3

According to Wikipedia, Plouffe gives the series $$\begin{align} \zeta(5)&=\frac1{294}\pi^5-\frac{72}{35}\sum_{n\ge1}\frac1{n^5(e^{2\pi n}-1)}-\frac2{35}\sum_{n\ge1}\frac1{n^5(e^{2\pi n}+1)}\\ &=12\sum_{n\ge1}\frac1{n^5\sinh(\pi n)}-\frac{39}{20}\sum_{n\ge1}\frac1{n^5(e^{2\pi n}-1)}-\frac1{20}\sum_{n\ge1}\frac1{n^5(e^{2\pi n}+1)}, \end{align}$$ and $$\zeta(7)=\frac{19}{56700}\pi^7-2\sum_{n\ge1}\frac1{n^7(e^{2\pi n}-1)}.$$ And in general, it seems to be true that $$0=A_n\zeta(n)-B_n\pi^n+C_nS_-(n)+D_nS_+(n),$$ where $$S_{\pm}(s)=\sum_{n\ge1}\frac{1}{n^s(e^{2\pi n}\pm 1)},$$ and $A_n,B_n,C_n,D_n$ are non-negative integers.

In fact, Plouffe provides much more, but all without any links to proofs.

So, I am requesting any or all of the following:

  • Proofs of the above identities involving $\zeta(5),\zeta(7)$
  • Sources (containing proofs, sorry Ramanujan) of the theory or techniques behind Plouffe's identities in the link above
  • any other sources that you think would be relevant to this investigation.

Thank you!

clathratus
  • 17,161

2 Answers2

5

Let $n$ be a positive integer such that $n\equiv 3\pmod{4}$. Then Ramanujan says that $$\zeta(n) =\frac{(2\pi)^n}{2(n+1)!}\sum_{k=0}^{(n+1)/2}(-1)^{k+1}\binom{n+1}{2k}B_{n+1-2k}B_{2k}-2S_{-}(n)$$ Further one should check the easily verifiable identity $$S_{-} (n, \alpha) - S_{+} (n, \alpha) =2S_{-}(n, 2\alpha)$$ where $$S_{\pm} (n, \alpha) =\sum_{k=1}^{\infty} \frac{1}{k^n(e^{2\alpha k} \pm 1)}$$ Using these results one can verify the identities given by Plouffe. The formula for the case when $n\equiv 1\pmod{4}$ is more complicated.

Ramanujan deals with sums of type $\sum_{k\geq 1}k^nq^k/(1-q^k)$ for odd positive integer values of $n$ in great detail with simple proofs based on algebraic manipulation. Unfortunately he does not deal with negative odd values of $n$ in same manner. I believe he did have a proof based on algebraic manipulation which was far simpler than the later proofs based on Mellin Transform.

You can also have a look at this paper by Bruce Berndt which gives some details and approaches to prove Ramanujan's formula. However a few key results have been provided here without proof.

Paramanand Singh
  • 87,309
1

Too long for a comment: note that Zacky's link is very sloppy, p.6 for integer $m$ it is applying the residue theorem to the inverse Mellin transform integral $$\frac1{2i\pi}\int_{2m+1-i\infty}^{2m+1+i\infty} \Gamma(s)\zeta(s)\zeta(s-2m+1)x^{-s}ds=\sum_{n\ge 1}\sigma_{2m-1}(n)e^{-2\pi nx}$$ Since $\zeta(s)\zeta(s-2m+1)$ vanishes at negative integers he gets that $$x^{2m}\sum_{n\ge 1}\sigma_{2m-1}(n)e^{-2\pi nx}$$ is a polynomial which is obviously incorrect due to the exponential decay.

It fails because $\Gamma(s)\zeta(s)\zeta(s-2m+1)$ isn't decaying as $\Re(s)\to -\infty$, in fact it is invariant under $s\to 2m-s$

For a fixed $2m$ OP's formula can be obtained from the theory of modular forms and elliptic curves with CM from which we can evaluate $E_{2m}(i),E_{2m}(2i)$ in term of $E_4(i),E_6(i)$. I don't know if there is a simple formula giving the constants for every $m$.

reuns
  • 77,999