1. Is it already known whether the $\zeta(4):=\sum_{n=1}^{\infty}1/n^4$ accelerated convergence series $(1)$, proved for instance in [1, Corollaire 5.3], could be obtained by a similar technique to the ones explained by Alf van der Poorten in [2, section 1] for $\zeta(3)$ and $\zeta(2)$? $$\zeta(4)=\frac{36}{17}\sum_{n=1}^{\infty}\frac{1}{n^{4}\binom{2n}{n}}.\tag{1}$$
  2. (a) In other words, does there exist a pair of functions $F(n,k), G(n,k)$ obeying equation $$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k)\tag{$\ast$}$$ from which $(1)$ can be proved? That is, is it possible to transform the defining series for $\zeta(4):=\sum_{n=1}^{\infty}1/n^4$ by means of the Wilf-Zeilberger method (or the Markov-WZ Method) into the faster series $(1)$? (b) Most likely there isn't any such a pair $(F, G)$, but I do not have the means to use these methods on my own.

Short description of section 1 of Alf van der Poorten's paper

The defining series for $\zeta(3):=\sum_{n=1}^{\infty}1/n^3$ and $\zeta(2):=\sum_{n=1}^{\infty}1/n^2$ are accelerated resulting in

\begin{equation*} \zeta (2)=3\sum_{n=1}^{\infty } \frac{1}{n^{2}\binom{2n}{n}},\tag{2} \end{equation*}

\begin{equation*} \zeta (3)=\frac{5}{2}\sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}\tag{3}. \end{equation*}

For instance, $(3)$ follows from the identity

\begin{equation*} \sum_{n=1}^{N}\frac{1}{n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}}=\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}-\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}\tag{4}, \end{equation*}

by letting $N\rightarrow \infty $ and noticing that

\begin{equation*} \lim_{N\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}}=0. \end{equation*}

Equality $(4)$ can be explained as follows:

  1. Write \begin{equation*} X_{n,k}=\frac{(-1)^{k-1}}{k^{2}\binom{n+k}{k}\binom{n-1}{k}},\qquad D_{n,k}=\frac{(-1)^{k}}{n^{2}\binom{n+k}{k}\binom{n-1}{k}}\qquad k<n. \end{equation*}
  2. Notice that $$X_{n,k}=D_{n,k-1}-D_{n,k}.\tag{5}$$ Hence \begin{eqnarray*} \sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{k=1}^{n-1}\left( \frac{D_{n,k-1}}{ n}-\frac{D_{n,k}}{n}\right) =\frac{D_{n,0}}{n}-\frac{D_{n,n-1}}{n} \\ &=&\frac{1}{n^{3}}-2\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}},\qquad\frac{D_{n,0}}{n} =\frac{1}{n^{3}},\quad \frac{D_{n,n-1}}{n}=2\frac{ \left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}} \end{eqnarray*}
  3. Sum over $k$, $1\leq k\leq n-1$ \begin{equation*} \sum_{k=1}^{n-1}X_{n,k}=\sum_{k=1}^{n-1}\left( D_{n,k-1}-D_{n,k}\right) =D_{n,0}-D_{n,n-1}. \end{equation*}
  4. Now, summing over $n$, $1\leq n\leq N$, and noticing that \begin{equation*} \frac{X_{n,k}}{n}=E_{n,k}-E_{n-1,k},\qquad E_{n,k}=\frac{(-1)^{k}}{2k^{3}\binom{n+k}{k}\binom{n}{k}},\tag{6} \end{equation*} we obtain \begin{equation*} \sum_{k=1}^{N-1}\sum_{n=k+1}^{N}\frac{X_{n,k}}{n}=\sum_{k=1}^{N-1} \sum_{n=k+1}^{N}\left( E_{n,k}-E_{n-1,k}\right) =\sum_{k=1}^{N}\left( E_{N,k}-E_{k,k}\right). \end{equation*}
  5. So, on the one hand \begin{eqnarray*} \sum_{n=1}^{N}\sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{n=1}^{N}\frac{1}{ n^{3}}-2\sum_{n=1}^{N}\frac{\left( -1\right) ^{n-1}}{n^{3}\binom{2n}{n}},\tag{7} \end{eqnarray*} and on the other hand \begin{eqnarray*} \sum_{n=1}^{N}\sum_{k=1}^{n-1}\frac{X_{n,k}}{n} &=&\sum_{k=1}^{N}E_{N,k}-\sum_{k=1}^{N}E_{k,k} \\ &=&\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{N+k}{k}\binom{N}{k}} -\sum_{k=1}^{N}\frac{(-1)^{k}}{2k^{3}\binom{2k}{k}}.\tag{8} \end{eqnarray*} The identity $(4)$ follows.


  1. The combination of equations $(5)$ and $(6)$ forms an identity of the form $(\ast)$, which is equation $(6.1.2)$ of [3, chapter 6] (Zeilberger's Algorithm).
  2. As for $(2)$, [2, section 1] actually explains how to accelerate $\eta(2):=\sum_{n=1}^{\infty }(-1)^{n-1}/n^{2}$ and obtain $(2)$, using the relation $\eta(s) = \left(1-2^{1-s}\right) \zeta(s)$. As such, if feasible, I expect that accelerating $\eta(4)$ might be easier than $\zeta(4)$.


  1. Henri Cohen, Généralisation d'une Construction de R. Apéry
  2. Alfred van der Poorten, Some wonderful formulae... Footnotes to Apery's proof of the irrationality of $\zeta(3)$
  3. Marko Petkovsek, Herbert Wilf, Doron Zeilberger, A = B
  • $\begingroup$ what is the question? $\endgroup$
    – Masacroso
    May 14, 2017 at 19:42
  • $\begingroup$ @Masacroso In short, can (1) be derived from an identity similar to (4)? $\endgroup$ May 14, 2017 at 19:45
  • 1
    $\begingroup$ See also this fast converging series en.wikipedia.org/wiki/… $\endgroup$
    – reuns
    May 19, 2017 at 20:53


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