proof of values of the Riemann zeta function in terms of Bernoulli numbers

This article proves part of the theorem given in the article.

Theorem 1.

For any positive integer $n$

$$\zeta (2n)=\frac{{(2\pi )}^{2n}|B_{2n}|}{2(2n)!}$$

where $B_{\mathrm{2}\mathit{}n}$ is the $\mathrm{2}\mathit{}{n}^{\mathrm{th}}$ Bernoulli number.

Proof.The method is as follows. Using Fourier series together with induction on $n$, we derive a formula for the Bernoulli periodic function $B_{2n}(x)$ involving an infinite sum. On setting $x$ to $0$, this sum reduces to a constant times the appropriate zeta function, and the result follows.

We first compute the Fourier series for $B_2(x)$. $B_2(x)$ is periodic with period $1$, so

$$c_n={\int }_0^1{B}_2(x)e^{-2\pi inx}𝑑x={\int }_0^1{x}^2{e}^{-2\pi inx}𝑑x-{\int }_0^{1}x{e}^{-2\pi inx}𝑑x+\frac{1}{6}{\int }_0^1{e}^{-2\pi inx}𝑑x$$

We have

$${\int }_0^1{e}^{-2\pi inx}𝑑x=0$$

$${\int }_0^{1}x{e}^{-2\pi inx}dx=\frac{-1}{2\pi n}x{e}^{-2\pi inx}{|}_0^1+\frac{1}{2\pi in}{\int }_0^1{e}^{-2\pi inx}dx=\frac{i}{2\pi n}$$

$${\int }_0^1{x}^2{e}^{-2\pi inx}dx=\frac{-1}{2\pi in}{x}^2{e}^{-2\pi inx}{|}_0^1+\frac{2}{2\pi in}{\int }_0^{1}x{e}^{-2\pi inx}dx=\frac{1}{2{\pi }^2{n}^2}+\frac{i}{2\pi n}$$

so that

$$c_n=\frac{1}{2{\pi }^2{n}^2}$$

But then $b_n=c_n-c_{-n}=0$ for all $n$, $a_0=0$, and for $n>0$, $a_n=c_n+c_{-n}={\displaystyle \frac{1}{{\pi }^2{n}^2}}$ (where $a_n$ are the coefficients of $\cos$ and $b_n$ the coefficients of $\sin$ in the Fourier series). Thus

$$B_2(x)=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{\pi }^2{k}^2}\cos (2\pi kx)=\frac{1}{{\pi }^2}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^2}\cos (2\pi kx)$$

Using this case as an inductive hypothesis, assume that for some $n\ge 2$

$$B_{2(n-1)}(x)=\frac{{(-1)}^{n}2\cdot (2(n-1))!}{{(2\pi )}^{2(n-1)}}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2(n-1)}}\cos (2\pi kx)$$

Then on $(0,1)$

$$B_{2n}^{\prime \prime }(x)=(2n)(2n-1)B_{2(n-1)}(x)=\frac{{(-1)}^{n}2\cdot (2n)!}{{(2\pi )}^{2(n-1)}}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2(n-1)}}\cos (2\pi kx)$$

and thus

$$B_{2n}(x)=\frac{{(-1)}^{n}2\cdot (2n)!}{{(2\pi )}^{2(n-1)}}\int \int \sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2(n-1)}}\cos (2\pi kx)dxdx$$

Since $n\ge 2$, the sum converges absolutely, so we can move the sum outside the integrals, and we get

$B_{2n}(x)$$={\displaystyle \frac{{(-1)}^{n}2\cdot (2n)!}{{(2\pi )}^{2(n-1)}}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{k^{2(n-1)}}}{\displaystyle \int \int \cos (2\pi kx)𝑑x𝑑x}$

$={\displaystyle \frac{{(-1)}^{n}2\cdot (2n)!}{{(2\pi )}^{2(n-1)}}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{k^{2(n-1)}}}{\displaystyle \frac{-1}{4{\pi }^2{k}^2}}\cos (2\pi kx)$

$={\displaystyle \frac{{(-1)}^{n+1}2\cdot (2n)!}{{(2\pi )}^{2n}}}{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{\displaystyle \frac{1}{k^{2n}}}\cos (2\pi kx)$

Thus we have established this formula for all $n\ge 1$. Setting $x=0$, then, we get

$$B_{2n}=\frac{{(-1)}^{n+1}2\cdot (2n)!}{{(2\pi )}^{2n}}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^{2n}}=\frac{{(-1)}^{n+1}2\cdot (2n)!}{{(2\pi )}^{2n}}\zeta (2n)$$

or, trivially rewriting,

$$\zeta (2n)=\frac{{(-1)}^{n+1}{(2\pi )}^{2n}{B}_{2n}}{2(2n)!}$$

But clearly $\zeta (2n)>0$ for $n\ge 1$, so it must be that the $B_{2n}$ alternate in sign, and thus

$$\zeta (2n)=\frac{{(2\pi )}^{2n}|B_{2n}|}{2(2n)!}$$

Note that as a effect of this proof, we see that the even-index Bernoulli numbers alternate in sign!