6
$\begingroup$

I need to find a power series $\sum a_n z^n$ that converges for $|x| \leq 1$ and diverges otherwise.

I think I have one I just want to be sure.

So, the series:

$\sum \frac{z^n}{n^2}$

has radius of convergence of 1. So it converges when $|z| <1$ and diverges when $|z| >1$, correct?

And we know it converges at $z= \pm 1$ by the comparison test, correct? This part is where I'm having trouble with. Could someone explain in detail how to use the comparison test with this? I know the comparison test says, "if you have two series $\sum a_n$ and $\sum b_n$ with $a_n, b_n \geq0$ and $a_n \leq b_n$, then if $\sum b_n$ converges, then $\sum a_n$ converges." But what other series would you use in the comparison test. I also know that $|\frac{z^n}{n^2}|= \frac{1}{n^2}$. Can you use this fact?

Please help! This series would work correct?

$\endgroup$
  • $\begingroup$ You don't want the comparison test, you want the integral test or the p-test. $\endgroup$ – vadim123 Dec 16 '14 at 4:49
5
$\begingroup$

You don't need the comparison test. If $|z|=1$ then $\sum z^2/n^2$ converges absolutely so it converges.

$\endgroup$
  • $\begingroup$ Brillant! So, since $\sum |\frac{z^n}{n^2}| = \sum \frac{1}{n^2}$ and we know the latter converges, then $\sum \frac{z^n}{n^2}$ converges as well? $\endgroup$ – user201483 Dec 16 '14 at 4:52
  • $\begingroup$ @user201483 Yes. $\endgroup$ – Hanul Jeon Dec 16 '14 at 4:53
  • $\begingroup$ And that goes for any series? If the absolute value of the series converges, then the original series converges? $\endgroup$ – user201483 Dec 16 '14 at 4:54
  • $\begingroup$ @user201483 Yes. $\endgroup$ – Hanul Jeon Dec 16 '14 at 4:56
  • $\begingroup$ Nice. Also, I love your hat choice! $\endgroup$ – Prism Dec 16 '14 at 5:06

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.