(revision of 2012) In this post, we evaluate even values of the Riemann zeta function.
Calculation 19. For $n\in \N$, we have $$ \zeta(2n) = \frac{|B_{2n}|2^{2n-1} \pi^{2n}}{(2n)!}, $$ where $B_n$’s are Bernoulli numbers. |
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.
Comment
Euler의 증명방식이군요.