# Prove $\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$

This series represents the sum of reciprocal pentagonal numbers (multiplied by $3$). I got its value from Wolfram Alpha:

$$\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$$

My attempt: Turn the series into an integral:

$$\sum_{n=1}^\infty \frac{x^{3n}}{n}=-\ln(1-x^3)$$

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

$$\sum_{n=1}^\infty \frac{1}{n(3n-1)}=-\int_0^1 \frac{\ln(1-x^3)}{x^2} dx$$

I don't know how to solve this integral.

We can also transform the common term into two parts:

$$\frac{1}{n(3n-1)}=\frac{3}{3n-1}-\frac{1}{n}$$

But this gives us two diverging series, so it doesn't help.

• It seems like integration by parts would reduce that last integral to the integral of a rational function. – Matthew Leingang Jul 12 '16 at 15:36
• $1-x^3=(1-x)(x^2+x+1)$. Then we can use integration by parts. – Mc Cheng Jul 12 '16 at 15:43
• Thank you for the comments! I've got it, should've figure out myself – Yuriy S Jul 12 '16 at 15:44

$$\sum_{n\geq 1}\frac{1}{3n(3n-1)}=\int_{0}^{1}\sum_{n\geq 1}\left(x^{3n-2}-x^{3n-1}\right)\,dx = \int_{0}^{1}\frac{x-x^2}{1-x^3}\,dx$$ and the integral $$\int_{0}^{1}\frac{x-x^2}{1-x^3}\,dx = \int_{0}^{1}\frac{x}{1+x+x^2}\,dx$$ is straightforward to compute.

• I am starting to feel annoyed by random downvotes these days. Please explain what is wrong here. – Jack D'Aurizio Jul 12 '16 at 22:40

We can also continue your way. We have $$-\int\frac{\log\left(1-x^{3}\right)}{x^{2}}dx=\frac{\log\left(1-x^{3}\right)}{x}-\int\frac{3x}{1-x^{3}}dx$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\int\frac{1-x}{x^{2}+x+1}dx-\int\frac{1}{x-1}dx$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\int\frac{2x+1}{x^{2}+x+1}dx-\frac{3}{2}\int\frac{1}{x^{2}+x+1}dx$$ $$=\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\log\left(x^{2}+x+1\right)-\frac{3}{2}\int\frac{1}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}dx$$

$$=\color{red}{\frac{\log\left(1-x^{3}\right)}{x}-\log\left(x-1\right)-\frac{1}{2}\log\left(x^{2}+x+1\right)-\sqrt{3}\arctan\left(\frac{2x+1}{\sqrt{3}}\right)+C.}$$

• @You'reInMyEye You're welcome. – Marco Cantarini Jul 12 '16 at 17:10

We can do the indefinite integral. Start by the obvious guess $$\frac{d}{dx} \frac{-\log(1-x^3)}{x} = \frac{\log(1-x^3)}{x^2}+\frac{3x}{1-x^3}$$ Now we have to wipe out that second term.
$$\frac{3x}{1-x^3}=\frac{1}{1-x}+\frac{x-1}{1+x+x^2} \\ \int\frac{1}{1-x}= -\log(1-x)$$ leaving the term $\frac{x-1}{1+x+x^2}$ to deal with. A reasonable try would be subtracting off $$\frac{d}{dx} (\log{\sqrt{1+x+x^2}}) = \frac{x+\frac12}{1+x+x^2}$$ And we are left with $$\frac12\int\frac1{1+x+x^2}dx$$ If you know that $$\frac{d}{dx} \tan^{-1}(a+b*x) = \frac{b}{1+(a+bx)^2}$$ and set that equal to $\zeta/(1+x+x^2)$ you can find $a, b$ and the multiplicative constant, to get $$\frac{d}{dx}\left( \frac{\tan^{-1}(\frac{1+2x}{\sqrt{3}})} {\sqrt{3}} \right) =\frac12\int\frac1{1+x+x^2}dx$$ I may have a factor of $3$ error in this chain. At any rate, you can now see that the $\sqrt{3}\pi$ in the answer comes from an arctan of $\frac{1+2\cdot 1}{\sqrt{3}}$, which is $\pi/3$.

Glue all these parts together and you have your indefinite integral.