Calculus 2 : Alternating Series

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #51 : Types Of Series

Determine if the following series is convergent or divergent: 

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

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 \(\displaystyle (-1)^{n}\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

\(\displaystyle b_{n}=\frac{e^{n}}{n}\)

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{e^{n}}{n} \rightarrow \infty\)

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

Example Question #52 : Types Of Series

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

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

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 \(\displaystyle (-1)^{n}\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

\(\displaystyle b_{n}=\frac{n^{7}+6}{n^{7}+8}\)

1.

\(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{n^{7}+6}{n^{7}+8} = \lim_{n \rightarrow \infty}\frac{n^{7}}{n^{7}} = \lim_{n \rightarrow \infty}\frac{1}{1} = 1\)

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.

Example Question #231 : Series In Calculus

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

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

Possible Answers:

Inconclusive according to the alternate series test

Divergent according to the ratio test

Divergent according to the alternate series test

Convergent 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 \(\displaystyle (-1)^{n}\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

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

1.\(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

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

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

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

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

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

Since the 2 tests pass, this series is convergent.

Example Question #11 : Alternating Series

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

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

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:

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 \(\displaystyle cos(n\pi)\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

\(\displaystyle b_{n}=\frac{10}{n^{4}}\)

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{10}{n^{4}} =0\)

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

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

\(\displaystyle b_{n}'=-40\frac{1}{n^5}\)

\(\displaystyle b_{1}'=-40\)

Since the 2 tests pass, this series is convergent.

Example Question #54 : Types Of Series

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

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

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 \(\displaystyle cos(n\pi)\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

 \(\displaystyle b_{n}=\frac{e^{2n}}{n^{3}}\)

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{e^{2n}}{n^{3}} \rightarrow \infty\)

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

Example Question #54 : Types Of Series

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

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

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 \(\displaystyle cos(n\pi)\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

\(\displaystyle b_{n}=\frac{n^3+n^2}{e^{3n}}\)

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{n^3+n^2}{e^{3n}} \rightarrow 0\)

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

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

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

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

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

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

Since the 2 tests pass, this series is convergent.

Example Question #55 : Types Of Series

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

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

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 \(\displaystyle cos(n\pi)\)

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

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

1.\(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

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

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

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

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

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

Since the 2 tests pass, this series is convergent.

Example Question #61 : Types Of Series

Determine how many terms need to be added to approximate the following series within \(\displaystyle \frac{1}{1000}\)

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

Possible Answers:

\(\displaystyle 10\)

\(\displaystyle 7\)

\(\displaystyle 8\)

\(\displaystyle 9\)

Correct answer:

\(\displaystyle 7\)

Explanation:

This is an alternating series test.

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

Step 1:

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

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

\(\displaystyle b_{n}=\frac{10}{n!}\)

\(\displaystyle 1. \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{10}{n!} =0\)

2. {\(\displaystyle b_{n}\)} is a decreasing sequence

\(\displaystyle b_{1} = \frac{10}{1},b_{2}=\frac{10}{2}, b_{3} = \frac{10}{3*2}...\)

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 \(\displaystyle b_{n+1} < \frac{1}{1000}\)

\(\displaystyle b_{7} = \frac{1}{504} > \frac{1}{1000}\)

\(\displaystyle b_{8} = \frac{1}{4032} < \frac{1}{1000}\)

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

 

Example Question #3021 : Calculus Ii

Determine how many terms need to be added to approximate the following series within \(\displaystyle \frac{1}{1000}\)

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

Possible Answers:

\(\displaystyle 3\)

\(\displaystyle 5\)

\(\displaystyle 4\)

\(\displaystyle 6\)

Correct answer:

\(\displaystyle 4\)

Explanation:

This is an alternating series test.

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

Step 1:

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

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

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

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

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

2. {\(\displaystyle b_{n}\)} 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 \(\displaystyle b_{n+1} < \frac{1}{1000}\)

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

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

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

Example Question #19 : Alternating Series

Determine how many terms need to be added to approximate the following series within \(\displaystyle \frac{1}{1000}\)

\(\displaystyle \sum_{n=1}^{\infty} cos(n\pi)\frac{13}{n^5}\)

Possible Answers:

\(\displaystyle 9\)

\(\displaystyle 7\)

\(\displaystyle 6\)

\(\displaystyle 8\)

Correct answer:

\(\displaystyle 7\)

Explanation:

This is an alternating series test.

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

Step 1:

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

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

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

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

  1.        \(\displaystyle \lim_{n \rightarrow \infty} b_{n} = 0\)
  2.        {\(\displaystyle b_{n}\)} is a decreasing sequence, or in other words \(\displaystyle b_{n=1}' < 0\)

Solution:

\(\displaystyle b_{n}=\frac{13}{n^5}\)

1. \(\displaystyle \lim_{n \rightarrow \infty}b_{n}\)

\(\displaystyle \lim_{n \rightarrow \infty}\frac{13}{n^5} =0\)

2.\(\displaystyle b_{n}'=(\frac{13}{n^5})'\)

\(\displaystyle b_{n}'=13(\frac{1}{n^5})' = -13(\frac{(n^5)'}{n^{10}})\)

\(\displaystyle b_{n}'=-13*5\frac{n^4}{n^{10}} = -65\frac{1}{n^6}\)

\(\displaystyle b_{1}'=-65\frac{1}{1^6} = -65 < 0\)

Since the 2 tests pass, this series is convergent.

Step 2:

Plug in n values until \(\displaystyle b_{n+1} < \frac{1}{1000}\)

\(\displaystyle b_{6}=\frac{13}{6^5} = \frac{13}{7776} > \frac{1}{1000}\)

\(\displaystyle b_{7}=\frac{13}{7^5} = \frac{13}{16807} < \frac{1}{1000}\)

7 terms are needed to approximate the sum within .001

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