3+1
diverges.
Question 26
Question
Determine whether the series ∞
X
n=1
n2+ 1
n3+ 1 converges or diverges.
Solution
To determine the convergence of the series, we need to examine the limit of the
general term n2+1
n3+1 as napproaches infinity.
Step 1: Find the limit of the general term.
lim
n→∞
n2+ 1
n3+ 1 = lim
n→∞
1 + 1
n2
n+1
n2
=1+0
∞+ 0 = 0
Step 2: Apply the Limit Comparison Test. Since limn→∞
n2+1
n3+1 = 0, we can
compare the given series with the series ∞
X
n=1
1
n, which is a p-series with p= 1.
Step 3: Compare the two series. We have:
n2+ 1
n3+ 1 <n2+n2
n3=2n2
n3=2
n
Step 4: Apply the Comparison Test. Since ∞
X
n=1
2
nis a divergent p-series
with p= 1, and the given series is less than this divergent series, the given
series ∞
X
n=1
n2+ 1
n3+ 1 also diverges by comparison test.
Therefore, the series ∞
X
n=1
n2+ 1
n3+ 1 diverges.
19
Question 27
Question
Determine if the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Apply the ratio test.
Consider the limit:
L= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Find the limit L.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 3: Analyze the limit.
Since L= 1, the ratio test is inconclusive.
Step 4: Use another test.
The series P∞
n=1 n!
nnis similar to the series P∞
n=1 1
nwhich diverges by the
p-series test with p= 1. Therefore, the series P∞
n=1 n!
nnalso diverges.
Hence, the series P∞
n=1 n!
nndiverges.
Question 28
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n2+ 2n
n3+ 3n+ 1.
20
Solution
To determine the convergence of the series, we can use the limit comparison
test. We will compare the given series to a known convergent series.
Step 1: Find a known convergent series Let’s consider the series
P∞
n=1 1
n. This is the harmonic series, which is known to diverge.
Step 2: Determine the limit We will calculate the limit of the ratio
between the general term of the given series and the general term of the harmonic
series:
lim
n→∞
n2+2n
n3+3n+1
1
n
= lim
n→∞
n3+ 2n2
n3+ 3n+ 1.
Step 3: Simplify the limit Dividing by the highest power of n, we get:
lim
n→∞
n3+ 2n2
n3+ 3n+ 1 = lim
n→∞
1 + 2
n
1 + 3
n+1
n3
=1
1= 1.
Step 4: Interpret the limit Since the limit is a finite positive number,
by the limit comparison test, the given series P∞
n=1 n2+2n
n3+3n+1 has the same con-
vergence behavior as the harmonic series, which diverges. Therefore, the given
series also diverges.
Question 29
Question
Determine the convergence or divergence of the series
∞
X
n=1
n+ 1
n2+ 3.
Solution
To determine the convergence or divergence of the series, we will use the Com-
parison Test.
Step 1: We write an=n+1
n2+3 .
Step 2: First, we observe that as napproaches infinity, the term n+1
n2+3
behaves like n
n2=1
n.
Step 3: We know that the harmonic series P∞
n=1 1
ndiverges. Thus, we
compare our series to the harmonic series by considering
lim
n→∞
an
1
n
= lim
n→∞
n+ 1
n2+ 3 ·n.
Step 4: Simplifying the limit, we get
lim
n→∞
n+ 1
n2+ 3 ·n= lim
n→∞
1 + 1
n
1 + 3
n2
= 1.
21
Step 5: Since the limit is a finite positive number, by the Comparison Test,
we conclude that the series P∞
n=1 n+1
n2+3 diverges.
Question 30
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1
√an
nalso converges.
Solution
Given that P∞
n=1 anis a convergent series with positive terms, we know that
limn→∞ an= 0 since the terms of the series anmust tend to zero for convergence
to occur.
We want to show that the series P∞
n=1
√an
nalso converges.
Step 1: Establish that √an
n≤anfor all n≥1.
Since anare all positive, we have an≥0 for all n≥1. Taking the square
root of both sides gives us √an≤anfor all n≥1. Dividing both sides by n
(where n≥1 is positive) gives us √an
n≤anfor all n≥1.
Step 2: Apply the comparison test to show convergence of P∞
n=1
√an
n.
Since √an
n≤anfor all n≥1, and P∞
n=1 anconverges, by the comparison
test, we have that P∞
n=1
√an
nconverges as well.
Therefore, the series P∞
n=1
√an
nconverges.
Question 31
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. 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
22
lim
n→∞
n!(n+ 1)
n!(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1
Step 3: Compute the limit.
lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 32
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
an
n.
Solution
To determine the convergence of the series P∞
n=1
an
n, we will use the Comparison
Test.
