Sum of Reciprocals of Squares Alternating in Sign/Proof 1

Proof
Let $f \left({x}\right)$ be the real function defined on $\left({0 \,.\,.\, 2 \pi}\right)$ as:


 * $f \left({x}\right) = \begin{cases}

\left({x - \pi}\right)^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end{cases}$

From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:


 * $(1): \quad f \left({x}\right) \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos n x} {n^2} + \left({\frac {\left({-1}\right)^n \pi} n + \frac {2 \left({\left({-1}\right)^n - 1}\right)} {\pi n^3} }\right) \sin n x}\right)$

We have that:

where $f \left({\pi - 0}\right)$ and $f \left({\pi + 0}\right)$ denote the limit from the left and limit from the right respectively of $f \left({\pi}\right)$.

It is apparent that $f \left({x}\right)$ satisfies the Dirichlet conditions:
 * $(\mathrm D 1): \quad f$ is bounded on $\left({0 \,.\,.\, 2 \pi}\right)$


 * $(\mathrm D 2): \quad f$ has a finite number of local maxima and local minima.


 * $(\mathrm D 3): \quad f$ has $1$ of discontinuity, which is finite.

Hence from Fourier's Theorem:

Thus setting $x = \pi$ in $(1)$: