Originally Posted by
mr fantastic $\displaystyle \displaystyle 1 + \frac{1}{3^6} + \frac{1}{5^6} + .... = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \left(\frac{1}{2^6} + \frac{1}{4^6} + .... \right) = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{1}{2^6} \left(1 + \frac{1}{2^6} + .... \right)$
That is,
$\displaystyle \displaystyle S = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{S}{2^6}$
and $\displaystyle \displaystyle 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ....$ is a well known series .... (Google it).
A more rigorous proof can be constructed using Fourier analysis.