## Reciprocal of a number raised to s-1 as a sum of reciprocals raised to s

Mathematica 8

 Clear[s, n, k] s = 2; Table[Sum[n/(k + n - 1)^s - n/(k + n)^s, {k, 1, Infinity}], {n, 1, 12}] 
{1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10, 1/11, 1/12}

$\frac{1}{n^{s-1}} = \sum\limits_{k=1}^{\infty} \frac{n}{(k+n-1)^{s}} - \frac{n}{(k+n)^{s}}$