Prove that:
$\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{k^q} = (1 + \frac{q}{2})\zeta(q + 1) - \frac{1}{2}\cdot \sum_{n=1}^{q-2}\zeta(k+1)\zeta(q-k)$
It looks tough just to start off with.
Any ideas on approach?
Prove that:
$\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{k^q} = (1 + \frac{q}{2})\zeta(q + 1) - \frac{1}{2}\cdot \sum_{n=1}^{q-2}\zeta(k+1)\zeta(q-k)$
It looks tough just to start off with.
Any ideas on approach?