Can the value of the Riemann zeta function at $n=2$ be derived from the Wallis formula for $\pi$?

Question

It is well known that the Riemann zeta function, defined for all positive integers $n>1$ by $$ \zeta(n) = \sum_{m=1}^{\infty} m^{-n} $$ takes the value $\displaystyle \frac{\pi^2}{6}$ at $n=2$. On the other hand, a product by Wallis converges to $\displaystyle \frac{\pi}{2}$: $$ \prod _{n=1}^{\infty }\left({\frac {2n}{2n-1}}\cdot {\frac {2n}{2n+1}}\right)={\frac {2}{1}}\cdot {\frac {2}{3}}\cdot {\frac {4}{3}}\cdot {\frac {4}{5}}\cdot {\frac {6}{5}}\cdot {\frac {6}{7}}\cdot {\frac {8}{7}}\cdot {\frac {8}{9}}\cdots ={\frac {\pi }{2}}. $$

There exist many proofs of $\zeta(2)=\displaystyle \frac{\pi^2}{6}$, but I have found none that involves the Wallis product.

1. Can $\zeta(2)$ be derived from the Wallis product?
2. If so, can the general formula for $\zeta(2n)$ by derived in a similar manner?

Answer 1

There are several proofs for the sine product, e.g. look for “sine product formula proof”.

You can also take a look here, there is a link to the old proof of Euler.

For a proof of the Wallis product it’s enough to look e.g. here.

It’s $~\displaystyle \frac{\pi}{2} = \frac{\pi x}{\sin(\pi x)}|_{x=\frac{1}{2}} = \Gamma(1-x)\Gamma(1+x)|_{x=\frac{1}{2}} = \prod\limits_{n=1}^\infty \frac{2n}{2n-1}\frac{2n}{2n+1}~$ .

The connection to the Riemann zeta function can be seen by using the logarithm.

$\displaystyle \ln\frac{\pi x}{\sin(\pi x)} = \ln(\Gamma(1-x)\Gamma(1+x)) = \sum\limits_{n=1}^\infty\frac{x^{2n}}{n}\zeta(2n)$

The last equation comes from $~\displaystyle \ln\prod\limits_{k=1}^n (1-zb_k)^{a_k} = \sum\limits_{k=1}^\infty\frac{z^k}{k}\sum\limits_{v=1}^n a_v b_v^k$

with $~\displaystyle (a_v;b_v;n;z):=\left(1;\frac{1}{v^2};\infty;x^2\right)~$ .

To get the formula for $~\zeta(2n)~$ one should use the Bernoulli numbers $\,B_k\,$,

traditionelly introduced by $\displaystyle \frac{x}{e^x-1} = \sum\limits_{k=0}^\infty \frac{x^k}{k!}B_k$ .

With the first derivation of $~\displaystyle \ln\frac{\pi x}{\sin(\pi x)}~$ follows:

$\displaystyle \frac{1 -\pi x\cot(\pi x)}{x} =\frac{d}{dx}\ln\frac{\pi x}{\sin(\pi x)} = \frac{d}{dx}\sum\limits_{n=1}^\infty\frac{x^{2n}}{n}\zeta(2n) = 2\sum\limits_{n=1}^\infty x^{2n-1}\zeta(2n)$

with $\enspace\displaystyle \cot(z) = \frac{\cos z}{\sin z} = i\frac{ e^{iz}+e^{-iz} }{ e^{iz}-e^{-iz} } = i\frac{ e^{iz}-e^{-iz}+2e^{-iz} }{ e^{iz}-e^{-iz} } = i\left(1+\frac{1}{iz}\frac{i2z}{e^{i2z}-1}\right)$

$\hspace{2.3cm}\displaystyle = i\left(1+\frac{1}{iz}\sum\limits_{k=0}^\infty \frac{(i2z)^k}{k!}B_k\right) = \frac{1}{z}+\frac{1}{z}\sum\limits_{k=1}^\infty (-1)^k\frac{(2z)^{2k}}{(2k)!}B_{2k}$

We get $\displaystyle \enspace\frac{1 -\pi x\cot(\pi x)}{x} = \sum\limits_{k=1}^\infty (-1)^{k-1}\frac{(2\pi)^{2k}x^{2k-1}}{(2k)!}B_{2k}$

and comparing the coefficients of $~x^{2k-1}~$ it follows $~\displaystyle \zeta(2n)=(-1)^{k-1}\frac{(2\pi)^{2k}}{2(2k)!}B_{2k}~$ .