[Calculation 19] Even values of the Zeta Function

(revision of 2012)  In this post, we evaluate even values of the Riemann zeta function.

 
Proof. We have the general formula for the generating function:
\begin{align}
\frac{z}{e^z -1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!}
\end{align}
for $|z|<2\pi$, and the following identity:
\begin{align}
\frac{z}{e^z -1} + \frac{z}{2} = \frac{z}{2} \coth \frac{z}{2}.
\end{align}
Combining (1) and (2), we get
\begin{align}
z \coth z = 1 + \sum_{n=1}^\infty B_{2n} \frac{(2z)^{2n}}{(2n)!},
\end{align}
for $|z|< \pi$ where we replaced $B_0 = 1$, $B_1 = - \frac{1}{2}$, and $B_{2k+1} = 0$, $k\in \N$.   On the other hand, we have $$\sin z = z \prod_{k=1}^\infty \left( 1 - \left(\frac{z}{k\pi}\right)^2\right).$$ Taking logarithm both sides and differentiating with respect to $z$, we obtain $$ \cot z = \frac{1}{z} - 2 \sum_{k=1}^\infty \frac{z}{k^2 \pi^2 -z^2} $$ or \begin{align} z\coth z &= 1 + 2\sum_{k=1}^\infty \frac{z^2}{k^2 \pi^2 + z^2} \\ &=1 - 2 \sum_{k=1}^\infty \sum_{n=1}^\infty \left( \frac{-z^2}{k^2 \pi^2}\right)^n\\ &= 1 + 2 \sum_{n=1}^\infty \frac{\zeta(2n) (-1)^{n+1}}{\pi^{2n}} z^{2n} \end{align} for $|z|<\pi$. Comparing the coefficients of $z^{2n}$ of (3) and (6), we finally obtain $$ \zeta(2n) = \frac{(-1)^{n+1} B_{2n} 2^{2n-1} \pi^{2n}}{(2n)!} = \frac{|B_{2n}|2^{2n-1} \pi^{2n}}{(2n)!}, $$ which completes the proof.$\square$

 
References.
[1] Bruce C. Berndt, Elementary Evaluation of ζ(2n), Math. Magazine 48, No.3 (1975), 148–154.