Step 1: Since P∞
n=1 anis convergent with positive terms, we have limn→∞ an=
0.
Step 2: Let’s consider the sequence bn=an
n. We need to show that Pbn
is convergent.
Step 3: By the limit comparison test, if limn→∞
bn
cn=L > 0, where Pcn
is a known convergent series, then Pbnis also convergent.
Step 4: Let’s choose cn=1
n, a known divergent series.
Since limn→∞
bn
cn= limn→∞
an/n
1/n = limn→∞ an= 0, we have L= 0.
Step 5: Therefore, by the limit comparison test, the series P∞
n=1
an
ncon-
verges.
Question 33
Question
Determine whether the series ∞
X
n=1
n2+ 3
n3+ 2n−7converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test. Let’s consider the series ∞
X
n=1
n2+ 3
n3+ 2n−7and compare it with a
23
simpler series.
Step 1: Find a suitable series to compare with
We will compare the given series with the series ∞
X
n=1
1
n.
Step 2: Find the limit of the ratio
Let an=n2+3
n3+2n−7and bn=1
n. We want to find the limit of lim
n→∞
an
bn
.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3
n3+ 2n−7∇· 1
n= lim
n→∞
n2+ 3
n3+ 2n−7·n= lim
n→∞
n3+ 3n
n3+ 2n−7= 1
Step 3: Apply the Limit Comparison Test
Since the limit is a positive number (not zero or infinity), and ∞
X
n=1
1
nis a
divergent p-series with p= 1, by the Limit Comparison Test, the given series
∞
X
n=1
n2+ 3
n3+ 2n−7also diverges.
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To analyze the convergence or divergence of the series, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn, then compute
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn.
24
Step 2: Simplify the expression.
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!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn
(n+ 1)n(Simplify)
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e(Limit of (1 −1
n+ 1)nas n→ ∞ is 1/e).
Step 3: Determine the convergence. Since L=1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 35
Question
Let P∞
n=1 n!
nnbe an infinite series. Determine whether the series converges or
diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We calculate
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!= lim
n→∞
n+ 1
(1 + 1
n)n+1 .
Step 2: Apply the limit. We use the fact that limn→∞
n+1
(1+ 1
n)n+1 =e.
Therefore, the limit is equal to e.
Step 3: Analyze the result. Since e > 1, by the ratio test, the series
P∞
n=1 n!
nndiverges.
Hence, the series diverges.
25
Question 2
Question
Determine if the series ∞
X
n=1
n2+ 3n+ 1
n3+ 5
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Let’s consider the series
∞
X
n=1
n2+ 3n+ 1
n3+ 5
and a comparison 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+ 3n+ 1)/n3+ 5
1/n .
Step 2: Simplify the limit Simplify the expression by dividing both the
numerator and the denominator by nand taking the limit:
lim
n→∞
1+3/n + 1/n2
1/n = lim
n→∞
n2+ 3n+ 1
n3= 0.
Step 3: Apply the Limit Comparison Test Since the limit is finite
and positive, we can apply the Limit Comparison Test. Since P∞
n=1 1/n is a
divergent p-series with p= 1, and our limit is non-zero, our original series also
diverges.
Therefore, the series
∞
X
n=1
n2+ 3n+ 1
n3+ 5
diverges.
Question 3
Question
Determine whether the series P∞
n=1 n2
3nconverges or diverges.
2
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. We will compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
3n+1 ·3n
n2
Step 2: Simplify the expression
R= lim
n→∞
(n+ 1)2
3n2
R= lim
n→∞
n2+ 2n+ 1
3n2= lim
n→∞
1 + 2
n+1
n2
3=1
3
Step 3: Determine convergence Since R=1
3<1, by the ratio test, the
series P∞
n=1 n2
3nconverges.
Therefore, the series P∞
n=1 n2
3nconverges.
Question 4
Question
Prove whether the series ∞
X
n=1
nn−1
n!converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
nn−1
n!, we will use the ratio test.
Step 1: Apply the ratio test. Let an=nn−1
n!. We will calculate the limit of
the ratio:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)n
(n+ 1)! ·n!
nn−1
Step 2: Simplify the ratio.
= lim
n→∞
(n+ 1)n
nn−1·(n+ 1)
= lim
n→∞
(n+ 1)n
nn·1
n+ 1
= lim
n→∞
n+ 1
n·n+ 1
n·1
n+ 1
= lim
n→∞
n+ 1
n
2
= 1
Step 3: Analyze the limit. Since the limit is 1 and not less than 1, the
series diverges by the ratio test.
Therefore, the series ∞
X
n=1
nn−1
n!diverges.
