## Borwein and Bradley’s Apery-like formulas for zeta(4n+3)

Apery gave,

\begin{aligned} \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom {2k}k}\end{aligned}

J. Borwein and D. Bradley found this can be generalized to $\zeta(4n+3)$. Define the functions,

\begin{aligned} &B(a_0)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\\ &B(a_0,a_1,a_2,\dots)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{a_0}\,\binom {2k}k}\; \sum_{p=1}^{k-1} \frac{1}{p^{a_1}}\;\sum_{q=1}^{k-1} \frac{1}{q^{a_2}}\;\dots \end{aligned}

then,

\begin{aligned} \frac{2}{5}\,\zeta(3) &= B(3)\\ \frac{2}{5}\,\zeta(7) &= B(7)+5B(3,4)\\ \frac{2}{5}\,\zeta(11) &= B(11)+5B(7,4)-\frac{15}{2}B(3,8)+\frac{25}{2}B(3,4,4)\\ \frac{2}{5}\,\zeta(15) &= B(15)+5B(11,4)-\frac{15}{2}B(7,8) +\frac{25}{2}B(7,4,4)+\frac{130}{6}B(3,12)\\&-\frac{225}{6}B(3,8,4)+ \frac{125}{6}B(3,4,4,4) \end{aligned}

and so on.  Beautiful, aren’t they? Notice that all the $a_i$ (excepting $a_0$) are all divisible by 4. This infinite family has a generating function. Let z $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^3(1-z^4/k^4)}&=\frac{5}{2}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^3\;\binom {2k}k} \frac{1}{1-z^4/k^4}\prod_{j=1}^{k-1}\frac{1+4z^4/j^4}{1-z^4/j^4}\end{aligned}

On the other hand, for s = 4n+1,

\begin{aligned} 2\,\zeta(5) &= 4B(5)-5B(3,2)\\[2.5mm] 4\,\zeta(9) &= 9B(9)-5B(7,2)+20B(5,4)+45B(3,6)-25B(3,4,2)\\[2.5mm] 12\,\zeta(13) &= 28B(13)-10B(11,2)+90B(9,4)\\&+90B(7,6)-50B(7,4,2)-60B(5,8)+100B(5,4,4)\\&-310B(3,10)+75B(3,8,2)+450B(3,6,4)-125B(3,4,4,2)\end{aligned}

with this version for $\zeta(13)$ found by Jim Cullen.  There are various versions for both s = 4n+1 and 4n+3.  For example, for $\zeta(7)$, we have the relations,

\begin{aligned} 4\zeta(7) &= 8B(7,0)\,-\,5B(3,4)\,-\,8B(5,2)+5B(3,2,2)\\ 0 &= -2B(7,0) - 55B(3,4)-8B(5,2)+5B(3,2,2)\end{aligned}

Eliminating the last two terms will yield the shorter relation given by Borwein and Bradley. There is a generating function for all s = 2n+1, but none is known that is only for s = 4n+1. See Apery-Like Formulae for $\zeta(4n+3)$ for more details.