# Closed form of $$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{H_n}{kn (k+n)^3}$$

Prove that: $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{H_n}{kn (k+n)^3}=\frac{215}{48}\zeta(6) - 3\zeta^2(3)$

Notations:

This is a part of the set Formidable Series and Integrals.

2 years, 5 months ago

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