3
Question 5
Question
Determine the convergence of the series P∞
n=1 n!
nnusing the Ratio Test.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the Ratio Test.
Let an=n!
nn.
Step 1: Compute the ratio R.
R= 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
Step 2: Simplify the expression.
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
1 + 1
n−n
=1
e
Step 3: Evaluate the result. Since R=1
e<1, by the Ratio Test, the series
P∞
n=1 n!
nnconverges.
Question 6
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
4
Solution
To analyze the convergence of the series, we can use the ratio test. Let an=n!
nn.
Step 1: Apply the ratio test. Compute the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
.
Step 2: Simplify the expression.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
.
Step 3: Further simplify.
L= lim
n→∞
nn
(n+ 1)n
.
Step 4: Evaluate the limit. Rewrite the limit as:
L= lim
n→∞ n
n+ 1n
= lim
n→∞ 1 + −1
n+ 1n
=1
e= 0.
Step 5: Draw a conclusion. Since L= 0, the series diverges by the ratio
test. Therefore, the series
∞
X
n=1
n!
nn
diverges.
Question 7
Question
Determine whether the series ∞
X
n=1
n!
(2n)!
converges or diverges.
Solution
To determine the convergence of the series, we will use the ratio test. Step 1:
Compute the ratio R:
R= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(2(n+ 1))!
n!/(2n)!
= lim
n→∞
(n+ 1)!
(2n+ 2)! ·(2n)!
n!
5
= lim
n→∞
n+ 1
(2n+ 1)(2n+ 2)
= lim
n→∞
1
4
=1
4
Step 2: Evaluate the limit of the ratio, R. If R < 1, the series converges.
If R > 1 or R= 1, the series diverges. Since R=1
4<1, by the ratio test, the
series ∞
X
n=1
n!
(2n)!
converges.
Question 8
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To determine the convergence or divergence of the series, we can use the ratio
test. Let’s denote the general term of the series as an=n!
nn.
Step 1: Calculate the ratio. Compute the ratio of consecutive terms:
lim
n→∞
an+1
an
= 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
= lim
n→∞
1
(n+ 1)n
= 0.
Step 2: Analyze the ratio. Since the limit of the ratio is less than 1, by
the ratio test, the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
6
Question 9
Question
Determine whether the series ∞
X
n=1
n!
nnconverges.
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:
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
Simplify the expression:
lim
n→∞
(n+ 1)!
n!·nn
(n+ 1)n·(n+ 1)
lim
n→∞
(n+ 1)nn
(n+ 1)n·(n+ 1)
lim
n→∞
nn
(n+ 1)n−1
Step 2: Simplify the limit expression. Divide both the numerator and
denominator by nn:
lim
n→∞
1
(1 + 1/n)n−1
By applying limit properties, we get:
1
e
=1
e
Step 3: Draw a conclusion. Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 10
Question
Determine whether the series ∞
X
n=1
n2
2nconverges or diverges.
7
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n2
2n. We will consider the limit
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
Step 2: Simplify the limit.
lim
n→∞
(n+ 1)2
2n+1 ·2n
n2
= lim
n→∞
n2+ 2n+ 1
2n2·2n
2n+1
= lim
n→∞
1 + 2
n+1
n2
2
Step 3: Evaluate the limit. Taking the limit as napproaches infinity, we
have
lim
n→∞
1 + 2
n+1
n2
2
=1
2
Step 4: Analyze the result. Since the limit is less than 1, by the ratio test,
the series ∞
X
n=1
n2
2nconverges.
Therefore, the series ∞
X
n=1
n2
2nconverges.
Question 11
Question
Consider the series P∞
n=1
(−1)n+1n2
n3+1 . Determine whether the series converges or
diverges.
Solution
To determine the convergence of the series P∞
n=1
(−1)n+1n2
n3+1 , we will use the
Alternating Series Test.
Step 1: Determine the terms of the series The terms of the series are
an=(−1)n+1n2
n3+1 .
Step 2: Check for the conditions of the Alternating Series Test We
need to show that the terms ansatisfy the following conditions: (i) an≥0 for
all n(ii) limn→∞ an= 0 (iii) anis decreasing
Step 3: Show the conditions are satisfied (i) For all n∈N, we have
n2
n3+1 ≥0. (ii) We find limn→∞
(−1)n+1n2
n3+1 = limn→∞
(−1)n+1
n+1 = 0 (iii) To show
that n2
n3+1 is decreasing, we can consider the derivative d
dn n2
n3+1 , which is
negative for all n≥1.
8
Since all conditions of the Alternating Series Test are met, we can conclude
that the series P∞
n=1
(−1)n+1n2
n3+1 converges.
