Calculating $\sum_{n=1}^\infty {\frac{1}{2^nn(3n-1)}}$

Question

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.

Answer 1

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.

Answer 2

$$\log\frac{1}{1-x}=\sum_{n=1}^\infty \frac{x^n}{n}$$ $$\log\frac{1}{1-x^3}=\sum_{n=1}^\infty \frac{x^{3n}}{n}$$ $$\frac{1}{x^2}\log\frac{1}{1-x^3}=\sum_{n=1}^\infty \frac{x^{3n-2}}{n}$$ $$\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx=\sum_{n=1}^\infty \int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{x^{3n-2}}{n}dx$$ $$\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx=\sum_{n=1}^\infty \frac{\sqrt[3]{2}}{n2^n(3n-1)}$$ so $$\sum_{n=1}^\infty \frac{1}{n2^n(3n-1)}=\frac{1}{\sqrt[3]{2}}\int_{0}^{\frac{1}{\sqrt[3]{2}}}\frac{1}{x^2}\log\frac{1}{1-x^3}dx$$ $$=-\log 2+\frac{1}{2\sqrt[3]{2}}\log(\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{2}}+1)-\frac{1}{\sqrt[3]{2}}\log(1-\frac{1}{\sqrt[3]{2}})-\frac{\sqrt{3}}{\sqrt[3]{2}}\tan^{-1}(\frac{\sqrt[3]{4}+1}{\sqrt{3}})+\frac{\pi}{2\sqrt[3]{2}\sqrt{3}}$$ $$=0.282319554....$$

Answer 3

The approach with hypergeometric functions is $$ \sum_{n\ge 1}\frac{1}{n(3n-1)} \frac{1}{2^n} = \frac{1}{2}\sum_{n\ge 0}\frac{1}{(n+1)(3n+2)} \frac{1}{2^n} = \frac{1}{6}\sum_{n\ge 0}\frac{1}{(n+1)(n+2/3)} \frac{1}{2^n} $$ $$ = \frac{1}{6}\sum_{n\ge 0}\frac{(1)_n(2/3)_n}{(2)_n(2/3)(5/3)_n} \frac{1}{2^n} = \frac{1}{4}\sum_{n\ge 0}\frac{(1)_n(2/3)_n(1)_n}{(2)_n(5/3)_n} \frac{1}{2^nn!} = \frac{1}{4}{}_3F_2(1,1,2/3;2,5/3;1/2) $$ By the known contiguous relations, see e.g. Gottschalk and Maslen "Reduction formulae for generalised hypergeometric functions of one variable" in J. Phys A 21 (1988) 1983-1998: $$ (1-2/3){}_3F_2(1,1,2/3;2,5/3;1/2) = {}_3F_2(1,2,2/3;2,5/3;1/2) - (2/3) {}_3F_2(1,1,5/3;2,5/3;1/2) $$ $$ = {}_2F_1(1,2/3;5/3;1/2) - (2/3) {}_2F_1(1,1;2;1/2). $$

There is a well known (Abramowitz-Stegun 15.1.4) $$ {}_2F_1(1,1;2;z)= -\frac{1}{z}\log(1-z) $$ for the second term on the right hand side.

For the first term on the right hand side we have the integral representation (Abramowitz-Stegun) $$ {}_2F_1(1,2/3;5/3;1/2) =\frac{\Gamma(5/3)}{\Gamma(2/3)}\int_0^1 t^{-1/3} (1-t/2)^{-1} dt =\frac23\int_0^1 t^{-1/3} (1-t/2)^{-1} dt $$ Substituting $t=x^3$, $dt/dx=3x^2$ this is $$ =\frac23\int_0^1 \frac{1}{x}\frac{1}{1-x^3/2} 3x^2dx =2\int_0^1 \frac{x}{1-x^3/2} dx $$ which is known by the Gradsteyn-Gyzhik formula 2.126 $$ \int \frac{x dx}{a+bx^3} = -\frac{1}{3b\alpha}\left[\frac12 \ln\frac{(x+\alpha)^2}{x^2-\alpha x +\alpha^2}-\surd 3 \arctan\frac{2x-\alpha}{\surd 3 \alpha}\right] $$ where $\alpha\equiv \sqrt{3}(a/b)$.