Let $\zeta(n)$ denote the Riemann Zeta function for positive integers $n>1$ as usual by:

$$ \zeta(n)=\sum_{m=1}^{\infty}m^{-n}. $$

There are fast-converging series for $\zeta(2)$ and $\zeta(3)$, but not others. In the spirit of Apéry's

$$ {\displaystyle {\begin{aligned}\zeta (3)&={\frac {5}{2}}\sum _{k=1}^{\infty }{\frac {(-1)^{k-1}}{{\binom {2k}{k}}k^{3}}}\end{aligned}}}, $$

quick numerical computations show that

Conjecture 1. $$ {\displaystyle {\begin{aligned}\zeta (4)&={\frac {36}{17}}\sum _{k=1}^{\infty }{\frac {1}{{\binom {2k}{k}}k^{4}}}\end{aligned}}}.$$

However, I am unable to prove this statement. I have tried using

$$ 2(\sin^{-1}x)^2 =\sum_{k=1}^{\infty}{\frac{(2x)^{2k}}{{\binom {2k}{k}}k^{2}}}, $$

but to no avail. Any help is appreciated.

  • 2
    $\begingroup$ This conjecture was proved over 30 years ago by Bombieri and van der Poorten, and the auxiliary series you mention played a role. See pages 2 and 3 of their paper "Continued Fractions of Algebraic Numbers", including the footnote on page 3. This paper is freely available online. $\endgroup$
    – KCd
    Oct 8 '18 at 13:37
  • $\begingroup$ @KCd Thank you. You may post this as answer and I will accept it. $\endgroup$
    – Klangen
    Oct 8 '18 at 14:21

$$\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}x^k}{k^2\binom{2k}{k}}=2\,\text{arcsinh}^2\left(\frac{x}{2}\right)\tag{1}$$ due to the Maclaurin series of the squared arcsine, hence the computation of the series appearing in your conjecture is equivalent to the computation of the following integral: $$ I = \int_{0}^{1}\frac{2}{x}\text{arcsinh}^2\left(\frac{x}{2}\right)\log^2(x)\,dx.\tag{2} $$ In 1981 Leschiner pointed out a nice consequence of creative telescoping: $$\begin{eqnarray*}\sum_{n\geq 0}\left(1-\frac{1}{2^{2n+1}}\right)a^{2n+2}\zeta(2n+2)&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2-a^2}\\&=&\frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2\binom{2k}{k}}\cdot\frac{3k^2+a^2}{k^2-a^2}\prod_{m=1}^{k-1}\left(1-\frac{a^2}{m^2}\right)\tag{3}\end{eqnarray*}$$ and in 2006 Bailey, Borwein and Bradley pointed out that similarly $$\begin{eqnarray*}\sum_{n\geq 0}\zeta(2n+2)a^{2n}&=&\sum_{k\geq 1}\frac{1}{k^2-a^2}\\&=&3\sum_{k\geq 1}\frac{1}{\binom{2k}{k}(k^2-a^2)}\prod_{m=1}^{k-1}\frac{m^2-4a^2}{m^2-a^2}\tag{4}\end{eqnarray*} $$ holds. By comparing the coefficients of $a^0$ in the LHS and RHS of $(3)$ we get the well-known $$ \frac{1}{2}\zeta(2)=\frac{3}{2}\sum_{k\geq 1}\frac{1}{k^2\binom{2k}{k}}.$$ The coefficient of $a^4$ in the LHS of $(3)$ is $$ \frac{7}{8}\zeta(4) = \frac{1}{2}\sum_{k\geq 1}\frac{1}{k^2\binom{2k}{k}}\left[\frac{4}{k^2}-3 H_{k-1}^{(2)}\right]$$ and the series $$ \sum_{k\geq 1}\frac{H_{k-1}^{(2)}}{k^2\binom{2k}{k}} $$ can be computed from $(4)$ or from the Maclaurin series of $\arcsin^4(x)$ (identity $(20)$ here), proving $$ \sum_{k\geq 1}\frac{\color{red}{1}}{k^4 \binom{2k}{k}} = \frac{17}{36}\zeta(4).\tag{5}$$

  • $\begingroup$ Quality answer, as always Jack $\endgroup$
    – Klangen
    Oct 8 '18 at 17:22
  • $\begingroup$ For info, the Conjecture is wrong (it should have $1$ instead of $(-1)^{k-1}$ in the numerator within the sum...) $\endgroup$
    – Klangen
    Oct 9 '18 at 16:37
  • 1
    $\begingroup$ @PierreTheFermented: correct, this answer is about $\sum_{k\geq 1}\frac{1}{k^4 \binom{2k}{k}}$. $\endgroup$ Oct 9 '18 at 16:38

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