<s>[INST] Deriving formula from Fourier series: $\frac{\pi^2}{12} = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ The equation/formula $$ \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$ is to be derived.
I know that the Fourier expansion of $f(x)=x$ for $x \in (-\pi,\pi)$ is $$f(x)=x=\sum_{n=1}^{\infty} \frac{2\,(-1)^{n+1}}{n}\,\sin{nx}$$ and there was an example of how for $x=\frac{\pi}{2}$ you'd get $\frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}- \cdots = \sum_{l=0}^\infty \frac{(-1)^{l}}{2l+1}$, but here it's not the square of $n$. I got a Fourier series for $$f_1(x)=x^2=\sum_{n=1}^{\infty} \frac{4\,(-1)^n}{n^2}\,\cos{nx}$$ but then the exponent of the $(-1)$ isn't the same. 
I'm obviously missing something, so any hints are appreciated.
 [/INST] 
This does not directly answer the question posed in the OP regarding the use of Fourier Series to prove that 
$$\frac{\pi^2}{12}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} \tag 1$$
But, I thought it might be instructive to present an approach that relies only on the Basel Problem
$$\frac{\pi^2}{6}=\sum_{n=1}^\infty \frac{1}{n^2} \tag 2$$
which was proven by Euler without use of Fourier Series.  To that end, we proceed.

Note that we can write the series in $(1)$ as 
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right) \tag 3$$
Alongside this result, we can write the series in $(2)$ as 
$$\begin{align}
\frac{\pi^2}{6}&=\sum_{n=1}^\infty \frac{1}{n^2} \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}+\frac{1}{(2n)^2}\right) \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)+2\sum_{n=1}^\infty\frac{1}{(2n)^2} \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)+\frac12\sum_{n=1}^\infty\frac{1}{n^2} \\\\
&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)+\frac{\pi^2}{12}\\\\
\frac{\pi^2}{12}&=\sum_{n=1}^\infty \left(\frac{1}{(2n-1)^2}-\frac{1}{(2n)^2}\right)\tag 4
\end{align}$$
Substituting $(4)$ into $(3)$ yields the coveted result.
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