# Contradicting statements about the Riemann zeta function at positive odd integers

I have found two contradicting statements about the value of $\zeta(k)$ when $k=2n+1$ and $n\in\mathbb{Z_0^+}$. Which one is correct?

1. "The Riemann zeta function for odd integers has no known closed-form formula", that is, you can't find $\zeta(k)=f(k)$ such that $f$ holds for all positive odd integers. This is apparently a "well-known" fact.
2. Wikipedia states here that $\zeta(k)$ can be expressed for all positive integers, even or odd, using the formula:

$\zeta(k)=\frac{2^k}{2^k-1}+\sum_{r=2}^\infty\frac{(p_{r-1}\#)^k}{J_k(p_r\#)}\quad(k=2,3,\dots)$

which involves primorials and the Jordan totient function.

EDIT: I understand the above is not a closed formula. But then what about this paper, which proposes an explicit closed-form formula?

• I wouldn't call the second expression a "closed form", so for me both are true. – TZakrevskiy Nov 14 '14 at 8:20
• Agreed. The second is no more a closed form than the definition $\zeta(k) = \sum_{n=1}^{\infty} \frac{1}{n^k}$ is a closed form. Actually I would say it is even more opaque to me than the definition. – Qiaochu Yuan Nov 14 '14 at 8:22
• What about the second link then? arxiv.org/abs/1211.5033 – Klangen Nov 14 '14 at 8:23
• Isn't $\zeta(k)$ a closed formula? – bof Nov 14 '14 at 9:57

The paper you link to proposes "closed forms" of the type

$$\zeta(3)=\frac{2^3}{2^3-1}\beta(3)-\dots$$

with

$$\beta(s)=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(2k-1)^s}$$

Thus the infinite term is just hidden away in the function $\beta$ (which occurs in every expression they give) and the claim of having obtained closed forms is a tall one indeed.

• Is it then safe to say that there is still no closed-form formula for $\zeta(2k+1)$? – Klangen Nov 17 '14 at 8:29
• @Pickle: I'm no expert, but to me this sounds like safe bet, yes (it depends on your definition of "closed form", of course). If there is a closed form, it's certainly not to be found in that paper. – user139000 Nov 17 '14 at 8:30
• Simple and concise answer. Thanks a lot man! – Klangen Nov 17 '14 at 9:38