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In this paper on section [5], Recently J. Choi [4, Corollary 3] proved a sequence of identities:

$$\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=2\tag1$$

Let just generalize $(1)$

$$\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)\cdots(n+k)}\tag2$$

where $k\ge 2$

We conjectured the closed form of $(2)$ to be

$$\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)\cdots(n+k)}=\frac{2^k}{(2k-2)!!}\cdot \frac{1}{(k-1)^3}=\frac{2}{(k-1)^3(k-1)!}\tag3$$

How may we prove $(3)$?

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Just realize you are computing $$\int_{0}^{1}(1-x)^n \log^2(1-x)\,dx,\qquad \int_{0}^{1}(1-x)^m\text{Li}_2(x)\,dx $$ which are elementary integrals.

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just a little contribution

we have $$\displaystyle \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$ integrate both sides w.r.t $x$ from $x=0$ to $t$, we get: $$\displaystyle \sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)\frac{t^{n+1}}{n+1}=-\frac13\ln^3(1-t)$$ integate both sides w.r.t $t$ from $t=0$ to $1$, we get: $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=-\frac13\int_0^1\ln^3(1-t)\ dt=\frac13\int_0^1\ln^3(t)\ dt=-\frac13(-6)=2$$

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