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

Question

The goal is to prove for equality: \begin{equation*} \sum_{n = 1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6} \end{equation*} To do this, follow the steps below:

1. Study the poles of the function $\phi(z) = \frac{\pi\cos(\pi z)}{\sin(\pi z)}$.
2. Find an expression for the integral $$\int_{C_{N}} \frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}$$, where $C_{N}$ is a curve that encloses the integer up until $\pm N$.
3. Considering $C_{N}$ as the rectangle with vertices $\pm(N+1/2)\pm(N+1/2)i$, prove that $$\lim_{N\to\infty}\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz=0$$
4. Conclude.

What i try is

The dist thing is to note that the set of poles on $C_{N}$ is: \begin{equation*} \mathcal{P}(N) = \{k : |k|\leq N\} \end{equation*} then I note that the residue on $k=0$ is: \begin{equation*} \begin{split} \text{Res}\left(\phi,k=0\right) &= \lim_{z\to 0}(z-0)\phi(z)\\ &= \lim_{z\to0}(z-0)\cdot\frac{\pi\cos(\pi z)}{\sin(\pi z)}\\ &= \lim_{z\to0}\cos(\pi z)\cdot\frac{\pi x}{\sin(\pi z)}=1\\ \end{split} \end{equation*} To calculate the integral \begin{equation*} \int_{C_{N}}\frac{\phi(z)}{z^{2}}dz = \int_{C_{N}}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}dz \end{equation*} whe can see that the pole is of order, meanwhile the other poles are simple, then: \begin{eqnarray*} \frac{1}{2\pi i}\int_{C_{N}}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}dz & = & \sum_{k=-N}^{N}\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right)\\ & = & \text{Res}\left(\frac{\phi(z)}{z^{2}},0\right)+\sum_{k=-N,k\neq0}^{N}\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right)\\ & = & \lim_{z\to0}\frac{1}{2!}\frac{d^{2}}{dz^{2}}\left[(z-0)^{3}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}\right]+\sum_{k=-N,k\neq0}^{N}\text{Res}\left(\frac{\phi(z)}{z^{2}},k\right) \end{eqnarray*} \begin{eqnarray*} \text{Res}\left(\frac{\phi(z)}{z^{2}},k\right) & = & \lim_{z\to k}(z-k)\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}\\ & = & \frac{1}{k^2}\lim_{z\to k}(z-k)\frac{\pi\cos(\pi z)}{\sin(\pi z)}\\ & = & \frac{1}{k^2}\lim_{z\to k}\frac{\pi\cos(\pi z) - (z-k)\pi^{2}\sin(\pi z)}{\pi\cos(\pi z)}\\ & = & \frac{1}{k^2}\lim_{z\to k} \left(\frac{\pi\cos(\pi z)}{\pi\cos(\pi z)}-\frac{(z-k)\pi^{2}\sin(\pi z)}{\pi\cos(\pi z)}\right) = \frac{1}{k^{2}}\\ \end{eqnarray*} Entonces \begin{equation*} \frac{1}{2\pi i}\int_{C_{N}}\frac{\pi\cos(\pi z)}{z^{2}\sin(\pi z)}dz = \text{Res}\left(\frac{\phi(z)}{z},0\right) + \sum_{\begin{subarray}{c}k=-N\\k\neq0\end{subarray}}^{N}\frac{1}{k^{2}} \end{equation*} and \begin{equation*} \cot(z) = \frac{1}{z} - \frac{z}{3}-\frac{z^{3}}{45}+o(z^{5}) \end{equation*} with this we calculate $$\text{Res}\left(\frac{\phi(z)}{z^{2}},0\right)=\text{Res}\left(\frac{\pi\cot(z)}{z^{2}},0\right)$$ On the other side: \begin{equation*} \cot(\pi z) = \frac{\cos(\pi z)}{\sin(\pi z)} = i\frac{e^{\pi i z} + e^{-\pi i z}}{e^{\pi i z} - e^{-\pi i z}} = i\frac{e^{2\pi i z} + 1}{e^{2\pi i z} - 1} \end{equation*} as $|\cot(\pi z)|<2$, then \begin{eqnarray*} \left|\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz\right| & \leq & \int_{C_{N}}\left|\frac{\phi(z)}{z^{2}}\right|dz\\ & \leq & \int_{C_{N}}\left|\frac{2\pi}{z^{2}}\right|dz\\ & \leq & |2\pi|\int_{C_{N}}\frac{1}{\left|z^{2}\right|}dz\\ & \leq & 2\pi\max_{z\in C_{N}}\left(\frac{1}{z^{2}}\right)\cdot L(C_{N})\\ \end{eqnarray*}

