$$\sum_\limits{n=1}^\infty\frac{(-1)^{n+1}}{n^6}$$
With the error being $\vert \mathbf{error} \vert \lt \ .00005$
In order for the series to undergo the Alternating Series Estimation Theorem
According to the James Stewart Textbook Essential Calculus Early Transcendentals Second Edition states that the theorem goes like this:
Theorem
If $s = \sum (-1)^{n-1}b_n$ is the sum of the an alternating series that satisfies
$$(\mathbf{i})\ \ \ \ 0 \le b_{n+1} \le b_n \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\mathbf{ii}) \ \lim_\limits{n \to \infty} b_n = 0$$
then
$$\vert R_n \vert = \vert s – s_n \vert \le b_{n+1}$$
My first step was then to verify the first condition which was then as follows:
$$b_n = \frac{1}{n^6}$$
$$b_{n+1} = \frac{1}{(n+1)^6}$$
$$0 \le \frac{1}{(n+1)^6} \le \frac{1}{n^6}$$
Which proved out to be true in this case. From there I went for the second condition and verified the limit.
$$\lim_\limits{n \to \infty} \frac{1}{n^6}= 0$$
Afterwards from verifying these two conditions I then began the procedure to find the error.
$$\sum_\limits{n=1}^\infty\frac{(-1)^{n+1}}{n^6}= 1 – \frac{1}{64}+ \frac{1}{729} – \frac{1}{4096}+ \frac{1}{15625} – \frac{1}{46656}$$
At this term the error turns out to be:
$$b_6 \lt .00005$$
$$.00002 \lt .00005$$
Therefore I concluded this statement:
$$\vert R_5 \vert = \vert s – s_5 \vert \le b_6$$
Boiling it down to the following:
$\vert s – 0.98557 \vert \le .00002$
From here I get a bit lost with absolute value but using Wolfram Alpha solution. If I break that absolute value and add the $s_5$ to both sides I get the approximation is that right? Or because I would have to split the inequality into a positive and negative side?
Best Answer
The process is right, though it is not really mentioned why you chose to stop there. Note that we have:
$$\frac1{n^6}<0.00005\iff n^6>20000\iff n>\sqrt[6]{20000}\simeq5.2$$
Hence taking the 6th term as the error term would have sufficed.
Without absolute value bars, we have:
$$0.98555\le s\le0.98559$$
As a sidenote, the actual value is given by:
$$s=0.98555109\dots$$
and that only 4 terms are required for the desired accuracy.