# Convergence acceleration technique for $\zeta(4)$ (or $\eta(4)$) via creative telescoping?

Question

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 $$\begin{equation*} X_{n,k}=\frac{(-1)^{k-1}}{k^{2}\binom{n+k}{k}\binom{n-1}{k}},\qquad D_{n,k}=\frac{(-1)^{k}}{n^{2}\binom{n+k}{k}\binom{n-1}{k}}\qquad k
2. Notice that $$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.

Remarks

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 $$\zeta(4)$$.

References

• what is the question? May 14, 2017 at 19:42
• @Masacroso In short, can (1) be derived from an identity similar to (4)? May 14, 2017 at 19:45
• See also this fast converging series en.wikipedia.org/wiki/… May 19, 2017 at 20:53