How to find $\sum \left(1 + 1/2 + \dots + 1/(n + k + 1)\right)\frac{1}{n(n + 1)...(n + k + 1)}$? I've got two series that look alike, 
\begin{align}
S_1 = \sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}\right).\frac{1}{n(n + 1)}\tag{1}\\
S_2 =\sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \dots + \frac{1}{n+ 1}\right).\frac{1}{n(n + 1)}.\tag{2}\\ 
\end{align}
For the first one I can find the sum, it equals to $\pi^2/6$ by writing $$S_1 = \sum_{n = 1}^\infty \left(\frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n} -\frac{1 + \frac{1}{2} + \dots + \frac{1}{n+ 1}}{n + 1}+ \frac{1}{(n + 1)^2}\right)$$
and the sum telesope.
As for the second one, by substracting the two series we got 
$$S_2 - S_1 = \sum_{n = 1}^\infty \frac{1}{n(n + 1)^2} = 2 - \frac{\pi^2}{6},$$  hence $S_2 = 2$.
I got curious  if there exist any general formula for the series $$ \sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \dots + \frac{1}{n + k}\right)\frac{1}{n(n + 1)...(n + k+1)}$$
and
$$ \sum_{n = 1}^\infty \left(1 + \frac{1}{2} + \dots + \frac{1}{n + k+ 1}\right)\frac{1}{n(n + 1)...(n + k+1)}$$ with $k \geq 0$.
I think it would be similar to the case $k = 0$ above. But I just can't find my way.

$$\begin{eqnarray*}\sum_{n\geq 1}\frac{H_{n+k+1}}{n(n+1)\cdots(n+k+1)}&=&\frac{1}{k!}\sum_{n\geq 1}\left(\frac{\binom{k}{0}}{n}-\frac{\binom{k}{1}}{n+1}+\ldots\pm\frac{\binom{k}{k}}{n+k}\right)\frac{H_{n+k+1}}{n+k+1}\\&=&\frac{1}{k!}\sum_{n\geq 1}\frac{H_{n+k+1}}{n+k+1}\int_{0}^{1}\sum_{h=0}^{k}\binom{k}{h}(-1)^h x^{n-1+h}\,dx\\&=&\frac{1}{k!}\int_{0}^{1}(1-x)^k\sum_{n\geq 1}\frac{H_{n+k+1}}{n+k+1}x^{n-1}\,dx\end{eqnarray*}$$
can be computed from Euler's Beta function,
$$ \sum_{m\geq 1}\frac{H_m}{m}\, x^m =\frac{1}{2}\log^2(1-x)+\text{Li}_2(x),$$
$$ \int_{0}^{1}(1-x)^k \log^2(1-x)\,dx = \frac{2}{(k+1)^3}, $$
$$ \int_{0}^{1}(1-x)^k\text{Li}_2(x)\,dx \stackrel{\text{IBP}}{=}\frac{\zeta(2)}{k+1}-\frac{H_{k+1}}{(k+1)^2}.$$ 