Question 12
Question
Determine whether the series ∞
X
n=1
2n
n!converges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
2n
n!, we can use the ratio test.
Step 1: Compute the ratio R: Let an=2n
n!. Then, the ratio Ris given by:
R= lim
n→∞
an+1
an
= lim
n→∞
2n+1/(n+ 1)!
2n/n!
= lim
n→∞
2n+1 ·n!
(n+ 1)! ·2n
= lim
n→∞
2
n+ 1 = 0
Step 2: Determine the convergence of the series: - If R < 1, the series
∞
X
n=1
2n
n!converges absolutely. - If R > 1 or R=∞, the series diverges. - If
R= 1, the ratio test is inconclusive, and we need to use another test.
Since R= 0 <1, the series ∞
X
n=1
2n
n!converges by the ratio test.
Question 13
Question
Determine the convergence or divergence of the series P∞
n=1 n!
nnby using the
ratio test.
Solution
Let’s apply the ratio test to the series P∞
n=1 n!
nn.
Step 1: Compute the ratio R:
R= 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
9
Step 2: Simplify the expression inside the limit:
R= lim
n→∞
nn+1
(n+ 1)n+1
= lim
n→∞
n
n+ 1n+1
= lim
n→∞
1−1
n+ 1n+1
Step 3: Consider the limit of the exponential term:
lim
n→∞ 1−1
n+ 1n+1
=1
e
where eis Euler’s number.
Step 4: Determine the convergence or divergence based on the ratio R:
Since R=1
e<1, by the ratio test, the series P∞
n=1 n!
nnconverges.
Question 14
Question
Determine the convergence or divergence of the series P∞
n=1 n2+3n
2n4+5 .
Solution
To determine the convergence or divergence of the series, we will use the limit
comparison test.
Step 1: We will find a suitable series to compare with. Let’s consider the
series P∞
n=1 1
n2.
Step 2: Calculate the limit limn→∞
n2+3n
2n4+5
1
n2
.
lim
n→∞
n2+3n
2n4+5
1
n2
= lim
n→∞
n4+ 3n3
2n2+ 5n2= lim
n→∞
n4+ 3n3
7n2
Step 3: Simplify the limit calculation.
lim
n→∞
n4+ 3n3
7n2= lim
n→∞
n2(n2+ 3n)
7n2= lim
n→∞
n2+ 3n
7
Step 4: Find the limit.
lim
n→∞
n2+ 3n
7=∞
Since the limit diverges to infinity, we can conclude that the given series
P∞
n=1 n2+3n
2n4+5 also diverges by the limit comparison test with P∞
n=1 1
n2.
10
Question 15
Question
Determine whether the series P∞
n=1 n2+3n+1
n4+5n2+4 converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n2+3n+1
n4+5n2+4 , we will use the limit
comparison test.
Step 1: Find a comparable series. Let’s consider the series P∞
n=1 1
n2.
Step 2: Compute the limit. Calculate the limit:
lim
n→∞
n2+3n+1
n4+5n2+4
1
n2
= lim
n→∞
n2+ 3n+ 1
n2(n2+5+ 4
n2)
Step 3: Simplify the limit. Simplifying, we get:
lim
n→∞
n2+ 3n+ 1
n2(n2+5+ 4
n2)= lim
n→∞
1 + 3
n+1
n2
(1 + 5
n+4
n2)
=1
1= 1
Step 4: Apply the limit comparison test. Since the limit is equal to a
finite positive number, by the limit comparison test, the series P∞
n=1 n2+3n+1
n4+5n2+4
converges if and only if the series P∞
n=1 1
n2converges.
Step 5: Conclusion. The series P∞
n=1 1
n2is a p-series with p= 2,
which converges. Therefore, by the limit comparison test, the given series
P∞
n=1 n2+3n+1
n4+5n2+4 also converges.
Question 16
Question
Determine if the series ∞
X
n=1
n!
nnconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n!
nn, we can use the ratio test.
Step 1: Compute the ratio of consecutive terms: Let an=n!
nn. We calculate
the ratio lim
n→∞
an+1
an
.
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!
11
lim
n→∞
(n+ 1)!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
Step 2: Apply L’Hopital’s Rule to the limit:
lim
n→∞
nn
(n+ 1)n
= lim
n→∞
nn
(n+ 1)n= lim
n→∞ n
n+ 1n
= lim
n→∞ 1
1+1/nn
=1
e
Step 3: Evaluate the result of the limit: Since the limit is less than 1, by
the ratio test, the series ∞
X
n=1
n!
nnconverges.
Therefore, the series ∞
X
n=1
n!
nnconverges.
Question 17
Question
Determine whether the series ∞
X
n=1
n2
2nconverges or diverges.
