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[parent] proof of values of the Riemann zeta function in terms of Bernoulli numbers (Proof)

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

Theorem 1   For any positive integer $ n$
$\displaystyle \zeta(2n)=\frac{(2\pi)^{2n}\lvert B_{2n}\rvert}{2(2n)!}$
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

$\displaystyle c_n=\int_0^1B_2(x)e^{-2\pi i n x}dx = \int_0^1x^2e^{-2\pi i n x}dx - \int_0^1 xe^{-2\pi i n x}dx + \frac{1}{6}\int_0^1 e^{-2\pi i n x}dx$

We have

$\displaystyle \int_0^1 e^{-2\pi i n x}dx=0$    
$\displaystyle \int_0^1 xe^{-2\pi i n x}dx = \frac{-1}{2\pi n}xe^{-2\pi i n x}\big\lvert_0^1 + \frac{1}{2\pi i n}\int_0^1e^{-2\pi i n x}dx = \frac{i}{2\pi n}$    
$\displaystyle \int_0^1 x^2e^{-2\pi i n x} dx = \frac{-1}{2\pi i n}x^2e^{-2\pi i... ...{2}{2\pi i n}\int_0^1 xe^{-2\pi i n x}dx = \frac{1}{2\pi^2n^2}+\frac{i}{2\pi n}$    

so that
$\displaystyle c_n = \frac{1}{2\pi^2n^2}$
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^2n^2}$ (where $ a_n$ are the coefficients of $ \cos$ and $ b_n$ the coefficients of $ \sin$ in the Fourier series). Thus
$\displaystyle B_2(x)=\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2}\cos(2\pi k x)=\frac{1}{\pi^2}\sum_{k=1}^{\infty}\frac{1}{k^2}\cos(2\pi k x)$

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

$\displaystyle B_{2(n-1)}(x)=\frac{(-1)^n2\cdot(2(n-1))!}{(2\pi)^{2(n-1)}}\sum_{k=1}^{\infty}\frac{1}{k^{2(n-1)}}\cos(2\pi k x)$
Then on $ (0,1)$
$\displaystyle B_{2n}''(x)=(2n)(2n-1)B_{2(n-1)}(x) = \frac{(-1)^n2\cdot(2n)!}{(2\pi)^{2(n-1)}}\sum_{k=1}^{\infty}\frac{1}{k^{2(n-1)}}\cos(2\pi k x)$
and thus
$\displaystyle B_{2n}(x) = \frac{(-1)^n2\cdot(2n)!}{(2\pi)^{2(n-1)}}\int\int\sum_{k=1}^{\infty}\frac{1}{k^{2(n-1)}}\cos(2\pi k x)dx dx$
Since $ n\geq 2$, the sum converges absolutely, so we can move the sum outside the integrals, and we get
$\displaystyle B_{2n}(x)$ $\displaystyle = \frac{(-1)^n2\cdot(2n)!}{(2\pi)^{2(n-1)}}\sum_{k=1}^{\infty}\frac{1}{k^{2(n-1)}}\int\int\cos(2\pi k x)dx dx$    
  $\displaystyle =\frac{(-1)^n2\cdot(2n)!}{(2\pi)^{2(n-1)}}\sum_{k=1}^{\infty}\frac{1}{k^{2(n-1)}}\frac{-1}{4\pi^2k^2}\cos(2\pi k x)$    
  $\displaystyle =\frac{(-1)^{n+1}2\cdot(2n)!}{(2\pi)^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}\cos(2\pi k x)$    

Thus we have established this formula for all $ n\geq 1$. Setting $ x=0$, then, we get
$\displaystyle B_{2n} = \frac{(-1)^{n+1}2\cdot(2n)!}{(2\pi)^{2n}}\sum_{k=1}^{\infty}\frac{1}{k^{2n}}=\frac{(-1)^{n+1}2\cdot(2n)!}{(2\pi)^{2n}}\zeta(2n)$
or, trivially rewriting,
$\displaystyle \zeta(2n) = \frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!}$
But clearly $ \zeta(2n)>0$ for $ n\geq 1$, so it must be that the $ B_{2n}$ alternate in sign, and thus
$\displaystyle \zeta(2n) = \frac{(2\pi)^{2n}\lvert B_{2n}\rvert}{2(2n)!}$

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



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Cross-references: integrals, converges absolutely, inductive hypothesis, coefficients, period, periodic, zeta function, sum, infinite, Bernoulli periodic function, induction, Fourier series, Bernoulli number, integer, positive

This is version 2 of proof of values of the Riemann zeta function in terms of Bernoulli numbers, born on 2008-02-05, modified 2008-02-05.
Object id is 10235, canonical name is ProofOfValuesOfTheRiemannZetaFunctionInTermsOfBernoulliNumbers.
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Classification:
AMS MSC: 11M99 (Number theory :: Zeta and $L$-functions: analytic theory :: Miscellaneous)
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