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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Alternating Series

Determine if the following series is convergent or divergent:

$\sum_{n=1}^{\infty} (-1)^{n}\frac{1}{n+n^2}$

Possible Answers:

Divergent according to the ratio test

Divergent according to the alternating series test

Inconclusive according to the alternating series test

Convergent according to the alternating series test

Correct answer:

Convergent according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $(-1)^{n}$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^{n} * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{1}{n+n^2}$

$\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{1}{n+n^2}=0$

$b_{n}'=(\frac{1}{n+n^2})'$

$b_{n}'=\frac{-(n^{2}+n)'}{(n^2+n)^2}$

$b_{n}'=\frac{-(2n+1)}{(n^2+n)^{2}}$

$b_{1}'=\frac{-(2+1)}{(1+1)^2}=\frac{-3}{4} < 0$

Since the 2 tests pass, this series is convergent.

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Example Question #51 : Types Of Series

Determine if the following series is convergent or divergent:

$\sum_{n=1}^{\infty} (-1)^{n}\frac{e^{n}}{n}$

Possible Answers:

Convergent according to the alternating series test

Inconclusive according to the alternating series test

Divergent according to ratio test

Divergent according to the alternating series test

Correct answer:

Inconclusive according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $(-1)^{n}$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^{n} * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{e^{n}}{n}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{e^{n}}{n} \rightarrow \infty$

Our tests stop here. Since the limit of $b_{n}$ goes to infinity, we can say that this function does not converge according to the alternating series test.

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Example Question #52 : Types Of Series

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} (-1)^{n}\frac{n^{7}+6}{n^{7}+8}$

Possible Answers:

Convergent according to the alternating series test

Inconclusive according to the alternating series test

Divergent according to the ratio test

Divergent according to the alternating series test

Correct answer:

Inconclusive according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $(-1)^{n}$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^{n} * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{n^{7}+6}{n^{7}+8}$

$\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{n^{7}+6}{n^{7}+8} = \lim_{n \rightarrow \infty}\frac{n^{7}}{n^{7}} = \lim_{n \rightarrow \infty}\frac{1}{1} = 1$

This concludes our testing. The limit does not equal to 0, so we cannot say that this series converges according to the alternate series test.

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Example Question #233 : Series In Calculus

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} (-1)^{n}\frac{8}{n^{3}+\sqrt{n}}$

Possible Answers:

Divergent according to the ratio test

Divergent according to the alternate series test

Convergent according to the alternate series test

Inconclusive according to the alternate series test

Correct answer:

Convergent according to the alternate series test

Explanation:

Conclusive according to the alternating series test

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $(-1)^{n}$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^{n} * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{8}{n^{3}+\sqrt(n)}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{8}{n^{3}+\sqrt(n)} =0$

2. $b_{n}'=(\frac{8}{n^{3}+\sqrt(n)})'$

$b_{n}'=8(\frac{1}{n^{3}+\sqrt{n}})' = 8(-\frac{(n^3+\sqrt{n})'}{(n^3+\sqrt{n})^2})$

$b_{n}'=-8\frac{3n^2+\frac{1}{2\sqrt{n}}}{(n^3+\sqrt{n})^2}$

$b_{1}'=-8\frac{3*1^2+\frac{1}{2\sqrt{1}}}{(1^3+\sqrt{1})^2} = -8\frac{3+\frac{1}{2}}{4} < 0$

Since the 2 tests pass, this series is convergent.

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Example Question #53 : Types Of Series

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} cos(n\pi)\frac{10}{n^{4}}$

Possible Answers:

Divergent according to the ratio test

Convergent according to the alternating series test

Divergent according to the alternating series test

Inconclusive according to the alternating series test

Correct answer:

Convergent according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{10}{n^{4}}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{10}{n^{4}} =0$

2. $b_{n}'=(\frac{10}{n^{4}})'$

$b_{n}'=10(\frac{1}{n^4})' = -10*4(\frac{1}{n^5})$

$b_{n}'=-40\frac{1}{n^5}$

$b_{1}'=-40$

Since the 2 tests pass, this series is convergent.

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Example Question #54 : Types Of Series

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} cos(n\pi)\frac{e^{2n}}{n^{3}}$

Possible Answers:

Divergent according to the alternating series test

Inconclusive according to the alternating series test

Divergent according to the ratio test

Convergent according to the alternating series test

Correct answer:

Inconclusive according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{e^{2n}}{n^{3}}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{e^{2n}}{n^{3}} \rightarrow \infty$

Conclude with tests. The limit of $b_{n}$ goes to infinity and so we cannot use the alternating series test to reach a convergence/divergence conclusion.

