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I was looking at this wolfram site on section [25]

A general identity due to B. Cloitre (pers. comm., Jan. 7, 2006)

where $k\ge1$

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

I was trying to generalize $(1)$

Let generalize $(1)$,

where $x\ge0$

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

and hoping to find a closed form but I could only got a partial of it

$k=1$

$$\sum_{n=1}^{\infty}\frac{H_n}{(n+1+x)(n+2+x)}=-\frac{H_x}{x+1}-(H_x)^2+(H_{x+1})^2\tag3$$

How do we go about to find the closed form for $(2)$?

I am assuming $(2)$, may take the closed form of

$$\sum_{n=1}^{\infty}\frac{H_n}{(n+1+x)(n+2+x)\cdots(n+k+1+x)}=G(x)-\sum_{j=0}^{k}(-1)^j{k \choose j}(H_{x+j})^2\tag4$$

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By summation by parts the computation of $(2)$ boils down to the computation of $$ \sum_{n\geq 1}\frac{1}{(n+1)(n+1+x)(n+2+x)\cdots(n+k+x)}=\sum_{n\geq 1}\frac{\Gamma(n+1+x)}{(n+1)\Gamma(n+k+x)} $$ or $$ \frac{1}{\Gamma(k-1)}\sum_{n\geq 1}\frac{B(n+1+x,k-1)}{n+1}=\frac{1}{\Gamma(k-1)}\int_{0}^{1}\sum_{n\geq 1}\frac{(1-z)^{k-2}z^{n+x+2}}{n+1}\,dz $$ or $$ \frac{1}{\Gamma(k-1)}\int_{0}^{1}(1-z)^{k-2}z^{1+x}(-z-\log(1-z))\,dz $$ which only depends on the Beta function and its partial derivatives. In particular the last expression equals

$$ -\frac{\Gamma(x+2)\Gamma(k-1)}{\Gamma(x+k+2)}\left[2+x+(x+k+1)(H_{k-2}-H_{k+x})\right]. $$

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