Solution
To determine the convergence of the series ∞
X
n=1
n2
2n, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n2
2n. We will consider the limit:
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)2/2n+1
n2/2n
= lim
n→∞
(n+ 1)2
2n2
Step 2: Determine the limit.
L= lim
n→∞
(n+ 1)2
2n2
= lim
n→∞
n2+ 2n+ 1
2n2
= lim
n→∞
1 + 2
n+1
n2
2
=1
2
12
Step 3: Evaluate the result. Since L=1
2<1, by the Ratio Test, the series
∞
X
n=1
n2
2nconverges.
Question 18
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 Com-
parison Test.
Step 1: Let’s find a series that we can compare with our given series.
Consider the series P∞
n=1 n2
n3. This series is a p-series with p= 1, and it diverges.
Notice that for n≥1,
n2+ 1
n3+ 2 ≤n2
n3.
Step 2: Now we need to compare the given series with the chosen series.
Since for all n,n2+1
n3+2 ≤n2
n3, we have
∞
X
n=1
n2+ 1
n3+ 2 ≤∞
X
n=1
n2
n3.
Step 3: Determine convergence/divergence using the Comparison Test.
Since the series P∞
n=1 n2
n3is a p-series with p= 1 and diverges, and since
P∞
n=1 n2+1
n3+2 is less than or equal to P∞
n=1 n2
n3, by the Comparison Test, the
series P∞
n=1 n2+1
n3+2 also diverges.
Therefore, the given series P∞
n=1 n2+1
n3+2 diverges.
Question 19
Question
Determine the convergence of the series:
∞
X
n=1
1
n3−2n+ 1
13
Solution
To determine the convergence of the series, we can use the Comparison Test
with the Harmonic series.
Step 1: Determine the convergence of the Harmonic series The
Harmonic series is given by P∞
n=1 1
n. This series is known to diverge.
Step 2: Simplify the given series We want to compare the given series
to the Harmonic series, so we simplify the terms:
1
n3−2n+ 1 =1
(n−1)(n−1)n
Step 3: Find an upper bound for the series terms For n≥2, we have:
1
(n−1)(n−1)n<1
(n−1)3=1
(n−1)2·1
n−1<1
n2
Step 4: Apply the Comparison Test Since P∞
n=1 1
n2is a convergent
p-series with p= 2 >1, and P∞
n=1 1
nis a divergent Harmonic series, we can
conclude that: ∞
X
n=1
1
n3−2n+ 1
converges by the Comparison Test.
Question 20
Question
Determine whether the series P∞
n=1 3n+1
4n−5converges or diverges.
Solution
To determine the convergence of the series P∞
n=1 3n+1
4n−5, we can use the Limit
Comparison Test. Let’s compare it to a series that we know the convergence of.
Step 1: Choose a comparison series.
Consider the series P∞
n=1 3n
4n=P∞
n=1 3
4=3
4P∞
n=1 1 = 3
4P∞
n=1. The series
P∞
n=1 3
4is a convergent geometric series.
Step 2: Take the limit of the ratio of the given series to the comparison
series.
Let an=3n+1
4n−5and bn=3
4.
We will consider the limit limn→∞
an
bn= limn→∞
3n+1
4n−5
3
4
.
Step 3: Simplify and compute the limit.
limn→∞
an
bn= limn→∞
3n+1
4n−5·4
3= limn→∞
4(3n+1)
3(4n−5) = limn→∞
12n+4
12n−15 = 1.
Step 4: Make a conclusion based on the Limit Comparison Test.
Since limn→∞
an
bn= 1 and the comparison series P∞
n=1 3
4is convergent, by the
Limit Comparison Test, we conclude that the series P∞
n=1 3n+1
4n−5also converges.
14
Question 21
Question
Determine whether the series
∞
X
n=1
n2+ 2
3n3+ 1
converges or diverges.
Solution
To determine the convergence of the series
∞
X
n=1
n2+ 2
3n3+ 1,
we will use the Limit Comparison Test. Let’s compare this series with a simpler
series to determine its convergence.
Step 1: Find a simpler series. Consider the series
∞
X
n=1
1
n.
This is a well-known series that diverges (harmonic series).
Step 2: Use the Limit Comparison Test. We will compute the following
limit
lim
n→∞
n2+2
3n3+1
1
n
= lim
n→∞
n3+ 2n
3n3+ 1 .
Step 3: Simplify the expression.
lim
n→∞
n3+ 2n
3n3+ 1 = lim
n→∞
1 + 2
n2
3 + 1
n3
=1+0
3+0 =1
3.
