Method for solving radius of convergence problem

Question

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.

Answer 1

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 $$ |x| < \frac{1}{\limsup_n \left|\frac{a_{n+1}}{a_n}\right|}. $$

Answer 2

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.

In fact this is basically how the criterion you used first is obtained in the first place.

Thus, what you try to do is fine. However, it is somewhat more complicated since in fact you redo the proof of a criterion you already know.