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### Two powerful alternating sums $\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}$ and $\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}$

where $H_n$ is the harmonic number and can be defined as: $H_n=1+\frac12+\frac13+...+\frac1n$ $H_n^{(2)}=1+\frac1{2^2}+\frac1{3^2}+...+\frac1{n^2}$ these two sums are already solved by Cornel using ...
### How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$?
I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$, where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$. Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ ...