Prove this Ramanujan series 
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 $$\sum\frac{1}{n^3}\cdot\frac{1}{2^n}$$

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
$$
\text{Li}_{3}\left(\frac{-z}{1-z}\right)+\text{Li}_{3}(z)+\text{Li}_{3}(1-z)\\
=\zeta(3)+\zeta(2)\log(1-z)-\frac{1}{2}\log(z)\log^2(1-z)+\frac{1}{3}\log^3(1-z)\\
$$
