Sum of Reciprocals of Squares Alternating in Sign/Proof 1

Proof
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:


 * $\map f x = \begin {cases}

\paren {x - \pi}^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:


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

We have that:

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

It is apparent that $\map f x$ satisfies the Dirichlet conditions:
 * $(\text D 1): \quad f$ is bounded on $\openint 0 {2 \pi}$


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


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

Hence from Fourier's Theorem:

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