Prove or disprove that $S=\displaystyle\sum_{n=1}^{\infty}\frac{{H_n^{2}}~{H_n^{(2)}}+3{H_n^{(4)}}}{n~2^n}=\frac{25}{16}\zeta(5)+\frac{7}{8}\zeta(2)\zeta(3)$.
2026-03-26 02:44:39.1774493079
Hard sum with harmonics numbers
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