Deriving formula from Fourier series: $\frac{\pi^2}{12} = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$

We have that: $$ \sum_{n\geq 1}\frac{2(-1)^{n+1}}{n}\,\sin(nx) $$ is the Fourier series of the $2\pi$-periodic extension of $f(x)=x$, so, by termwise integration, $$\begin{eqnarray*} \forall x\in(-\pi,\pi),\qquad x^2&=&\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}\left(1-\cos(nx)\right)\\&=&\color{blue}{\left(\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}\right)}-\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}\cos(nx)\end{eqnarray*}$$ and it follows that the blue term is the mean value of the function $g(x)=x^2$ over the interval $(-\pi,\pi)$, so: $$\color{blue}{\sum_{n\geq 1}\frac{4(-1)^{n+1}}{n^2}}=\frac{1}{2\pi}\int_{-\pi}^{\pi}x^2\,dx = \color{blue}{\frac{\pi^2}{3}}$$ and

$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}=\color{red}{\frac{\pi^2}{12}}$$

readily follows.

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.

I believe you have an error in the Fourier series for the function $ f(x)=x^2 $.

Recall that the real Fourier series can be written

$$ \frac{a_0}2+\sum_{n=1}^\infty[a_n\cos(nx)+b_n\sin(nx) ] $$

Where $$ a_n=\frac{1}\pi \int_{-\pi}^\pi f(x)\cos(nx)dx $$

$$ b_n=\frac{1}\pi \int_{-\pi}^\pi f(x)\sin(nx)dx $$

For even functions $f(-x)=f(x), b_n=0 $, for all $n\in\mathbb{N} $.

For $f(x)=x^2$, $a_0$ is not equal to $ 0 $, as you have:

$$ a_0=\frac{1}\pi \int_{-\pi}^\pi x^2dx = \frac{1}{3\pi}[x^3]_{-\pi}^{\pi}=\frac{2}{3}\pi^2 $$

Hence (and by using your correct formula for $a_n$ for positive $n$):

$$f(x)=x^2=\frac{\pi^2}3+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\cos(nx) $$

Now, evaluate this expression at $x=0$ to get the formula for $\frac{\pi^2}{12}$:

$$0=\frac{\pi^2}3+4\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2} $$

$$\Rightarrow \frac{\pi^2}{12}=-\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} $$