# [FOM] Some questions regarding irrational numbers

Timothy Y. Chow tchow at alum.mit.edu
Tue Mar 3 17:13:20 EST 2009

I checked with the experts about the status of the multiple zeta value
identity I posted.  David Broadhurst wrote:

>       Jos Vermaseren at NIKHEF has a database of /proven/
>       reductions (from shuffles and stuffles)
>       to a minimal basis. So adding up his entries for these
>       three at weight=10 and depth=4
>       one may surely construct the brute-force proof.

On the other hand, Jon Borwein suggested a different example; see below,
as well as the paper "About a new kind of Ramanujan-type series," by Jesus
Guillera, Experim. Math. 12 (2003), 507-510.  For those who don't want to
cope with the LaTeX, Gourevitch's conjecture is

32  =  pi^3 sum_{n>=0} r(n)^7 (1 + 14n + 76n^2 + 168n^3)/64^n

where r(n) = ( (1/2)*(3/2)*...*(2n-1)/2 ) / n!

Tim

Jon Borwein wrote:
> Dear Tim: I striking formula that is certainly unproven is the
> formula fdor 32/pi^3 below discovered a few years ago  by Gourevich.
> Guillera proved the first two using the Wilf-Zeiberger algorithm.
> Cheers, Jon

\begin{eqnarray*}
\frac{128}{\pi^2} &=& \sum_{n=0}^\infty (-1)^n
r(n)^5 (13 + 180 n + 820 n^2)\left(\frac{1}{32}\right)^{2n}
\\
\frac{8}{\pi^2} &
=& \sum_{n=0}^\infty (-1)^n
r(n)^5 (1 + 8 n + 20 n^2)\left(\frac{1}{2}\right)^{2n} \\
\label{guillera3} \frac{32}{\pi^3}
&\stackrel{?}{=}&\sum_{n=0}^\infty
r(n)^7 (1 + 14 n + 76 n^2 + 168
n^3)\left(\frac{1}{8}\right)^{2n}.
\end{eqnarray*}
where
\begin{eqnarray*}
r(n) &=& \frac{(1/2)_{n}}{n!}
\; = \; \frac{1/2 \cdot 3/2 \cdot \, \cdots \, \cdot (2n-1)/2}{n!}
\; = \; \frac {\Gamma(n+1/2)}{\sqrt{\pi }\,\Gamma(n+1)}
\end{eqnarray*}