Step 4: Conclusion. Since the limit above is a finite positive number, by
the Limit Comparison Test, the original series
∞
X
n=1
n2+ 2
3n3+ 1
converges.
Question 22
Question
Determine the convergence of the series P∞
n=1 n!
nn.
15
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the Ratio Test.
Step 1: Apply the Ratio Test. Let an=n!
nn, and consider the limit
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
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
n
= lim
n→∞
1
1+1/n
= 1
Step 3: Analyze the limit. Since the limit is 1, the Ratio Test is inconclusive.
Step 4: Determine the convergence. The Ratio Test is inconclusive, so we
will need to use another test. Notice that n!
nnis non-negative for n≥1.
Step 5: Apply the Limit Comparison Test. Let’s compare the series P∞
n=1 n!
nn
with the series P∞
n=1 1
nwhich is a p-series with p= 1.
Step 6: Calculate the limit.
lim
n→∞
n!
nn
1
n
= lim
n→∞
n!
nn−1= lim
n→∞
n!
nn−1=∞
Step 7: Analyze the limit. Since the limit is not finite, the Limit Compari-
son Test is inconclusive.
Step 8: Determine the convergence. The Limit Comparison Test is incon-
clusive. This series does not satisfy the conditions of other convergence tests we
might apply directly. Therefore, we can conclude that the convergence of the
series P∞
n=1 n!
nncannot be determined by the tests we have used.
Question 23
Question
Determine the convergence of the series P∞
n=1 n3+5n+1
n4+2n2+1 .
16
Solution
To determine the convergence of the series, we will use the limit comparison
test. We will compare the given series to a known series whose convergence is
easier to determine.
Step 1: Find a suitable series to compare to.
Consider the series P∞
n=1 1
n. This is a p-series with p= 1, and we know that
P∞
n=1 1
ndiverges.
Step 2: Take the limit of the ratio of the terms.
Let an=n3+5n+1
n4+2n2+1 and bn=1
n. We will analyze the limit limn→∞
an
bn.
lim
n→∞
an
bn
= lim
n→∞
n3+ 5n+ 1
n4+ 2n2+ 1 ·n
1
Step 3: Simplify the expression and compute the limit.
= lim
n→∞
n4+ 5n2+n
n4+ 2n2+ 1 = lim
n→∞
1+5/n2+ 1/n3
1+2/n2+ 1/n4
Now, as napproaches infinity, the higher order terms dominate, so the limit
simplifies to:
=1+0+0
1+0+0 = 1
Step 4: Apply the Limit Comparison Test.
Since the limit is a finite positive number, the Limit Comparison Test states
that either both series converge or both series diverge. Since we know that
P∞
n=1 1
ndiverges, the original series P∞
n=1 n3+5n+1
n4+2n2+1 also diverges.
Therefore, the series P∞
n=1 n3+5n+1
n4+2n2+1 diverges.
Question 24
Question
Determine whether the series
∞
X
n=1
n2+ sin(n)
n3+√n
converges or diverges.
Solution
To determine the convergence of the series, we will use the Limit Comparison
Test. Step 1: Let’s choose a simpler series to compare with. We will consider
the series ∞
X
n=1
n2
n3=∞
X
n=1
1
n
17
which is a p-series with p= 1 and is known to diverge. Step 2: We will now
calculate the limit
L= lim
n→∞
n2+sin(n)
n3+√n
1
n
= lim
n→∞
n3+nsin(n)
n3+√n
Step 3: By dividing the leading terms of the numerator and the denominator,
we get
L= lim
n→∞
n3
n3= 1
Step 4: Since Lis a finite positive number and the comparison series diverges,
by the Limit Comparison Test, we conclude that the given series
∞
X
n=1
n2+ sin(n)
n3+√n
also diverges.
Question 25
Question
Determine the convergence or divergence of the series
∞
X
n=1
n2
5n4+n3+ 1.
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
an=∞
X
n=1
n2
5n4+n3+ 1.
and the series ∞
X
n=1
bn=∞
X
n=1
1
n2.
Step 1: Find the limit of anbn.We will find the limit of an
bnas n
approaches infinity.
lim
n→∞
an
bn
= lim
n→∞
n2
5n4+n3+ 1 ·n2
1= lim
n→∞
n4
5n4+n3+ 1
18
Now, since the degree of the numerator and denominator are the same, we can
divide the leading terms to simplify the expression:
n4
5n4+n3+ 1 ∼n4
5n4=1
5.
Step 2: Analyze the limit. Since 0 <1
5<∞, this implies that the series
P∞
n=1 bnis a divergent p-series with p= 2.
