User remainder theorem to estimate $ \sum^{\infty}_{k=1} \frac{12(-1)^{k+1}}{k^2} $ within $\frac{2}{5}$ I am trying to use the alternate series test to estimate:
$ \sum^{\infty}_{k=1} \frac{12(-1)^{k+1}}{k^2}  $
The question asks: 
By the alternating series test, which of the following is known to be an estimate of L to within an error of $\frac{2}{5}$? 
choices: 10, 9, 12, 11, 31/3.
In doing the work, I did not get an answer. The first terms of the series are:
12 - 3 + $\frac{4}{3} - \frac{3}{4} + \frac{12}{25} - \frac{1}{3} + \frac{12}{49}$. Based on the remainder theorem for alternating series, I need to add up the first 5 terms, which gives me a value of 10.06333 and the remainder should be less than the absolute value of the 6th term (1/3). However, my answer does not match any of the multiple choice possibilities above. Did I do something wrong? 
 A: In order to estimate
$$ \eta(2) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} $$
we may quote Euler. He noticed (in his third or fourth proof of the Basel problem, $\approx 1741$) that we may simply exploit the trigonometric integral
$$ \int_{0}^{1}\frac{\arcsin(x)}{\sqrt{1-x^2}}\,dx = \frac{1}{2}\arcsin^2(1) = \frac{\pi^2}{8}. $$
Since the Taylor series of $\arcsin(x)$ and $\frac{1}{\sqrt{1-x^2}}$ in a neighbourhood of the origin are well-known, we have:
$$\begin{align*}
\frac{\pi^2}{6}=\frac{4}{3}\int_0^1\frac{\arcsin x}{\sqrt{1-x^2}}\,dx&=\frac{4}{3}\int_0^1\frac{x+\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}}{\sqrt{1-x^2}}\,dx\\
&=\frac{4}{3}\int_0^1\frac{x}{\sqrt{1-x^2}}\,dx
+\frac{4}{3}\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}\int_0^1x^{2n}\frac{x}{\sqrt{1-x^2}}\,dx\\
&=\frac{4}{3}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!(2n+1)}\left[\frac{(2n)!!}{(2n+1)!!}\right]\\
&=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\\
&=\frac{4}{3}\left(\sum_{n=1}^{\infty}\frac{1}{n^2}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}\right)\\
&=\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2).\end{align*}$$
As a consequence, $\eta(2)=\frac{1}{2}\zeta(2)=\color{red}{\large\frac{\pi^2}{12}}$ can be approximated through the Taylor series of the squared arcsine:

$$ \eta(2)=\frac{3}{2}\sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}} $$

Since the terms of the last series decay pretty fast to zero, by considering just the first five terms of such series we get an approximation of $\eta(2)$ with an error term less than $10^{-4}$:

$$ \eta(2)=\frac{\pi^2}{12}\approx\color{red}{\frac{9211}{11200}}.$$

A: Let the remainder of your alternating sum $R$,$$R=\sum^{\infty}_{k=8} \frac{12(-1)^{k+1}}{k^2}$$By the correct use of remainder theorem on alternating series, it is true that $|R|\le \frac{12}{8^2}=0.1875$, here $\frac{12}{8^2}$ is the absolute value of the first term of remaining sum.
Can you make a better estimate from now on?
