Calculating $\sum_{n=1}^\infty {\frac{1}{2^nn(3n-1)}}$ I'd appreciate any help, I know it has something to do with the geometric series but I still can't figure out how. I thought about integration but couldn't find a way to do it.
$$\sum_{n=1}^\infty \frac{1}{2^nn(3n-1)}$$
Thanks.
 A: Outline: Let $f(x) = \sum_{n=1}^\infty \frac{x^n}{n(3n-1)}$, for $x \in (-1,1)$. Everything is well-defined and one can rewrite
$$
f(x) = 3\sum_{n=1}^\infty \frac{x^n}{3n-1} - \sum_{n=1}^\infty \frac{x^n}{n}.
$$
You are looking for $f(\frac12)$. Now, getting a closed-form expression for $h(x)=\sum_{n=1}^\infty \frac{x^n}{n}$ is not hard (and has been treated many times on this website), so let's focus on the first term (which is similar, albeit trickier). Setting $t=x^{1/3}$, we actually are interested in
$$
g(t)=\sum_{n=1}^\infty \frac{t^{3n}}{3n-1} = \frac{1}{t}\sum_{n=1}^\infty \frac{t^{3n-1}}{3n-1} = \frac{1}{t}\sum_{n=1}^\infty \int_0^t u^{3n-2} du = 
\frac{1}{t}\int_0^t \left( \sum_{n=1}^\infty u^{3n-2} \right)du
$$
which can be rewritten as follows (note that all above is "legit" for $t\in(0,1)$, by properties of power series within their radius of convergence):
$$
g(t) = \frac{1}{t} \int_0^t \frac{du}{u^2} \frac{u^3}{1-u^3} = 
\frac{1}{t} \int_0^t \frac{u}{1-u^3} du
$$
which can be computed by standard means (even though it's not my favourite activity, by far). This gives you a closed-form expression for $g(t)$; hence for $f(x) = 3g(x^{1/3}) - h(x)$. It remains to plug in $x=1/2$ to conclude.
A: $$\log\frac{1}{1-x}=\sum_{n=1}^\infty \frac{x^n}{n}$$
$$\log\frac{1}{1-x^3}=\sum_{n=1}^\infty \frac{x^{3n}}{n}$$
$$\frac{1}{x^2}\log\frac{1}{1-x^3}=\sum_{n=1}^\infty \frac{x^{3n-2}}{n}$$
$$\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx=\sum_{n=1}^\infty \int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{x^{3n-2}}{n}dx$$
$$\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx=\sum_{n=1}^\infty \frac{\sqrt[3]{2}}{n2^n(3n-1)}$$
so 
$$\sum_{n=1}^\infty \frac{1}{n2^n(3n-1)}=\frac{1}{\sqrt[3]{2}}\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx$$
$$=-\log 2+\frac{1}{2\sqrt[3]{2}}\log(\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{2}}+1)-\frac{1}{\sqrt[3]{2}}\log(1-\frac{1}{\sqrt[3]{2}})-\frac{\sqrt{3}}{\sqrt[3]{2}}\tan^{-1}(\frac{\sqrt[3]{4}+1}{\sqrt{3}})+\frac{\pi}{2\sqrt[3]{2}\sqrt{3}}$$
$$=0.282319554....$$
