Compute $\sum_{n=1}^{\infty } \frac{x^{3n+1}}{n(3n-1)}$ Compute $$\sum_{n=1}^{\infty } \frac{x^{3n+1}}{n(3n-1)}.$$
I've no idea what to do with it. Can you help me?
 A: For first, it is quite trivial that the radius of convergence is one. 
For any $z\in\mathbb{C}$ such that $|z|<1$ we have:
$$\sum_{n\geq 1}\frac{z^n}{n}=-\log(1-z)\tag{1}$$
and we may use the discrete Fourier transform in order to isolate monomials associated with $n\equiv 2\pmod{3}$. By taking $\omega=\exp\frac{2\pi i}{3}$ (a primitive third root of unity) we have:
$$ \mathbb{1}_{n\equiv 0\!\!\pmod{\!3}}=\frac{1}{3}\left(1^n+\omega^n+\omega^{2n}\right)\tag{2} $$
hence:
$$ \mathbb{1}_{n\equiv 2\!\!\pmod{\!3}}=\frac{1}{3}\left(1^n+\omega\cdot\omega^n+\omega^2\cdot\omega^{2n}\right)\tag{3} $$
from which:
$$\sum_{n\geq 1}\frac{z^{3n-1}}{3n-1}=\!\!\!\!\!\!\sum_{\substack{n\geq 1\\n\equiv2\!\!\pmod{3}}}\!\!\!\!\frac{z^n}{n}=-\frac{1}{3}\left(\log(1-z)+\omega\log(1-\omega z)+\omega^2\log(1-\omega^2 z)\right)\tag{4}$$
Now we just have to multiply $(4)$ by $z^2$ and consider that, from $(1)$, we have:
$$\sum_{n\geq 1}\frac{z^{3n+1}}{n}=-z\log(1-z^3).\tag{5}$$
Since $\frac{1}{n(3n-1)}=\frac{3}{3n-1}-\frac{1}{n}$ it follows that:

$$\sum_{n\geq 1}\frac{z^{3n-1}}{n(3n-1)}=-z^2\left(\log(1-z)+\omega\log(1-\omega z)+\omega^2\log(1-\omega^2 z)\right)+z\log(1-z^3).$$

A: First let's decide how to write $$F(x) = \sum_{n=1}^\infty \frac{1}{n(3n-1)} x^{3n-1}$$
The first derivative gives us:
$$F'(x) = \sum_{n=1}^\infty \frac{x^{3n-2}}{n} = x^{-2} \sum_{n=1}^\infty \frac{(x^3)^{n}}{n} = -x^{-2} \ln(1-x^3)$$
Your original series is $$\sum_{n=1}^\infty \frac{x^{3n+1}}{n(3n-1)} = x^2F(x)=-x^2\int_0^x \frac{\ln(1-t^3)}{t^2} dt$$

Also since you didn't ask any specific question about the series, we should also note that the radius of convergence of your series is 1.
This can be determined by computing:
$$\lim_{n\to \infty}\left| \frac{\frac{1}{(n+1)(3n+2)}x^{3n+4}}{\frac{1}{n(3n-1)}x^{3n+1}}\right|=\frac{n(3n-1)}{(n+1)(3n+2)} |x|^{3} = |x|^3.$$
If we wish to satisfy the ratio test, then the series converges provided $|x|^3 < 1$ which means $|x| <1$. Also the series diverges when $|x|^3 >1$. This means the radius of convergence is 1.
You can use the alternating series test to show that the series converges when $x=-1$ and you can use the limit comparison test with $\sum 1/n^2$ to show that the series converges when $x=1$.
Thus the interval of convergence for the series is $[-1,1]$.
