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? 

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}} 
$
