# How do I find the radius of convergence for this power series?

I stumbled on this power series and was wondering how could I evaluate the interval of convergence:

$$\sum_{n=1}^{\infty }(-1)^nx^{3n}\frac{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}{n!\cdot 3^{n}}$$

I first tried using the ratio test to determine the $x$ interval for which it's convergent, but ran into problems when I couldn't express the numerator in terms of a factorial or some kind of closed expression:

$$\lim_{n\rightarrow \infty }\frac{x^{3(n+1)}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)!\cdot 3^{n+1}}}{x^{3n}\cdot \frac{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}{n!\cdot 3^{n}}}$$ $$=\lim_{n\rightarrow \infty }\frac{x^{3n}\cdot x^{3}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)\cdot n!\cdot 3^{n}\cdot 3}}{x^{3n}\cdot \frac{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}{n!\cdot 3^{n}}}$$ $$=\lim_{n\rightarrow \infty }\frac{ x^{3}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)\cdot 3}}{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}$$ $$=\lim_{n\rightarrow \infty }{ x^{3}\cdot \frac{2\cdot 5\cdot 8\cdot ...\cdot (3n-1)}{(n+1)\cdot 3}}\cdot\frac{1}{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}$$

The best idea I could come up with was trying to write the numerator of the first fraction as $\frac{(3n!)}{3^{n}\cdot{n!}\cdot{1\cdot 4\cdot 7\cdot ...\cdot (3n-2)}}$ but that didn't get rid of the problem, and I'm not even sure if it's correct. Any help would be appreciated :)

• Does the Taylor series of $\frac{1}{\sqrt[3]{1+x^3}}$ ring some bell? :) – Jack D'Aurizio Feb 6 '18 at 14:02

You misunderstood the notation when you wrote down the "$n+1$" term. The ratio should be $$x^{3(n+1)}\frac{1\cdot4\cdot7\cdots(3n-2)\cdot(3(n+1)-2)}{(n+1)!\, 3^{n+1}} \bigg/x^{3n} \frac{1\cdot 4\cdot 7\cdots (3n-2)}{n!\,3^{n}}$$ which simplifies to $$\frac{(3n+1)x^3}{3(n+1)}\ ,$$ and I think you will find it easy enough from here.