Let $$ \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^\infty \frac{1}{(n+a)^2} - \frac{1}{a^2} \right) \right) = \sum_{n=1}^\infty \frac{1}{n^2}$$
I already know that $\sum_{n=-\infty}^\infty \frac{1}{(n+a)^2} = \frac{\pi^2}{\sin^2(\pi a)}$. With that in mind, we know that the series converges in for every $a\in\mathbb{R}$ (Is that right?)
Also, all terms are positive so the series converges absolutely. Hence,
$$= \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^{-1} \frac{1}{(n+a)^2} \right) + \left( \sum_{n=-1}^{\infty} \frac{1}{(n+a)^2} \right) + \frac{1}{a^2} - \frac{1}{a^2} \right) = \lim_{a\to 0} \frac{1}{2} \left( \left( \sum_{n=-\infty}^{-1} \frac{1}{(n+a)^2} \right) + \left( \sum_{n=-1}^{\infty} \frac{1}{(n+a)^2} \right) \right) = \\ \frac{1}{2} \left( \left( \sum_{n=1}^{\infty} \frac{1}{(-n)^2} \right) + \left( \sum_{n=1}^{\infty} \frac{1}{n^2} \right) \right) = \\ \sum_{n=1}^\infty \frac{1}{n^2}$$
Now, how do I justify the continuity of the series which enabled me to do this algebraic trick?
Since $$\sum_{n\in\mathbb{Z}}\frac{1}{(n+a)^2}=\frac{\pi^2}{\sin^2(\pi a)}$$ as a function of $a$, is an even meromorphic function with a double pole in zero, as the square of a meromorphic function with a simple pole in zero: $$ \frac{\pi}{\sin(\pi z)}=\frac{1}{z}+\zeta(2) z+\ldots $$ it happens that: $$ \lim_{z\to 0}\left(-\frac{1}{z^2}+\frac{\pi^2}{\sin^2(\pi z)}\right)=\lim_{z\to 0}\left(-\frac{1}{z^2}+\frac{1}{z^2}+2\zeta(2)+\ldots\right)=2\zeta(2)$$ as expected.