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Since Hjortnaes (and later Apéry), we know that

$$ \zeta(3)=\frac{5}{2} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {k^{3}\binom {2k}{k}}}. $$

I read somewhere that there might be a similar expression for $\zeta(5)$:

$$ \zeta(5)=\frac{a}{b} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {k^{5}\binom {2k}{k}}}, $$

where $a$ and $b$ are positive nonzero integers. I know that no such $a$ and $b$ have been found yet, otherwise Apéry's proof of the irrationality of $\zeta(3)$ could be extended to $\zeta(5)$. Are there any results suggesting that there exist similar sums for $\zeta(2n+1)$, $n>1$:

$$ \zeta(2n+1)=\frac{a}{b} \sum_{k=1}^{\infty}{\frac{ (-1)^{k-1}} {k^{2n+1}\binom {2k}{k}}}? $$

Conversely, are there any results that rule out the existence of such sums for certain odd positive integers?

Finally, what are the current known bounds on $a$ and $b$, if such sums can exist?

Klangen
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  • The paper by Nan-Yue and Williams [1] is worth consulting; equation (4.20) onwards involving the golden ratio $\Phi$.

    ( [1] Nan-Yue, Zhang; Williams, Kenneth S. Values of the Riemann zeta function and integrals involving $\log(2,{\rm sinh}(\theta/2))$ and $\log(2\sin(\theta/2))$. Pacific J. Math. 168 (1995), no. 2, 271--289. https://projecteuclid.org/euclid.pjm/1102620561 )

     – James Arathoon Jun 27 '18 at 15:45

1 Answers1

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The first terms of the continued fraction of $$ \frac{1}{\zeta(5)}\sum_{k\geq 1}\frac{(-1)^{k+1}}{k^5 \binom{2k}{k}} $$ are given by $$[0; 2, 10, 1, 1, 5, 1, 1, 1, 1, 2, 2, 1, 3, 3, 1, 66, 4, 135, 3, 1, \ 6, 1, 29,\ldots]$$ so if the previous number is $\frac{a}{b}\in\mathbb{Q}$, we have $\min(a,b)>3\cdot 10^{12}$.
In other terms, it looks pretty unlikely that the series $\sum_{k\geq 1}\frac{(-1)^{k+1}}{k^5 \binom{2k}{k}}$ provides a rational multiple of $\zeta(5)$.
The classical results $$ \sum_{n\geq 1}\frac{1}{n^2\binom{2n}{n}}=\frac{1}{3}\zeta(2),\quad \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^3\binom{2n}{n}}=\frac{2}{5}\zeta(3),\quad \sum_{n\geq 1}\frac{1}{n^4\binom{2n}{n}}=\frac{17}{36}\zeta(4)$$ (stating that in some cases the hypergeometric $\phantom{}_{p+1} F_p\left(1,1,1\ldots;\tfrac{3}{2},2,2,\ldots;\pm\tfrac{1}{4}\right)$ has a nice closed form in terms of the $\zeta$ function) can be seen as consequences of creative telescoping, or as consequences of symmetry relations for polylogarithms (see [1]). About $\frac{1}{k^5}$, creative telescoping produces spurious terms and the functional relations for $\text{Li}_4$ and $\text{Li}_5$ are pretty messy, so there are substantial obstructions in generalizing the classical approaches for producing a simple irrationality proof for $\zeta(5)$.

Jack D'Aurizio
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