0

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.

Stephen Muga
  • 501
  • See $\zeta(2n)$ at https://en.wikipedia.org/wiki/Riemann_zeta_function#Specific_values – Robert Z Oct 15 '19 at 14:59
  • It's $\pi^6/945$. – J.G. Oct 15 '19 at 16:23
  • @J.G. nice. That is indeed correct. I believe that for s=8 and s=10, the sum can also be cleanly expressed in terms of PI as you have done. When s=12, it seems we can't. When s=14, we can - I believe it's π^14/9121612.5. Why does this rather odd behavior occur? – Stephen Muga Oct 16 '19 at 17:25
  • @MugaS. Oh, but it can be done for $s=12$, or indeed any positive integer $s$ as per DietrichBurde's answer. The basic reason it happens is because you can express the coefficients in $\frac{\sin\pi z}{\pi z}=\prod_{n\ge1}\left(1-\frac{z^2}{n^2}\right)$ in terms of $\zeta(2n),,n\ge1$. – J.G. Oct 16 '19 at 17:27
  • @J.G. Really? We can do it for s=12? I dare you, then. Here's what I bring to the table: For s=8, I get π^8/9450 and for s=10, I get π^10/93555. For s=12, meh, nothing interesting like these two. – Stephen Muga Oct 16 '19 at 17:51
  • @MugaS. $\zeta(12)=\frac{691}{638512875}\pi^{12}$ (see here). In general, $\zeta(2n)/\pi^{2n}\in\Bbb Q^+$ for all $n\in\Bbb N$. – J.G. Oct 16 '19 at 18:09
  • @J.G. Of course. Perharps I just fell for the idea that all solutions will STRICTLY be of the form (π^s/n). – Stephen Muga Oct 16 '19 at 18:47

1 Answers1

2

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 $\zeta(2n)$

Dietrich Burde
  • 130,978