How can I derive this fast-converging series formula for $\zeta(4)$?

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

{\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. {\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.

• 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. – KCd Oct 8 '18 at 13:37
• @KCd Thank you. You may post this as answer and I will accept it. – 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 $$\sum_{k\geq 1}\frac{\color{red}{1}}{k^4 \binom{2k}{k}} = \frac{17}{36}\zeta(4).\tag{5}$$
• For info, the Conjecture is wrong (it should have $1$ instead of $(-1)^{k-1}$ in the numerator within the sum...) – Klangen Oct 9 '18 at 16:37
• @PierreTheFermented: correct, this answer is about $\sum_{k\geq 1}\frac{1}{k^4 \binom{2k}{k}}$. – Jack D'Aurizio Oct 9 '18 at 16:38