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Double Euler sum

Posted: Mon May 29, 2017 11:46 am
by Tolaso J Kos
Here is something that came up today while evaluating something else.

Prove that

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^6 \left ( m^2+n^2 \right )} = \frac{13 \pi^8}{113400}$$