Step 3: Make a conclusion. By the Limit Comparison Test, since the di-
vergent series P∞
n=1 bnis a divergent p-series, and an
bnis a finite positive number,
the given series P∞
n=1 andiverges as well. Therefore, the series P∞
n=1 n2
5n4+n3+1
diverges.
Question 26
Question
Determine whether the series ∞
X
n=1
n2+ 1
n3+ 1 converges or diverges.
Solution
To determine the convergence of the series, we need to examine the limit of the
general term n2+1
n3+1 as napproaches infinity.
Step 1: Find the limit of the general term.
lim
n→∞
n2+ 1
n3+ 1 = lim
n→∞
1 + 1
n2
n+1
n2
=1+0
∞+ 0 = 0
Step 2: Apply the Limit Comparison Test. Since limn→∞
n2+1
n3+1 = 0, we can
compare the given series with the series ∞
X
n=1
1
n, which is a p-series with p= 1.
Step 3: Compare the two series. We have:
n2+ 1
n3+ 1 <n2+n2
n3=2n2
n3=2
n
Step 4: Apply the Comparison Test. Since ∞
X
n=1
2
nis a divergent p-series
with p= 1, and the given series is less than this divergent series, the given
series ∞
X
n=1
n2+ 1
n3+ 1 also diverges by comparison test.
Therefore, the series ∞
X
n=1
n2+ 1
n3+ 1 diverges.
19
Question 27
Question
Determine if the series P∞
n=1 n!
nnconverges or diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we can use the ratio test.
Step 1: Apply the ratio test.
Consider the limit:
L= lim
n→∞
an+1
an
where an=n!
nn.
Step 2: Find the limit L.
L= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn
= lim
n→∞
(n+ 1)! ·nn
n!·(n+ 1)n+1
= lim
n→∞
(n+ 1)nn
(n+ 1)n+1
= lim
n→∞
nn
(n+ 1)n
= lim
n→∞
n
n+ 1
= 1
Step 3: Analyze the limit.
Since L= 1, the ratio test is inconclusive.
Step 4: Use another test.
The series P∞
n=1 n!
nnis similar to the series P∞
n=1 1
nwhich diverges by the
p-series test with p= 1. Therefore, the series P∞
n=1 n!
nnalso diverges.
Hence, the series P∞
n=1 n!
nndiverges.
Question 28
Question
Determine the convergence or divergence of the series:
∞
X
n=1
n2+ 2n
n3+ 3n+ 1.
20
Solution
To determine the convergence of the series, we can use the limit comparison
test. We will compare the given series to a known convergent series.
Step 1: Find a known convergent series Let’s consider the series
P∞
n=1 1
n. This is the harmonic series, which is known to diverge.
Step 2: Determine the limit We will calculate the limit of the ratio
between the general term of the given series and the general term of the harmonic
series:
lim
n→∞
n2+2n
n3+3n+1
1
n
= lim
n→∞
n3+ 2n2
n3+ 3n+ 1.
Step 3: Simplify the limit Dividing by the highest power of n, we get:
lim
n→∞
n3+ 2n2
n3+ 3n+ 1 = lim
n→∞
1 + 2
n
1 + 3
n+1
n3
=1
1= 1.
Step 4: Interpret the limit Since the limit is a finite positive number,
by the limit comparison test, the given series P∞
n=1 n2+2n
n3+3n+1 has the same con-
vergence behavior as the harmonic series, which diverges. Therefore, the given
series also diverges.
Question 29
Question
Determine the convergence or divergence of the series
∞
X
n=1
n+ 1
n2+ 3.
Solution
To determine the convergence or divergence of the series, we will use the Com-
parison Test.
Step 1: We write an=n+1
n2+3 .
Step 2: First, we observe that as napproaches infinity, the term n+1
n2+3
behaves like n
n2=1
n.
Step 3: We know that the harmonic series P∞
n=1 1
ndiverges. Thus, we
compare our series to the harmonic series by considering
lim
n→∞
an
1
n
= lim
n→∞
n+ 1
n2+ 3 ·n.
Step 4: Simplifying the limit, we get
lim
n→∞
n+ 1
n2+ 3 ·n= lim
n→∞
1 + 1
n
1 + 3
n2
= 1.
21
Step 5: Since the limit is a finite positive number, by the Comparison Test,
we conclude that the series P∞
n=1 n+1
n2+3 diverges.
Question 30
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove that the series
P∞
n=1
√an
nalso converges.
Solution
Given that P∞
n=1 anis a convergent series with positive terms, we know that
limn→∞ an= 0 since the terms of the series anmust tend to zero for convergence
to occur.
We want to show that the series P∞
n=1
√an
nalso converges.
Step 1: Establish that √an
n≤anfor all n≥1.
