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

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 $$(3)$$?

2 Answers

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.

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: $$\displaystyle \sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(n+1)(n+2)}=-\frac13\int_0^1\ln^3(1-t)\ dt=\frac13\int_0^1\ln^3(t)\ dt=-\frac13(-6)=2$$