$$\sum_\limits{n=1}^\infty\frac{(-1)^{n+1}}{n^6}$$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:
$\endgroup$ 4$$\vert R_5 \vert = \vert s - s_5 \vert \le b_6$$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?
2 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$Virtually, we can get the accurate result of the series sum.$$\text{First, we should know that }\zeta(6)=\sum_{n=1}^\infty \dfrac{1}{n^6}=\dfrac{\pi^6}{945}$$$$\text{Denote }A=\sum_{n=1}^\infty\dfrac{1}{(2n-1)^6}, B= \sum_{n=1}^\infty\dfrac{1}{(2n)^6}\text{ ,where $A$,$B$ are convegent.}$$Since $$A+B=\sum_{n=1}^\infty \dfrac{1}{n^6}=\dfrac{\pi^6}{945}$$$$B=\dfrac{\sum_{n=1}^\infty \dfrac{1}{n^6}}{64}=\dfrac{\pi^6}{60480}$$$$\sum_\limits{n=1}^\infty\dfrac{(-1)^{n+1}}{n^6}=A+B-2B=\dfrac{31\pi^6}{30240}$$
$\endgroup$