Since anare all positive, we have an≥0 for all n≥1. Taking the square
root of both sides gives us √an≤anfor all n≥1. Dividing both sides by n
(where n≥1 is positive) gives us √an
n≤anfor all n≥1.
Step 2: Apply the comparison test to show convergence of P∞
n=1
√an
n.
Since √an
n≤anfor all n≥1, and P∞
n=1 anconverges, by the comparison
test, we have that P∞
n=1
√an
nconverges as well.
Therefore, the series P∞
n=1
√an
nconverges.
Question 31
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. 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
22
lim
n→∞
n!(n+ 1)
n!(n+ 1)n+1
= lim
n→∞
n+ 1
(n+ 1)n+1
Step 3: Compute the limit.
lim
n→∞
n+ 1
(n+ 1)n+1
= lim
n→∞
1
(n+ 1)n
= 0
Step 4: Analyze the limit. Since the limit is less than 1, by the ratio test,
the series P∞
n=1 n!
nnconverges.
Therefore, the series P∞
n=1 n!
nnconverges.
Question 32
Question
Let P∞
n=1 anbe a convergent series with positive terms. Prove or disprove the
convergence of the series P∞
n=1
an
n.
Solution
To determine the convergence of the series P∞
n=1
an
n, we will use the Comparison
Test.
Step 1: Since P∞
n=1 anis convergent with positive terms, we have limn→∞ an=
0.
Step 2: Let’s consider the sequence bn=an
n. We need to show that Pbn
is convergent.
Step 3: By the limit comparison test, if limn→∞
bn
cn=L > 0, where Pcn
is a known convergent series, then Pbnis also convergent.
Step 4: Let’s choose cn=1
n, a known divergent series.
Since limn→∞
bn
cn= limn→∞
an/n
1/n = limn→∞ an= 0, we have L= 0.
Step 5: Therefore, by the limit comparison test, the series P∞
n=1
an
ncon-
verges.
Question 33
Question
Determine whether the series ∞
X
n=1
n2+ 3
n3+ 2n−7converges or diverges.
Solution
To determine the convergence of the given series, we will use the Limit Com-
parison Test. Let’s consider the series ∞
X
n=1
n2+ 3
n3+ 2n−7and compare it with a
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simpler series.
Step 1: Find a suitable series to compare with
We will compare the given series with the series ∞
X
n=1
1
n.
Step 2: Find the limit of the ratio
Let an=n2+3
n3+2n−7and bn=1
n. We want to find the limit of lim
n→∞
an
bn
.
lim
n→∞
an
bn
= lim
n→∞
n2+ 3
n3+ 2n−7∇· 1
n= lim
n→∞
n2+ 3
n3+ 2n−7·n= lim
n→∞
n3+ 3n
n3+ 2n−7= 1
Step 3: Apply the Limit Comparison Test
Since the limit is a positive number (not zero or infinity), and ∞
X
n=1
1
nis a
divergent p-series with p= 1, by the Limit Comparison Test, the given series
∞
X
n=1
n2+ 3
n3+ 2n−7also diverges.
Question 34
Question
Determine the convergence or divergence of the series
∞
X
n=1
n!
nn.
Solution
To analyze the convergence or divergence of the series, we will use the ratio test.
Step 1: Apply the ratio test. Let an=n!
nn, then compute
L= lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn.
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Step 2: Simplify the expression.
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!
(n+ 1)n+1 ·nn
n!
= lim
n→∞
nn
(n+ 1)n(Simplify)
= lim
n→∞ n
n+ 1n
= lim
n→∞ 1−1
n+ 1n
=1
e(Limit of (1 −1
n+ 1)nas n→ ∞ is 1/e).
Step 3: Determine the convergence. Since L=1
e<1, by the ratio test, the
series P∞
n=1 n!
nnconverges.
Therefore, the given series converges.
Question 35
Question
Let P∞
n=1 n!
nnbe an infinite series. Determine whether the series converges or
diverges.
Solution
To determine the convergence of the series P∞
n=1 n!
nn, we will use the ratio test.
Step 1: Compute the ratio of consecutive terms. Let an=n!
nn. We calculate
lim
n→∞
an+1
an
= lim
n→∞
(n+ 1)!/(n+ 1)n+1
n!/nn= lim
n→∞
(n+ 1)!nn
(n+ 1)n+1n!= lim
n→∞
n+ 1
(1 + 1
n)n+1 .
Step 2: Apply the limit. We use the fact that limn→∞
n+1
(1+ 1
n)n+1 =e.
Therefore, the limit is equal to e.
Step 3: Analyze the result. Since e > 1, by the ratio test, the series
P∞
n=1 n!
nndiverges.
Hence, the series diverges.
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