Ways to prove Euler's formula for $\zeta(2n)$

I recently, out of interest, tried to prove Euler's formula $\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}$ for all $n\in\mathbb{N}$. I adapted Euler's original proof for $\zeta(2)=\frac{\pi^2}{6}$:

We have the well known formulas $\sin(x)=\sum_{k=0}^{\infty}\frac{(-1)^kx^{2k+1}}{(2k+1)!}$ and $\sin(x)=x\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)$. The product formula can be transformed into a series expansion as well but in order to keep it neat, some definitions first:

For the real sequence $\left(a_i\right)_{i\in\mathbb N}$ with $\sum a_i$ convergent, define $$ A_n:=\sum_{i=1}^{\infty} a_i^n $$ $$ \alpha_n:=\sum_{(x_1,\dots,x_n)\in M_n}\prod_{r=1}^{n} a_{x_r} $$ Where $M_n:=\{(x_1,\dots,x_n)\in\mathbb N\space|\space x_1<\dots<x_n\}$. Now we write $c_i=-\frac{1}{i^2\pi^2}$. When we expand the product, we get only odd exponents of $x$ and the coefficient of $x^{2k+1}$ for $k≥1$ is $\gamma_k$. By comparison of the coefficients, we obtain $\gamma_k=\frac{(-1)^k}{(2k+1)!}$. Now, through telescoping, it can be shown algebraically that for every sequence $\left(a_i\right)_{i\in\mathbb N}$ with $\sum a_i$ convergent we have: $$\sum_{i=1}^{n}{(-1)^{n-i}\alpha_{n+1-i}\space A_{i}}=A_{n+1}+(-1)^{n+1}(n+1)\alpha_{n+1}$$ For all $n\in\mathbb N_0$. Since $C_n=\sum_{i=1}^{\infty} \left(-\frac{1}{i^2\pi^2}\right)^n=\frac{(-1)^n}{\pi^{2n}}\zeta{(2n)}$ we have: $$ \sum_{i=1}^{n}{(-1)^{n-i}\gamma_{n+1-i}\space C_{i}}=\sum_{i=1}^{n}{(-1)^{n-i}\frac{(-1)^{n+1-i}}{(2n-2i+3)!}\frac{(-1)^i}{\pi^{2i}}\zeta{(2i)}}=\sum_{i=1}^{n}{(-1)^{i-1}\frac{\zeta{(2i)}}{(2n-2i+3)!\pi^{2i}}}=\frac{(-1)^{n+1}}{\pi^{2n+2}}\zeta{(2n+2)}+(-1)^{n+1}(n+1)\frac{(-1)^{n+1}}{(2n+3)!}=\frac{(-1)^{n+1}}{\pi^{2n+2}}\zeta{(2n+2)}+\frac{(n+1)}{(2n+3)!} \iff \sum_{i=1}^{n+1}{(-1)^{i-1}\frac{\zeta{(2i)}}{(2n-2i+3)!\pi^{2i}}}=\frac{(n+1)}{(2n+3)!} $$ For all $n\in\mathbb N_0$. Now we are ready to prove the formula $\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n}$ with induction. With the equation above we can easily verify that it is true for $n=1$; therefore we assume that it holds for all $1≤i≤n$. The assertion $\zeta{(2n+2)}=(-1)^{n}\frac{(2\pi)^{2n+2}}{2(2n+2)!}B_{2n+2}$ is then equivalent to: $$ \sum_{i=1}^{n+1}{(-1)^{i-1}\frac{(-1)^{i-1}\frac{(2\pi)^{2i}}{2(2i)!}B_{2i}}{(2n-2i+3)!\pi^{2i}}}=\frac{(n+1)}{(2n+3)!}\iff \sum_{i=1}^{n+1}{\binom{2n+3}{2i}2^{2i-1}B_{2i}}=n+1\iff \sum_{i=0}^{2n+2} \binom{2n+3}{i}2^iB_i=0 $$ This identity can be proven by comparing the coefficients in the equation $\frac{e^x-1}{x}\frac{2x}{e^{2x}-1}=\frac{2}{e^x+1}$.

Is this proof valid? Nonetheless, if it is, it isn't very elegant. Are there faster ways to prove it?

Solution 1:

I usually prove it in the following way. Since: $$ \frac{t}{e^t-1} = \sum_{n\geq 0}\frac{B_n}{n!}t^n \tag{1}$$ it follows that: $$ \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}.\tag{2}$$ On the other hand, by taking the logarithmic derivative of the Weierstrass product for the $\sinh $ function it follows that: $$\begin{eqnarray*} \coth z -\frac{1}{z} &=& \sum_{n\geq 1}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right)\\&=&\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}\\&=&\sum_{n\geq 1}\sum_{m\geq 1}\frac{2(-1)^{m-1}z^{2m-1}}{\pi^{2m}n^{2m}} \\&=&\sum_{m\geq 1}\frac{2\,\zeta(2m)}{\pi^{2m}}(-1)^{m-1}z^{2m-1}\tag{3}\end{eqnarray*}$$ and we have the claim by comparing the coefficients in the RHSs of $(2)$ and $(3)$.

The intermediate identities are often very useful, too.

Solution 2:

In order to compute the values $\zeta(2n),$ for $n\in\mathbb{N}^+,$ use equation Riemann's functional equation for the Zeta function and the double angle sine formula to obtain

\begin{equation}\tag{1} \zeta(2n)=\frac{(2\pi)^{2n}\zeta(1-2n)}{2\Gamma(2n)\cos(n\pi)}=\frac{(-1)^n(2\pi)^{2n}}{2(2n-1)!}\zeta(1-2n). \end{equation}

Using the fact that, for $n\in\mathbb{N},$ \begin{equation}\tag{2} B_n=(-1)^{n+1}n\zeta(1-n). \end{equation}

We can use (2) to rewrite (1) as \begin{equation}\tag{3} \zeta(2n)=\frac{(-1)^n(2\pi)^{2n}}{2(2n-1)!}\zeta(1-2n)=\frac{(-1)^{n+1}(2\pi)^{2n}}{2(2n)!}B_{2n}. \end{equation}