0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ 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 $|\frac{a_{n}}{a_{n+1}}|$. $\endgroup$ – Alex Nov 8 '14 at 11:19
  • $\begingroup$ Is the radius of convergence sensitive to whether you take $\lim\limits|\frac{a_{n+1}}{a_{n}}|$ or $\lim\limits|\frac{a_{n}}{a_{n+1}}|$? $\endgroup$ – Alex Nov 8 '14 at 11:26
  • 1
    $\begingroup$ @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. $\endgroup$ – Rory Daulton Nov 8 '14 at 12:12
  • 1
    $\begingroup$ 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? $\endgroup$ – Rory Daulton Nov 8 '14 at 13:23
  • 1
    $\begingroup$ 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. $\endgroup$ – Rory Daulton Nov 8 '14 at 13:28
1
$\begingroup$

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 $1$.

$\endgroup$
  • $\begingroup$ 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 $|\frac{a_{n}}{a_{n+1}}|$. $\endgroup$ – Alex Nov 8 '14 at 11:18
  • $\begingroup$ if you are content with the answer, either upvote or check so users can know you question is answered. $\endgroup$ – Varun Iyer Nov 8 '14 at 11:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.