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Calculus 2 : Types of Series

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Example Questions

Example Question #8 : Alternating Series

Determine whether the series is convergent or divergent:

$\sum_{n=0}^{\infty }(-1)^n\left(\frac{n^3}{3n^3+n^2+1}\right)$

Possible Answers:

The series is divergent.

The series is (absolutely) convergent.

The series is conditionally convergent.

The series may be convergent, divergent, or conditionally convergent.

Correct answer:

The series is divergent.

Explanation:

To determine whether this alternating series converges or diverges, we must use the Alternating Series test, which states that for the series

$\sum a_{n}$ and $a_{n}=(-1)^n(b_{n})$ ,

where $b_{n}\geq n$ for all n, if $\lim_{n\rightarrow \infty }b_{n}=0$ and ${b_{n}}$ is a decreasing sequence, then the series is convergent.

First, we must identify ${b_{n}}$ , which is $\frac{n^3}{3n^3+n^2+1}$ . When we take the limit of ${b_{n}}$ as n approaches infinity, we get

$\lim_{n \to \infty }\frac{n^3}{3n^3+n^2+1}=\lim_{n \to \infty }\left(\frac{n^3}{n^3}\right)\left(\frac{1}{3+n^{-1}+n^{-3}}\right)=\frac{1}{3}\neq0$

Notice that for the limit, the negative power terms go to zero, so we are left with something that does not equal zero.

Thus, the series is divergent because the test fails.

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Example Question #1 : Alternating Series

Determine whether the series converges or diverges:

$\sum_{n=0}^{\infty }(-1)^{n+1}\left(\frac{n^4}{2n^4+1}\right)$

Possible Answers:

The series is conditionally convergent.

The series is divergent.

The series is (absolutely) convergent.

The series may be convergent, divergent, or conditionally convergent.

Correct answer:

The series is divergent.

Explanation:

To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for

$\sum a_{n}$ - and $a_{n}=(-1)^{n+1}b_{n}$ where $b_{n} \geq 0$ for all n - to converge,

$\lim_{n\rightarrow \infty }b_{n}$ must equal zero and $b_{n}$ must be a decreasing series.

For our series,

$\lim_{n\rightarrow \infty }b_{n}=\lim_{n\rightarrow \infty }\frac{n^4}{2n^4+1}=\frac{1}{2}$

because it behaves like

$\lim_{n\rightarrow \infty }\left(\frac{1}{2}\right)\frac{n^4}{n^4}=\frac{1}{2}$ .

The test fails because $\frac{1}{2}\neq0$ so we do not need to check the second condition of the test.

The series is divergent.

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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 #11 : Alternating 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

Divergent according to ratio test

Inconclusive according to the alternating series 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 #12 : Alternating 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:

Divergent according to the ratio test

Inconclusive according to the alternating series test

Divergent according to the alternating series 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 $(-1)^{n}$

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 #14 : Alternating 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:

Convergent according to the alternating series test

Divergent according to the ratio test

Inconclusive according to the alternating series test

Divergent 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 #15 : Alternating 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:

Convergent according to the alternating series test

Divergent according to the alternating series test

Inconclusive according to the alternating series test

Divergent according to the ratio 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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Calculus 2 : Types of Series

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #8 : Alternating Series

Example Question #1 : Alternating Series

Example Question #1 : Alternating Series

Example Question #11 : Alternating Series

Example Question #12 : Alternating Series

Example Question #233 : Series In Calculus

Example Question #14 : Alternating Series

Example Question #15 : Alternating Series

Example Question #54 : Types Of Series

Example Question #55 : Types Of Series

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