Are there identities for the sum of reciprocal powers similar to $\sum\frac{1}{k^2}=\frac{\pi^2}{6}$ and $\sum\frac{1}{k^4}=\frac{\pi^4}{90}$?

Question

I am well aware that

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}$$

and

$$\frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \frac{1}{4^4} + \cdots = \frac{\pi^4}{90}$$

I'm curious. Are there other known identities like the ones above? For instance, what is the following?

$$\frac{1}{1^6} + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + \cdots$$

EDIT: This is not a duplicate as I'm not just interested in even powers - it would be nice to obtain insights on odd powers as well, like @Dietrich Burde has done in one of their comments.

Answer 1

These are the famous formulas by Euler $$\zeta{(2n)}=(-1)^{n-1}\frac{(2\pi)^{2n}}{2(2n)!}B_{2n},$$ where $B_{2n}$ is the $2n$-th Bernoulli number. It starts with $$ \zeta(2)=\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}. $$

There are several proof given at this site:

Ways to prove Eulers formula for $\zeta(2n)$