# Radius of Convergence

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• Nov 14th 2013, 01:05 PM
iPod
Radius of Convergence
Hello,
I have to find the radius of convergence with this series

$\Sigma 2^{-n}z^{n^2}}$

I've tried the root test and the ratio test but the solution they give seem inconclusive, is the any clues as to what I should do?
Thank you
• Nov 14th 2013, 01:46 PM
Plato
Re: Radius of Convergence
Quote:

Originally Posted by iPod
Hello,
I have to find the radius of convergence with this series

$\Sigma 2^{-n}z^{n^2}}$

I've tried the root test

The root test works. For what values does ${\lim _{n \to \infty }}\frac{{{z^n}}}{2}$ converge?
• Nov 14th 2013, 01:56 PM
iPod
Re: Radius of Convergence
Oh I see, so basically it converges to 0 if |z|<1, meaning the series converges anywhere (provided |z|<1)
• Nov 14th 2013, 02:00 PM
iPod
Re: Radius of Convergence
Actually, |z|<2 because it's |z/2| < 1
• Nov 14th 2013, 03:20 PM
Plato
Re: Radius of Convergence
Quote:

Originally Posted by iPod
Actually, |z|<2 because it's |z/2| < 1

NO! you were right the first time.

Look at the webpage.

You can play around with the numerator to see how it works.