occasional meanderings in physics' brave new world

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Location: New Zealand

Marni D. Sheppeard

Thursday, June 05, 2008

POW Riemann II

A more respectable result using Riemann zeta values is

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

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

$\zeta (2k) = \frac{(-1)^{k+1} (2 \pi)^{2k}}{2 (2k)!} B_{2k}$

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

$\zeta (4m - 1) = - 2 \sum_{n} \textrm{Li}_{4m - 1} (e^{- 2 \pi n}) - \frac{1}{2} (2 \pi)^{4m - 1} \sum_{j=0}^{2m} (-1)^{j} \frac{B_{2j} B_{4m - 2j}}{(2j)!(4m - 2j)!}$

$\zeta (4m + 1) = (1 + (-4)^{m} - 2^{4m + 1})^{-1} [-2 \sum_{n} \textrm{Li}_{4m + 1} (e^{- 2 \pi n + \pi i})$
$+ 2(2^{4m+1} - (-4)^{m}) \sum_{n} \textrm{Li}_{4m + 1} (e^{- 2 \pi n})$
$+ (2 \pi)^{4m+1} \sum_{j=0}^{m} (-4)^{m+j} \frac{B_{4m - 4j + 2}B_{4j}}{(4m - 4j + 2)!(4j)!}$
$+ \frac{1}{2} (2 \pi)^{4m+1} \sum_{j=0}^{2m+1} (-4)^{j} \frac{B_{4m - 2j + 2}B_{2j}}{(4m - 2j + 2)!(2j)!} ]$

for $\textrm{Li}_{s}(x)$ 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.