determinant-prove-that-n-1-H-n-H-n-2-n-3-n-1-H-n-H-n-3-n-2-21-8-6-2-3-by-Math-Amin-11-fb-96-

Question Number 154544 by amin96 last updated on 19/Sep/21

\begin{array}{cc} prove that \\ \underset{n = 1}{\sum^{\infty}} \frac{H_{n} H_{n}^{(2)}}{n^{3}} + \underset{n = 1}{\sum^{\infty}} \frac{H_{n} H_{n}^{(3)}}{n^{2}} = \frac{21}{8} ζ (6) + ζ^{2} (3) \end{array}

b y M a t h . A m i n 11 . f b . 96