POW Riemann II

A more respectable result using Riemann zeta values is

$\left(\zeta \right(2)-1)+\left(\zeta \right(3)-1)+\left(\zeta \right(4)-1)+\cdots =1$

because the terms in this series start at 0.6449 and rapidly approach zero. It is well known that for even ordinals

$\zeta \left(2k\right)=\frac{(-1{)}^{k+1}(2\pi {)}^{2k}}{2\left(2k\right)!}{B}_{2k}$

for Bernoulli numbers $B_{2k}$. More recently, formulas for odd ordinals have been found by Linas Vepstas. From his 2006 paper we have

$\zeta (4m-1)=-2\sum _n{\text{Li}}_{4m-1}\left(e^{-2\pi n}\right)-\frac{1}{2}\left(2\pi {)}^{4m-1}\sum _{j=0}^{2m}\right(-1{)}^j\frac{B_{2j}{B}_{4m-2j}}{\left(2j\right)!(4m-2j)!}$

$\zeta (4m+1)=(1+(-4{)}^m-2^{4m+1}{)}^{-1}[-2\sum _n{\text{Li}}_{4m+1}(e^{-2\pi n+\pi i})$
$+2(2^{4m+1}-(-4{)}^m\left)\sum _n{\text{Li}}_{4m+1}\right(e^{-2\pi n})$
$+\left(2\pi {)}^{4m+1}\sum _{j=0}^m\right(-4{)}^{m+j}\frac{B_{4m-4j+2}{B}_{4j}}{(4m-4j+2)!\left(4j\right)!}$
$+\frac{1}{2}\left(2\pi {)}^{4m+1}\sum _{j=0}^{2m+1}\right(-4{)}^j\frac{B_{4m-2j+2}{B}_{2j}}{(4m-2j+2)!\left(2j\right)!}]$

for ${\text{Li}}_s\left(x\right)$ the polylogarithm function, which generalises the Riemann zeta function. In other words, one can think of $\zeta (4m-1)$ as the $n=0$ term in a formula relating polylogarithm values to the Bernoulli numbers.