Show $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ I know there are various methods showing that $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$, but I want to know how to derive it from letting $t\rightarrow 0^{+}$ for the following identity:
$$\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$$

Note that we have
$$\sum_{n=-\infty}^\infty \frac{1}{t^2+n^2}=\frac1{t^2}+2\sum_{n=1}^\infty\frac{1}{t^2+n^2}$$
Therefore, using $\sum_{n=-\infty}^\infty \frac{1}{t^2+n^2}=\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$, we find that 
$$\sum_{n=1}^\infty\frac{1}{t^2+n^2}=\frac12\left(\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}-\frac{1}{t^2}\right) \tag 1$$
The limit of the left-hand side of $(1)$ is the series of interest, $\sum_{n=1}^\infty\frac{1}{n^2}$.  The limit of the term on the right-hand side is 
$$\begin{align}
\frac12\lim_{t\to 0}\left(\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}-\frac{1}{t^2}\right)&=\frac12\lim_{t\to 0}\left(\frac\pi t \frac{1+1-2\pi t+2\pi^2t^2+O(t^3)}{2\pi t-2\pi^2t^2-\frac43 \pi^3t^3+O(t^4)}-\frac{1}{t^2}\right)\\\\
&=\frac12\lim_{t\to 0}\left(\frac{\frac23 \pi^3t^2+O(t^3)}{2\pi t^2\left(1+O(t)\right)}\right)\\\\
&=\pi^2/6
\end{align} $$ 
