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Question : Is $$\frac{\zeta (m+n)}{\zeta (m)\zeta (n)}$$ a rational number for $m,n\ge 2\in\mathbb N$ where $\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}$?

Motivation : We know that $$\zeta (2k)=(-1)^{k+1}\frac{B_{2k}(2\pi)^{2k}}{2(2k)!}$$ and that $B_{2k}$ is a rational number for any $k\in\mathbb N$ where $B_n$ is the Bernoulli numbers.

Hence, if both $m$ and $n$ are even, then we can see that $$\frac{\zeta (2a+2b)}{\zeta (2a)\zeta (2b)}=\frac{(-1)^{a+b+1}\frac{B_{2a+2b}(2\pi)^{2a+2b}}{2(2a+2b)!}}{(-1)^{a+1}\frac{B_{2a}(2\pi)^{2a}}{2(2a)!}\cdot (-1)^{b+1}\frac{B_{2b}(2\pi)^{2b}}{2(2b)!}}$$ is a rational number where $m=2a, n=2b$.

I know that we know little about $\zeta (2a+1)$. I'm asking this question just because I would like you to let me know something helpful.

Update : I've just been to able to prove the following theorem :

Theorem : If $(\star)$ is true, then $\frac{\zeta (5)}{\zeta (2)\zeta (3)}$ is an irrational number.

Here, supposing $a,b,c\in\mathbb Z$, $$\begin{align}\int_{0}^{\frac{\pi}{2}}(ax^4+b\pi x^3+c{\pi}^{2}x^2)\log(\sin x)dx=0\Rightarrow a=b=c=0\qquad(\star)\end{align}$$

However, I can't prove that $(\star)$ is true. I'm asking this question on MSE.

Community
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mathlove
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  • It's the odd cases that are going to be ursine. – ncmathsadist Oct 18 '13 at 14:50
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    Doubtful, since it's been long conjectured that odd zeta values are independent of $\pi$, so your function is unlikely to return rational values for m and n both odd, in which case their sum is obviously even; though if one were to be even, and the other odd, it would at least be plausible. – Lucian Oct 18 '13 at 15:12

1 Answers1

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It is very unlikely that $\zeta(m+n)/(\zeta(m) \zeta(n))$ is rational for any of the cases where $m$ or $n$ is odd. On the other hand, proving that any of those is irrational would get you a fair amount of fame.

Robert Israel
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