Show $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$

Question

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: $$\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}$$

Answer 1

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

$$\begin{align} \frac12\lim_{t\to 0}\left(\frac\pi t \frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}-\frac{1}{t^2}\right)&=\frac12\lim_{t\to 0}\left(\frac\pi t \frac{1+1-2\pi t+2\pi^2t^2+O(t^3)}{2\pi t-2\pi^2t^2-\frac43 \pi^3t^3+O(t^4)}-\frac{1}{t^2}\right)\\\\ &=\frac12\lim_{t\to 0}\left(\frac{\frac23 \pi^3t^2+O(t^3)}{2\pi t^2\left(1+O(t)\right)}\right)\\\\ &=\pi^2/6 \end{align} $$

Answer 2

Note we can write:

$$\frac{\pi}{t}-\frac{1}{t^2}\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{\pi}{t}+\frac{1}{t}\frac{2\pi e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{\pi}{t}\left(1+\frac{2e^{-2\pi t}}{1-e^{-2\pi t}}\right)=\frac{\pi}{t}\frac{1-e^{-2\pi t}+2e^{-2\pi t}}{1-e^{-2\pi t}}\\=\frac{\pi}{t}\frac{1+e^{-2\pi t}}{1-e^{-2\pi t}}=\sum_{n=-\infty}^{\infty}\frac{1}{t^2+n^2}=\frac{1}{t^2}+2\sum_{n=1}^{\infty}\frac{1}{t^2+n^2}=\frac{1}{t^2}+2\sum_{n=1}^{\infty}\frac{1/n^2}{(t/n)^2+1}\\=\frac{1}{t^2}-\frac{2}{t^2}\sum_{n=1}^{\infty}\frac{(ti/n)^2}{(ti/n)^2-1}=\frac{1}{t^2}-\frac{2}{t^2}\sum_{n=1}^{\infty}\left(\sum_{k=1}^{\infty}\frac{(ti)^{2k}}{n^{2k}}\right)=\frac{1}{t^2}-\frac{2}{t^2}\sum_{k=1}^{\infty}\left(\sum_{n=1}^{\infty}\frac{(ti)^{2k}}{n^{2k}}\right)\\\implies \frac{\pi}{t}-\frac{1}{t^2}\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\frac{1}{t^2}-\frac{2}{t^2}\sum_{k=1}^{\infty}(ti)^{2k}\zeta(2k)=\frac{1}{t^2}-\frac{2}{t^2}\sum_{k=1}^{\infty}t^{2k}(-1)^k\zeta(2k)\\\implies 1-\pi t+\frac{(-2\pi t)e^{-2\pi t}}{1-e^{-2\pi t}}=\sum_{k=1}^{\infty}t^{2k}\left(2(-1)^k\zeta(2k)\right)$$

Thus if we let $\frac{ze^{z}}{1-e^{z}}=f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}z^k$ for appropriately chosen $z$, we get that:

$$1-\pi t+\sum_{k=0}^{\infty}\frac{f^{(k)}(0)(-2\pi)^k}{k!}t^k=1-\pi t+f(-2\pi t)=\sum_{k=1}^{\infty}t^{2k}\left(2(-1)^k\zeta(2k)\right)$$

However since the summands on the right hand-side consist of only those terms at even indices we must have that all the terms of odd index on the left hand-side vanish as well, which gives us:

$$\sum_{k=0}^{\infty}\frac{f^{(2k)}(0)(2\pi)^{2k}}{(2k)!}t^{2k}=\sum_{k=1}^{\infty}t^{2k}\left(2(-1)^k\zeta(2k)\right)\implies \frac{f^{(2n)}(0)(-1)^n(2\pi)^{2n}}{2(2n)!}=\zeta(2n)$$

For all $n\in \mathbb{N}$ and thus in particular at $n=1$ we get $f^{(2)}(0)=-\frac{1}{6}$ so that $\zeta(2)=\frac{\pi^2}{6}$.