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?
 A: 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.
$$\sum_{n=1}^\infty \dfrac{H_n}{n^3} z^n = \int_0^z f_2(t)\; dt$$
A: $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n}=\frac{\pi^4}{720}+\frac{\ln^42}{24}-\frac{\ln2}8\zeta(3)+\operatorname{Li}_4\left(\frac12\right).$$
A: You can have instead the equivalent integral representation 

$$ I = \int_{0}^{1}\frac{\ln^2(u)\ln(1-u/2)}{u(u-2)}du \sim .5582373010. $$

Try to evaluate the above integral. See my answer. See also here.
A: By first finding the following integral by using the algebraic identity
$a^2b=\frac{1}{6}\left(a+b\right)^3-\frac{1}{6}\left(a-b\right)^3-\frac{1}{3}b^3$ one can easily prove avoiding Euler sums that:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=-\frac{1}{4}\zeta \left(4\right)+2\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{4}\ln ^4\left(2\right)$$
Now:
$$\int _0^1\frac{\ln ^2\left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx=\frac{1}{2}\ln \left(2\right)\int _0^1\frac{\ln ^2\left(x\right)}{1-\frac{x}{2}}\:dx+\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-\frac{x}{2}\right)}{1-\frac{x}{2}}\:dx$$
$$=2\ln \left(2\right)\sum _{k=1}^{\infty }\frac{1}{k^3\:2^k}-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\sum _{k=1}^{\infty }\frac{1}{k^4\:2^k}$$
$$=2\ln \left(2\right)\operatorname{Li}_3\left(\frac{1}{2}\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$
$$=\frac{7}{4}\ln \left(2\right)\zeta \left(3\right)-\ln ^2\left(2\right)\zeta \left(2\right)+\frac{1}{3}\ln ^4\left(2\right)-2\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}+2\operatorname{Li}_4\left(\frac{1}{2}\right)$$
By making use of the result we find:
$$\sum _{k=1}^{\infty }\frac{H_k}{k^3\:2^k}=\frac{1}{8}\zeta \left(4\right)+\operatorname{Li}_4\left(\frac{1}{2}\right)-\frac{1}{8}\ln \left(2\right)\zeta \left(3\right)+\frac{1}{24}\ln ^4\left(2\right)$$
