POW Riemann

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

$S(n) \equiv \sum_{k} B(n+k ; k)^{-1}$

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

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

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 $S(3) = \frac{3}{2}$. 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 $S(n)$ is a series of reciprocals of tetrahedral numbers in dimension $n$. Indeed

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

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

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

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

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

What kind of zeta sums do we get in general?