What is the value of the series $\sum\limits_{n=1}^\infty \frac{(-1)^n}{n^2}$? I am given the following series:
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$$ 
I have used the alternating series test to show that the series converges. 
However, how do I go about showing what it converges to? 
 A: As shown in this answer,
$$
\begin{align}
\zeta(2)
&=\sum_{k=1}^\infty\frac1{k^2}\\
&=\frac{\pi^2}{6}
\end{align}
$$
Now consider
$$
\begin{align}
&\hphantom{=\,}\frac1{1^2}\color{#C00000}{-\frac1{2^2}}+
\frac1{3^2}\color{#C00000}{-\frac1{4^2}}+
\frac1{5^2}\color{#C00000}{-\frac1{6^2}}+\dots\\
&=\frac1{1^2}+\frac1{2^2}+
\frac1{3^2}+\frac1{4^2}+
\frac1{5^2}+\frac1{6^2}+\dots\\
&-2\left(\,\,\hphantom{+}\frac1{2^2}\hphantom{+\frac1{3^2}\;}+\frac1{4^2}\hphantom{+\frac1{3^2}\:}+\frac1{6^2}+\dots\right)
\end{align}
$$
Now note that the stuff in the parentheses is $\frac14$ of the sum above it.
A: You may be expected to use the fact, not easy to prove, that the sum $\displaystyle\sum_1^\infty \frac{1}{n^2}$ is equal to $\dfrac{\pi^2}{6}$. 
The result was first proved by Euler. There are many proofs. In the usual undergraduate curriculum, a student is most likely to meet a proof when first dealing with Fourier series. 
Then note that the sum of the squares of the even terms of that series is $\frac{1}{4}$ of the full sum.
That should be enough to evaluate the alternating sum.   
A: Consider the Fourier series of $g(x)=x^2$ for $-\pi<x\le\pi$:
$$g(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(nx)+b_n\sin(nx)$$
note $b_n=0$ for an even function $g(t)=g(-t)$ and that:
$$a_n=\frac {1}{\pi} \int _{-\pi }^{\pi }\!{x}^{2}\cos \left( nx \right) {dx}
=4\,{\frac { \left( -1 \right) ^{n}}{{n}^{2}}},$$
$$\frac{a_0}{2}=\frac {1}{2\pi} \int _{-\pi }^{\pi }\!{x}^{2} {dx}
=\frac{1}{3}\pi^2,$$
$$x^2=\frac{1}{3}\,{\pi }^{2}+\sum _{n=1}^{\infty }4\,{\frac { \left( -1 \right) ^{n
}\cos \left( nx \right) }{{n}^{2}}},$$
$$x=0\rightarrow \frac{1}{3}\,{\pi }^{2}+\sum _{n=1}^{\infty }4\,{\frac { \left( -1 \right) ^{n
}}{{n}^{2}}}=0,$$
$$\sum _{n=1}^{\infty }{\frac { \left( -1 \right) ^{n}
}{{n}^{2}}}=-\frac{1}{12}\,{\pi }^{2}.$$
A: The series converges absolutely
$
\sum\limits_{n{\rm  = 1}}^\infty  {\dfrac{1}{{n^2 }}}  = \zeta\left( 2 \right) = \dfrac{{\pi ^2 }}{6}
$
the given series, to a greater extent, converges to
$
\sum\limits_{n=1}^\infty  {\dfrac{{\left( { - 1} \right)^{n + 1} }}{{n^2 }}}  = \dfrac{{\pi ^2 }}{{12}}
$
Fourier cosine series for $f(t)=\dfrac{\pi^2}{12}-\dfrac{t^2}{4}$ gives
$
\dfrac{{\pi ^2 }}{{12}} - \dfrac{{t^2 }}{4} = \sum\limits_{n = 1}^\infty  {\left( { - 1} \right)^{n + 1} } \dfrac{{\cos \left( {nt} \right)}}{{n^2 }}
$
plugging $t=0$ we get
$
 \sum\limits_{n = 1}^\infty  {\left( { - 1} \right)^{n + 1} } \dfrac{1}{n^2 } = \dfrac{{\pi ^2 }}{{12}} 
$
