# Method of finding radius of convergence

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?

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 $1$.

• Okay thanks, I just wanted to confirm that this is equivalent method as to where you simply take $a_{n} := (-1)^{n}n^{-\frac{2}{3}}$ and evaluate the radius of convergence by taking the limit of $|\frac{a_{n}}{a_{n+1}}|$. – Alex Nov 8 '14 at 11:19
• Is the radius of convergence sensitive to whether you take $\lim\limits|\frac{a_{n+1}}{a_{n}}|$ or $\lim\limits|\frac{a_{n}}{a_{n+1}}|$? – Alex Nov 8 '14 at 11:26
• @Alex: The usual way is to use the first ratio. I edited my answer to make that more clear (see the link). One reason this way is better is that you may get $\lim\limits|\frac{a_{n+1}}{a_{n}}|=0$; the reciprocal will get no limit. The other way would then be more complicated to allow that one lack of limit but disallow other lacks of limits. – Rory Daulton Nov 8 '14 at 12:12
• The line to which you refer is not the ratio test but rather a calculation of the radius of convergence. The very next line does refer to the ratio test and that line uses the standard $n+1$ on top. Is that clear? – Rory Daulton Nov 8 '14 at 13:23
• Yes, for that calculation put $a_{n+1}$ on the bottom. If the resulting limit is $+\infty$ then the radius of convergence is infinity, meaning that the series converges for all real numbers. – Rory Daulton Nov 8 '14 at 13:28

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 $1$.

• Okay thanks, I just wanted to confirm that this is equivalent method as to where you simply take $a_{n} := (-1)^{n}n^{-\frac{2}{3}}$ and evaluate the radius of convergence by taking the limit of $|\frac{a_{n}}{a_{n+1}}|$. – Alex Nov 8 '14 at 11:18
• if you are content with the answer, either upvote or check so users can know you question is answered. – Varun Iyer Nov 8 '14 at 11:19