TITLE: Complex Analysis Solution to the Basel Problem ($\sum_{k=1}^\infty \frac{1}{k^2}$)
QUESTION [10 upvotes]: Most of us are aware of the famous "Basel Problem":
$$\sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{6}$$
I remember reading an elegant proof for this using complex numbers to help find the value of the sum.  I tried finding it again to no avail.  Does anyone know a complex number proof for the solution of the Basel Problem?

REPLY [4 votes]: Observe that:
$\\ \displaystyle csc^2(\pi z)=\frac{1}{sin^2(\pi z)}=\frac{1}{4sin^2\left(\frac{\pi z}{2}\right)cos^2\left(\frac{\pi z}{2}\right)}=\frac{sin^2\left(\frac{\pi z}{2}\right)+cos^2\left(\frac{\pi z}{2}\right)}{4sin^2\left(\frac{\pi z}{2}\right)cos^2\left(\frac{\pi z}{2}\right)}=\frac{1}{4}\left(\frac{1}{cos^2\left(\frac{\pi z}{2}\right)}+\frac{1}{sin^2\left(\frac{\pi z}{2}\right)}\right)=\frac{1}{4}\left(sec^2\left(\frac{\pi z}{2}\right)+csc^2\left(\frac{\pi z}{2}\right)\right)= \frac{1}{4}\left(csc^2\left(\frac{\pi}{2}-\frac{\pi z}{2}\right)+csc^2\left(\frac{\pi z}{2}\right)\right)=\frac{1}{4}\left(csc^2\left(\frac{ (z-1)\pi}{2}\right)+csc^2\left(\frac{\pi z}{2}\right)\right) \\ \\$
We conclude:
\begin{equation}
     csc^2(\pi z)=\frac{1}{4}\left(csc^2\left(\frac{ (z-1)\pi}{2}\right)+csc^2\left(\frac{\pi z}{2}\right)\right) \tag{1}
\end{equation}
Take z=1/4, we get:
\begin{equation}
    2=\frac{1}{4}\left(csc^2\left(\frac{3\pi}{8}\right)+csc^2\left(\frac{\pi }{8}\right)\right) \tag{2}
\end{equation}
Let (2) our induction base, such as induction hypothesis valid for suppose that an n 
\begin{equation}
    1=\frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}csc^2\frac{(2k+1)\pi}{2^{n+1}} \tag{3}
\end{equation}
Applying (1) in (3) , we have:
\begin{equation}
    1=\frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}csc^2\frac{(2k+1)\pi}{2^{n+1}}=\frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}\frac{1}{4}\left(csc^2\left(\frac{ \left(\frac{(2k+1)}{2^{n+1}}-1\right)\pi}{2}\right)+csc^2\left(\frac{ \frac{(2k+1)\pi}{2^{n+1}}}{2}\right)\right)
\end{equation}
\begin{equation}
    =\frac{2}{4^{n+1}}\sum_{k=0}^{2^{n-1}-1}\left(csc^2\left(\frac{(2k+1-2^{n+1})\pi}{2^{n+2}}\right)+csc^2\left(\frac{(2k+1)\pi}{2^{n+2}}\right)\right)=
\end{equation}
\begin{equation}
    =\frac{2}{4^{n+1}}\sum_{k=0}^{2^{n-1}-1}csc^2\left(\frac{2^{n+1}-(2k+1))\pi}{2^{n+2}}\right)+\frac{2}{4^{n+1}}\sum_{k=0}^{2^{n-1}-1}csc^2\left(\frac{(2k+1)\pi}{2^{n+2}}\right)
\end{equation}
\begin{equation}
    =\frac{2}{4^{n+1}}\sum_{k=0}^{2^{n}-1}csc^2\left(\frac{(2k+1)\pi}{2^{n+2}}\right)
\end{equation}
Completing the proof by induction that equality below is true:
\begin{equation}
    1=\frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}csc^2\frac{(2k+1)\pi}{2^{n+1}} \tag{4}
\end{equation}
On the other hand, using that $\displaystyle csc^2x=cot^2x+1$, we get:
\begin{equation}
    1=\frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}cot^2\frac{(2k+1)\pi}{2^{n+1}}+\frac{2^{n}}{4^n}
\end{equation}
\begin{equation}
    \frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}cot^2\frac{(2k+1)\pi}{2^{n+1}}=1-\frac{1}{2^n} \tag{5}
\end{equation}
Using that $\displaystyle senx\leq x \leq tanx \Rightarrow cotx\leq \frac{1}{x} \leq cscx \Rightarrow cot^2x\leq \frac{1}{x^2} \leq csc^2x$, we have:
\begin{equation}
    \frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}cot^2\frac{(2k+1)\pi}{2^{n+1}}\leq \frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}\frac{2^{2n+2}}{((2k+1)\pi)^2} \leq \frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}csc^2\frac{(2k+1)\pi}{2^{n+1}}
\end{equation}
\begin{equation}
    \frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}cot^2\frac{(2k+1)\pi}{2^{n+1}}\leq \sum_{k=0}^{2^{n-1}-1}\frac{8}{((2k+1)\pi)^2} \leq \frac{2}{4^n}\sum_{k=0}^{2^{n-1}-1}csc^2\frac{(2k+1)\pi}{2^{n+1}} \tag{6}
\end{equation}
Substituting ( 4) and (5 ) in (6 ) , we have:
\begin{equation}
    1-\frac{1}{2^n}\leq \sum_{k=0}^{2^{n-1}-1}\frac{8}{((2k+1)\pi)^2} \leq 1
\end{equation}
Taking limit to infinity in all parts of inequality , we have:
\begin{equation}
    \lim_{n\rightarrow \infty}\left(1-\frac{1}{2^n}\right)\leq \lim_{n\rightarrow \infty} \sum_{k=0}^{2^{n-1}-1}\frac{8}{((2k+1)\pi)^2} \leq \lim_{n\rightarrow \infty} 1
\end{equation}
Hence:
\begin{equation}
     \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2} =\frac{\pi^2}{8}
\end{equation}
Observe that:
\begin{equation*}
     \sum_{k=1}^{\infty}\frac{1}{4k^2}= \lim_{n\rightarrow \infty} \left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^n}\right)\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2} =\lim_{n\rightarrow \infty}\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+...+\frac{1}{4^n}\right)\frac{\pi^2}{8}
\end{equation*}
\begin{equation}
     = \frac{1}{3}\frac{\pi^2}{8}\Rightarrow \sum_{k=1}^{\infty}\frac{1}{4k^2}=\frac{\pi^2}{24}
\end{equation}
It follows that:
\begin{equation}
     \sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}
\end{equation}