Find the radius of convergence $\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n+1}}{n+1}$ Find the radius of convergence $$\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n+1}}{n+1}$$
Normally, if I have to find the radius of convergence, I'll try to transform it to power series $\sum_{n=1}^{\infty}a_nx^{n}$ and calculate limit of $|\frac{a_{n+1}}{a_n}|$
For example, $\sum_{n=1}^{\infty} (-2)^{n}\frac{x^{3n}}{n+1}$, I will let $t=x^{3}$.
But in this problem, because the index number is 3n+1, I can't do the same. So I stuck. Hope everyone helps me. Thank you so much
 A: Note that, if $x\ne0$,$$\left|\frac{(-2)^{n+1}\frac{x^{3(n+1)+1}}{n+2}}{(-2)^n\frac{x^{3n+1}}{n+1}}\right|=2\frac{n+1}{n+2}|x|^3$$and that therefore$$\lim_{n\to\infty}\left|\frac{(-2)^{n+1}\frac{x^{3(n+1)+1}}{n+2}}{(-2)^n\frac{x^{3n+1}}{n+1}}\right|=2|x|^3.$$So, it follows from the ratio test that the radius of convergence is $\dfrac1{\sqrt[3]2}$.
A: Let $\sum a_n x^n$ denote any power series (at the origin). It is wise when asking a question here to take a moment to type the statement of the theorem you propose to use. I suspect you were trying to apply this "ratio test":

Theorem $1.\quad$ If $r^{-1}\,=\,\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|$, then the radius of convergence is $r$.

It is not clear how to use Theorem 1, since you have
$$\sum a_n x^n\,=\,a_4 x_4\,+\,0 x_5\,+\,0 x_6\,+\,a_7 x_7\,+\,\dots,$$
with infinitely many $a_j = 0.$
Now let $\sum c_n$ denote any series. We have another "ratio test" in this more general context:

Theorem $2.\quad$ Assume that $\lim_{n\to\infty}\frac{c_{n+1}}{c_{n}}\,=\,\rho$, and that $|\rho|<1.$ then $\sum c_n$ converges absolutely.

In the answer provided already, it is essentially this theorem that was applied, although Santos took absolute value from the start, which is no big deal.
By basic properties of power series (and of geometric series), we can see that the radius of convergence is at least $x=\xi=2^{-1/3}.$ We can show that the radius is exactly this (in agreement with the conclusion of Santos) if we show divergence of your series for either of $x=\pm \xi.$
You can check that $(-2)^n\,=\,(-\xi)^{-3n}.$ Then you will find that we get a harmonic series upon the choice $x=-\xi.$
