Question

consider the following series.

$$ sum _ { n = 1 } ^ { infty } \frac { ( - 1 ) ^ { n + 1 } } { n ^ { 7 } } ( | \text { error } | < 0.00005 ) $$

show that the series is convergent.

since this series is an alternating series, which condition(s) below show that it converges? (select all that apply.)

$$ \begin{array} { l } { \frac { 1 } { ( n + 1 ) ^ { 7 } } > \frac { 1 } { n ^ { 7 } } } \\ { lim _ { n ightarrow infty } \frac { 1 } { ( n + 1 ) ^ { 7 } } = infty } \\ { lim _ { n ightarrow infty } \frac { 1 } { n ^ { 7 } } = 0 } \\ { \frac { 1 } { ( n + 1 ) ^ { 7 } } < \frac { 1 } { n ^ { 7 } } } end{array} $$

impressive work!

how many terms of the series do we need to add in order to find the sum to the indicated accuracy?

8 terms

enhanced feedback

please try again, keeping in mind that you can apply the alternating series test. first you can rewrite the nth term in the form $$ ( - 1 ) ^ { n - 1 } b _ { n } $$ where $$ b _ { n } > 0 $$. then, check that $$ b _ { n + 1 } leq b _ { n } $$ for all sufficiently large n. then, check that

$$ lim _ { n ightarrow infty } b _ { n } = 0 $$

. if both conditions are satisfied, the series is convergent; otherwise, the alternating series test fails. to use the alternating series estimation theorem to estimate how many terms are needed to achieve the desired accuracy, recall that $$ | r _ { n } | = | s - s _ { n } | leq b _ { n + 1 } $$. set $$ b _ { n + 1 } < 0.00005 $$ and solve for n.

resources

Explanation:

Step1: Recall the Alternating Series Estimation Theorem

For an alternating series \(\sum_{n = 1}^{\infty}(-1)^{n - 1}b_{n}\) (where \(b_{n}>0\)), the error \(\vert R_{n}\vert=\vert S - S_{n}\vert\leq b_{n + 1}\), where \(S=\sum_{n = 1}^{\infty}(-1)^{n - 1}b_{n}\) and \(S_{n}=\sum_{k = 1}^{n}(-1)^{k - 1}b_{k}\). In our series \(\sum_{n = 1}^{\infty}\frac{(-1)^{n+1}}{n^{5}}\), we have \(b_{n}=\frac{1}{n^{5}}\).

Step2: Set up the inequality for the error

We want \(\vert R_{n}\vert<0.00005\). By the Alternating - Series Estimation Theorem, \(\vert R_{n}\vert\leq b_{n+1}\). So we set \(b_{n + 1}=\frac{1}{(n + 1)^{5}}<0.00005\).

Step3: Solve the inequality for \(n\)

First, rewrite the inequality \(\frac{1}{(n + 1)^{5}}<0.00005\) as \((n + 1)^{5}>\frac{1}{0.00005}=20000\). Then take the fifth - root of both sides: \(n+1>\sqrt[5]{20000}\).
We know that \(7^{5}=16807\) and \(8^{5}=32768\).
If \(n + 1 = 8\), then \(n=7\).

Answer:

\(7\)