A power series that converges for $|x| \leq 1$ and diverges otherwise.

Question

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?

You don't want the comparison test, you want the integral test or the p-test. — vadim123, Dec 16 '14 at 04:49

score 5 · Answer 1 · answered Dec 16 '14 at 04:49

5

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

answered Dec 16 '14 at 04:49

Hanul Jeon

27,376

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? – user201483 Dec 16 '14 at 04:52
@user201483 Yes. – Hanul Jeon Dec 16 '14 at 04:53
And that goes for any series? If the absolute value of the series converges, then the original series converges? – user201483 Dec 16 '14 at 04:54
@user201483 Yes. – Hanul Jeon Dec 16 '14 at 04:56
Nice. Also, I love your hat choice! – Prism Dec 16 '14 at 05:06

A power series that converges for $|x| \leq 1$ and diverges otherwise.

1 Answers1