Could someone help me with this part (3)?

Answer 1

You had the right idea using the series expansion to calculate the residue at 0. You can do the following:

In our case we have that $\cot(\pi z) = \frac{1}{\pi z} - \frac{\pi z}{3} - \frac{(\pi z)^3}{45} + o(z^5)$. Then: \begin{eqnarray*} \text{Res}\left(\frac{\phi(z)}{z^{2}},0\right) & = & \lim_{z\to0}\frac{1}{2!}\frac{d^{2}}{dz^{2}}\left[(z-0)^{3}\frac{\cot(\pi z)}{z^{2}}\right] \\ & = & \lim_{z\to0}\frac{1}{2!}\frac{d^{2}}{dz^{2}}\left[z\pi\cot(\pi z)\right] \\ & = & \lim_{z\to0}\frac{1}{2}\frac{d^{2}}{dz^{2}}\left[1 - \frac{(\pi z)^2}{3} - \frac{(\pi z)^3}{45} + o(z^5) \right] \\ & = & \lim_{z\to0}\frac{1}{2}\frac{d}{dz}\left[\frac{2 \pi^2 z}{3} - \frac{3 \pi z^2}{45} \right] \\ & = & \lim_{z\to0}\frac{1}{2}\left[\frac{ 2\pi}{3} - \frac{6 \pi z}{45}\right] \\ & = & \frac{- \pi }{3} \end{eqnarray*}

Note that if you write out the expansion with higher orders, they'll go to zero as $z \to 0$ just like the term $\frac{(\pi z)^3}{45}$.

Now we can see that $C_{N}\subseteq B(0,kN)$ for some $k\in\mathbb{R}$, so we have that $L(C_{N})\leq 2\pi kN$, then we have the upper bound: \begin{equation*} \frac{1}{z^{2}} \leq \frac{2\pi k}{N^{2}} \end{equation*} Now we can say that: \begin{equation*} 2\pi\max_{z\in C_{N}} \left(\frac{1}{z^{2}}\right) \cdot L(C_{N})\leq2\pi\max_{z\in C_{N}}\left(\frac{2\pi k}{N^{2}}\right)\cdot L(C_{N})\to0\text{ when }N\to\infty \end{equation*} Then it's easy to see the next limit: \begin{equation*} \left|\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz\right|\to0 \end{equation*}

Finally, considering we have: \begin{eqnarray*} 0=\lim_{N\to\infty}\int_{C_{N}}\frac{\phi(z)}{z^{2}}dz & = & 2\pi i\lim_{N\to\infty} \left( -\frac{\pi^{3}}{3} + \sum_{\begin{subarray}{c}k=-N\\k\neq0\end{subarray}}^{N}\frac{\pi}{k^{2}} \right)\\ & = & 2\pi i\lim_{N\to\infty} \left( -\frac{\pi^{3}}{3} + \sum_{k=1}^{N}\frac{2\pi}{k^{2}} \right)\\ \end{eqnarray*} We can conclude that \begin{equation*} \sum_{k=1}^{\infty}\frac{1}{k^{2}}=\frac{\pi^{2}}{6} \end{equation*}