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

###### Theorem 1.

For any positive integer $n$

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

where $B_{2n}$ is the $2n^{\mathrm{th}}$ 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

 $c_{n}=\int_{0}^{1}B_{2}(x)e^{-2\pi inx}dx=\int_{0}^{1}x^{2}e^{-2\pi inx}dx-% \int_{0}^{1}xe^{-2\pi inx}dx+\frac{1}{6}\int_{0}^{1}e^{-2\pi inx}dx$

We have

 $\displaystyle\int_{0}^{1}e^{-2\pi inx}dx=0$ $\displaystyle\int_{0}^{1}xe^{-2\pi inx}dx=\frac{-1}{2\pi n}xe^{-2\pi inx}\big{% \lvert}_{0}^{1}+\frac{1}{2\pi in}\int_{0}^{1}e^{-2\pi inx}dx=\frac{i}{2\pi n}$ $\displaystyle\int_{0}^{1}x^{2}e^{-2\pi inx}dx=\frac{-1}{2\pi in}x^{2}e^{-2\pi inx% }\big{\lvert}_{0}^{1}+\frac{2}{2\pi in}\int_{0}^{1}xe^{-2\pi inx}dx=\frac{1}{2% \pi^{2}n^{2}}+\frac{i}{2\pi n}$

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 $n>0$, $\displaystyle a_{n}=c_{n}+c_{-n}=\frac{1}{\pi^{2}n^{2}}$ (where $a_{n}$ are the coefficients of $\cos$ and $b_{n}$ the coefficients of $\sin$ in the Fourier series). Thus

 $B_{2}(x)=\sum_{k=1}^{\infty}\frac{1}{\pi^{2}k^{2}}\cos(2\pi kx)=\frac{1}{\pi^{% 2}}\sum_{k=1}^{\infty}\frac{1}{k^{2}}\cos(2\pi kx)$

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

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

Then on $(0,1)$

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

and thus

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

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

 $\displaystyle B_{2n}(x)$ $\displaystyle=\frac{(-1)^{n}2\cdot(2n)!}{(2\pi)^{2(n-1)}}\sum_{k=1}^{\infty}% \frac{1}{k^{2(n-1)}}\int\int\cos(2\pi kx)dxdx$ $\displaystyle=\frac{(-1)^{n}2\cdot(2n)!}{(2\pi)^{2(n-1)}}\sum_{k=1}^{\infty}% \frac{1}{k^{2(n-1)}}\frac{-1}{4\pi^{2}k^{2}}\cos(2\pi kx)$ $\displaystyle=\frac{(-1)^{n+1}2\cdot(2n)!}{(2\pi)^{2n}}\sum_{k=1}^{\infty}% \frac{1}{k^{2n}}\cos(2\pi kx)$

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

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

or, trivially rewriting,

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

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

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

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

Title proof of values of the Riemann zeta function in terms of Bernoulli numbers ProofOfValuesOfTheRiemannZetaFunctionInTermsOfBernoulliNumbers 2013-03-22 17:46:37 2013-03-22 17:46:37 rm50 (10146) rm50 (10146) 5 rm50 (10146) Proof msc 11M99