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

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