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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?

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  • $\begingroup$ You don't want the comparison test, you want the integral test or the p-test. $\endgroup$
    – vadim123
    Dec 16, 2014 at 4:49

1 Answer 1

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

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  • $\begingroup$ 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, 2014 at 4:52
  • $\begingroup$ @user201483 Yes. $\endgroup$
    – Hanul Jeon
    Dec 16, 2014 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, 2014 at 4:54
  • $\begingroup$ @user201483 Yes. $\endgroup$
    – Hanul Jeon
    Dec 16, 2014 at 4:56
  • $\begingroup$ Nice. Also, I love your hat choice! $\endgroup$
    – Prism
    Dec 16, 2014 at 5:06

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