# Method for solving radius of convergence problem

Hi I am interested if the following method for solving for the radius of convergence for power series problem is a valid method:

Find the radius of convergence of the following: $$\sum_{n=1}^{\infty}(-1)^{n}n^{-\frac{2}{3}}x^{n}$$

I know that one can solve this by taking $a_{n} := (-1)^{n}n^{-\frac{2}{3}}$ and then taking the limit to get the radius of convergence: $$R = \lim\limits_{n \rightarrow \infty}|\frac{a_{n}}{a_{n+1}}| = 1$$ Thus the radius of convergence is $1$. Can we use the following method where we instead take $a_{n} := (-1)^{n}n^{-\frac{2}{3}}x^{n}$ and then consider $$\lim\limits_{n \rightarrow \infty}|\frac{a_{n+1}}{a_{n}}| = |x|$$

In this method we take the radius of convergence to be the coefficient of $|x|$, which is once again $1$. Is this fine? Is it a well established method?

Thanks for any help.

Quick answer: yes, more or less it is correct! You are just applying the quotient rule to the series of general term $b_n = a_n x^n$, and it gives convergence provided that $\limsup_n \left|\frac{b_{n+1}}{b_n}\right|<1$, but this exactly means that $$|x| < \frac{1}{\limsup_n \left|\frac{a_{n+1}}{a_n}\right|}.$$
• Thanks for you response. Do you think it would be a valid method to consider $a_{n} = (-1)^{n}n^{-\frac{2}{3}}x^{n}$ take $\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}$ and then take the absolute value of the limit as the radius of convergence. – Alex Nov 8 '14 at 15:48
What you do is not unreasonable. When you show that the limit of $|a_{n+1}/a_n|=|x|$ you can continue by saying that therefore (this needs some justification, but is fine) the series converges for $|x|< 1$ and diverges for $|x|>1$, that is $1$ is its radius of convergence.