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Example Question #54 : Types Of Series

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} cos(n\pi) \frac{n^3+n^2}{e^{3n}}$

Possible Answers:

Divergent according to the alternating series test

Divergent according to the ratio test

Inconclusive according to the alternating series test

Convergent according to the alternating series test

Correct answer:

Convergent according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{n^3+n^2}{e^{3n}}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{n^3+n^2}{e^{3n}} \rightarrow 0$

2. $b_{n}'=(\frac{n^3+n^2}{e^{3n}})'$

$b_{n}'=e^{-3n}(n^3+n^2)'+(n^3+n^2)(e^{-3n})'$

$b_{n}'=e^{-3n}((n^3)' + (n^2)') + e^{-3n}(n^3+n^2)(-3n)'$

$b_{n}'= (3n^2+2n)e^{-3n} - 3e^{-3n}(n^3+n^2)$

$b_{n}'=\frac{-n(3n^2-2)}{e^{3n}}$

$b_{1}'=\frac{-(3*1^2-2)}{e^{3*1}} = - \frac{1}{e^3} < 0$

Since the 2 tests pass, this series is convergent.

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Example Question #55 : Types Of Series

Determine if the following series is convergent or divergent using the alternating series test:

$\sum_{n=1}^{\infty} cos(n\pi)\frac{7(n^{11}+n^7+n^3)}{n^{20}}$

Possible Answers:

Convergent according to the alternating series test

Divergent according to the alternating series test

Divergent according to the ratio test

Inconclusive according to the alternating series test

Correct answer:

Convergent according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of $cos(n\pi)$

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = cos(n\pi) * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{7(n^{11}+n^7+n^3)}{n^{20}}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{7(n^{11}+n^7+n^3)}{n^{20}} =0$

2. $b_{n}'=(\frac{7(n^{11}+n^7+n^3)}{n^{20}})'$

$b_{n}'=7(\frac{n^{11}+n^7+n^3}{n^{20}})'$

$b_{n}'= 7(\frac{n^{20}(n^{11}+n^7+n^3)' - (n^{20})'(n^11+n^7+n^3)}{n^{40}}$

$b_{n}'=\frac{7((11n^{10}+7n^{6}+3n^2)*n^{20} - 20n^{19}(n^{11}+n^7+n^3))}{n^{40}}$ $b_{1}'=\frac{7((11*1^{10}+7*1^{6}+3*1^2)*1^{20} - 20*1^{19}(1^{11}+1^7+1^3))}{1^{40}}$ $b_{1}'= \frac{7(11+7+3-20(1+1+1)}{1}=\frac{7(21-60)}{1} = -39*7 < 0$

Since the 2 tests pass, this series is convergent.

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Example Question #61 : Types Of Series

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

$\sum_{n=1}^{\infty} (-1)^n\frac{10}{n!}$

Possible Answers:

Correct answer:

Explanation:

This is an alternating series test.

In order to find the terms necessary to approximate the series within $\frac{1}{1000}$ first see if the series is convergent using the alternating series test. If the series converges, find n such that $b_{n+1} < \frac{1}{1000}$

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of (-1)^n

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^n * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{10}{n!}$

$1. \lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{10}{n!} =0$

2. { $b_{n}$ } is a decreasing sequence

$b_{1} = \frac{10}{1},b_{2}=\frac{10}{2}, b_{3} = \frac{10}{3*2}...$

Since the factorial is always increasing, the sequence is always decreasing.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{7} = \frac{1}{504} > \frac{1}{1000}$

$b_{8} = \frac{1}{4032} < \frac{1}{1000}$

so 7 terms are needed to approximate the sum within .001.

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Example Question #3021 : Calculus Ii

Determine how many terms need to be added to approximate the following series within $\frac{1}{1000}$ :

$\sum_{n=1}^{\infty} (-1)^n\frac{4}{(n!)^2}$

Possible Answers:

Correct answer:

Explanation:

This is an alternating series test.

Step 1:

An alternating series can be identified because terms in the series will “alternate” between + and –, because of (-1)^n

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

$a_{n} = (-1)^n * b_{n}$

$\lim_{n \rightarrow \infty} b_{n} = 0$
{ $b_{n}$ } is a decreasing sequence, or in other words $b_{n=1}' < 0$

Solution:

$b_{n}=\frac{4}{(n!)^2}$

1. $\lim_{n \rightarrow \infty}b_{n}$

$\lim_{n \rightarrow \infty}\frac{4}{(n!)^2} =0$

2. { $b_{n}$ } is a decreasing functon, since a factorial never decreases.

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until $b_{n+1} < \frac{1}{1000}$

$b_{4}=\frac{4}{(4!)^2}=\frac{1}{144} > \frac{1}{1000}$

$b_{5}=\frac{4}{(5!)^2}=\frac{1}{3600} < \frac{1}{1000}$

4 needs to be added to approximate the sum within $\frac{1}{1000}$ .

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Calculus 2 : Series in Calculus

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Alternating Series

Example Question #51 : Types Of Series

Example Question #52 : Types Of Series

Example Question #233 : Series In Calculus

Example Question #53 : Types Of Series

Example Question #54 : Types Of Series

Example Question #54 : Types Of Series

Example Question #55 : Types Of Series

Example Question #61 : Types Of Series

Example Question #3021 : Calculus Ii

All Calculus 2 Resources

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