Method of finding radius of convergence

Question

Hi is it acceptable to evaluate the radius of convergence $R$ of this power series $$\sum_{n=1}^{\infty}(-1)^{n}n^{-\frac{2}{3}}x^{n}$$ by instead of taking $a_{n} := (-1)^{n}n^{-\frac{2}{3}}$ we take $a_{n} := (-1)^{n}n^{-\frac{2}{3}}x^{n}$ which results in: $$\lim\limits_{n \rightarrow \infty}|\frac{a_{n+1}}{a_{n}}| = \lim\limits_{n \rightarrow \infty}|x(\frac{n}{n+1})^{\frac{2}{3}}| = x$$

The result being that the radius of convergence is always the coefficient of $x$. Is this an acceptable adaptation of the usual method of finding the radius of convergence?

Answer 1

The ratio test for convergence says that you need to get $$\lim\limits_{n \rightarrow \infty}|\frac{a_{n+1}}{a_{n}}| < 1$$

You already showed that the limit is $|x|$, so you just need $$|x|<1$$

Therefore, the radius of convergence is $1$.

Answer 2

You should always use this method to find the radius of convergence.

As you stated:$$\sum_{n=1}^{\infty}(-1)^{n}n^{-\frac{2}{3}}x^{n}$$

When conducting the root test, you must include the variable $x$. Otherwise your radius of convergence either will never convergence or be $\infty$

Therefore:

$$\lim_{n\to\infty}\left|(-1)^{n}n^{-\frac{2}{3}}x^{n}\right|^{1/n} = \left|x\right|$$

The limit only converges when it is $< 1$. Therefore, $|x| < 1$, and the radius of convergence is $1$.