Prove how this series: $\frac{1}{1^3}\cdot\frac{1}{2} + \frac{1}{2^3}\cdot\frac{1}{2^2} + \frac{1}{3^3}\cdot\frac{1}{2^3} + \frac{1}{4^3}\cdot\frac{1}{2^4} +...= \frac{1}{6}(\log 2)^3+\frac{\pi ^2}{12}(\log 2) + (\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+..)$
I don't know how to begin but I know that the left-hand side is of the form $$\sum\frac{1}{n^3}\cdot\frac{1}{2^n}$$
On
This is by no means a complete answer but really a long comment. We could also ask if for a fixed number $k$ is it true that $$ \sum_{n=1}^{\infty}\frac{1}{n^3k^n}=\text{Li}_{3}\left( \frac{1}{k}\right); $$ where $\text{Li}_{3}()$ is the trilogarithm. Really we are computing $\text{Li}_{3}(\frac{1}{2}).$
We have the following functional equation from Euler
$$ \text{Li}_{3}\left(\frac{-z}{1-z}\right)+\text{Li}_{3}(z)+\text{Li}_{3}(1-z)\\ =\zeta(3)+\zeta(2)\log(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{1}{3}\log^3(1-z)\\ $$
For $0 < x < 1$ we have
$$ \sum_{k=1}^{\infty}x^k = \frac{x}{1-x} $$
now calling $S_0 = \frac{x}{1-x}$ we have the relationships
$$ x S_3' = S_2\\ x S_2' = S_1\\ x S_1' = S_0 $$
after integration with null constants we have
$$ \left\{ \begin{array}{rcl} S_1 & = & -\log (1-x) \\ S_2 & = & -\log (1-x) \log (x)-\text{Li}_2(1-x) \\ S_3 & = & -\log (1-x) \log ^2(x)-\text{Li}_2(1-x) \log (x)-\text{Li}_2(x) \log (x)+\text{Li}_3(x) \\ \end{array} \right. $$
making now $x = \frac 12$
$$ \cases{ S_1(\frac 12)=\log (2) \\ S_2(\frac 12)= \frac{1}{12} \left(\pi ^2-6 \log ^2(2)\right) \\ S_3(\frac 12)=\frac{1}{24} \left(4 \log ^3(2)-2\pi ^2\log (2)+21 \zeta (3)\right) \\ } $$