Q:

Determining a radius convergence of a power series

Let $$ \sum_{n=0}^\infty \frac{(-1)^n}{3n+1} x^{3n+1} $$
Is there an immediate way to determine $R=1$? 

A:

There are some way how to determine $R=1$, for example:

Ratio test We have:

$$\lim_{n \to \infty}\frac{|a_{n+1}|}{|a_n|}=\lim_{n \to \infty}\frac{(3n+1)x^3}{(3n+4)}=|x^3|$$
So series converges for $|x|<1$.

Root test We have:

$$\limsup_{n \to \infty}\sqrt[n]{|a_n|}=\lim_{n \to \infty}|x|^3\sqrt[n]{3n+1}=|x|^3$$

