POW Riemann

Todd and Vishal's Problem of the Week number 3 (solution here) was to compute, for any $n>1$, the series (from $k=0$)

$S\left(n\right)\equiv \sum _{k}B(n+k;k{)}^{-1}$

where $B(n+k;k)$ is a binomial coefficient. In the case $n=2$ we see that the sum takes the form

$S\left(2\right)=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\cdots =2$

which is a sum of reciprocals of triangular numbers $\frac{1}{2}k(k+1)$ (from $k=1$). For $n=3$ we obtain the reciprocals of the tetrahedral numbers, and $S\left(3\right)=\frac{3}{2}$. The tetrahedral number $T_k=\frac{1}{6}k(k+1)(k+2)$ is the sum of the first $k$ triangular numbers. By the way, only three tetrahedral numbers are perfect squares, namely 1, 4 and $T_{48}=19600$. One guesses that in general $S\left(n\right)$ is a series of reciprocals of tetrahedral numbers in dimension $n$. Indeed

$S\left(n\right)=\frac{n}{n-1}$

But whenever discussing infinite series of simple polytopes, an M theorist cannot help thinking of the Riemann zeta function. Observe that for $n=2$

$S\left(2\right)=\sum _k\frac{2}{k^2+k}=2\zeta \left(2\right)-\sum _k\frac{2}{k^3+k^2}$
$=2\left[\zeta \right(2)-\zeta (3)+\zeta (4)-\zeta (5)+\cdots ]=2$

from which one deduces, allowing cancellation of infinities (!), that

$\zeta \left(2\right)-\zeta \left(3\right)+\zeta \left(4\right)-\zeta \left(5\right)+\cdots =1$

What kind of zeta sums do we get in general?