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We know that

$$\zeta(2)=\frac{\pi^2}{6}\tag1$$

$$\zeta(4)=\frac{\pi^4}{90} \tag2$$

$$\zeta(6)=\frac{\pi^6}{945}\tag3$$ $$.$$

$$.$$

$$.$$

My question is there a generalization for $\zeta(2n)$ in term of $\pi^{2n}$?

Thank you

Ali Shadhar
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    https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function – Arthur Dec 10 '19 at 19:17
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    It involves Bernoulli numbers – J. W. Tanner Dec 10 '19 at 19:32
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    The main point is that $$f(z)=\frac1{e^z-1}-\frac1z = \sum_{n\ge 1} \frac1{z+2i\pi n}+ \frac1{z-2i\pi n}$$ (you can look at the Fourier series of the periodic function $\sum_n \frac1{x+iy+2\pi n}$) differentiating we obtain $2 (2i\pi)^{-2k}\zeta(2k)=\frac{f^{(2k-1)}(0)}{(2k-1)!}= \frac{B_{2k}}{(2k)!}$ – reuns Dec 11 '19 at 13:25
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    @reuns thank you. why dont you post your comment as an answer? Its a good one. – Ali Shadhar Dec 11 '19 at 20:53

3 Answers3

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In this answer, it is shown that $$ \zeta(2n)=\frac{(-1)^{n-1}\pi^{2n}}{(2n+1)!}n+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}\pi^{2k}}{(2k+1)!}\zeta(2n-2k) $$ which can be used recursively to compute $\zeta(2n)$ for all $n\ge1$.

This can easily be modified to $$ \frac{\zeta(2n)}{\pi^{2n}}=\frac{(-1)^{n-1}n}{(2n+1)!}+\sum_{k=1}^{n-1}\!\frac{(-1)^{k-1}}{(2k+1)!}\frac{\zeta(2n-2k)}{\pi^{2n-2k}} $$ which shows that $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$.

robjohn
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Here is a generalization by @Américo Tavares where you can find the evaluation of $\zeta(2p)$ using the Fourier series for $x^{2p}$,

$$x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\left( \cos nx\cdot I_{2p}\right)$$

where

$$I_{2p}=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x$$

In the link, you can find the proof of $\zeta(2)$, $\zeta(4)$ and $\zeta(6)$.

Ali Shadhar
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Wikipedia tells us that $$ \zeta (2n)=(-1)^{{n+1}}{\frac {B_{{2n}}(2\pi )^{{2n}}}{2(2n)!}} $$ where $B_n$ is the $n$-th Bernoulli number, a rational number.

lhf
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