What's the generalization for $\zeta(2n)$?

Question

We know that

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

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

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

$$.$$

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

Thank you

Answer 1

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 $\frac{\zeta(2n)}{\pi^{2n}}\in\mathbb{Q}$.

Answer 2

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 $\zeta(6)$.

Answer 3

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.