Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^32^n}$ I'm trying to find a closed form for the following sum
$$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$
where $H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$ is a harmonic number.
Could you help me with it?

Start with the series $$\sum_{n=1}^\infty H_n z^n = - \dfrac{\ln(1-z)}{1-z} = f_0(z) $$   
Then (according to Maple 18)
$$ \sum_{n=1}^\infty \dfrac{H_n}{n} z^n = \int_0^z \dfrac{f_0(t)}{t}\; dt = \operatorname{Li}_{2}(1-z) + \dfrac{\ln(1-z)^2}{2}  = f_1(z)$$   
$$\displaystyle \sum_{n=1}^\infty \dfrac{H_n}{n^2} z^n = \int_0^z \dfrac{f_1(t)}{t} dt$$     
$$= 
\zeta  \left( 3 \right) +\dfrac{1}{2}\, \ln^2 (1-z) 
 \ln  \left( z \right) +\ln  (1-z) \operatorname{Li}_{2} (z) -\operatorname{Li}_{3}(1-z) + \operatorname{Li}_{3}(z) $$   
But for the next integration it fails to find a closed form.
$$\sum_{n=1}^\infty \dfrac{H_n}{n^3} z^n = \int_0^z f_2(t)\; dt$$
