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Radius of convergence

Find the radius of convergence of

\sum\limits_{n=0}^\infty 2^{-n}z^{n^2}

.

I used the ratio test to get the following where

a_n =2^{-n}z^{n^2}

\lim_{n \to \infty} \vert \dfrac{a_{n+1}}{a_n}\vert = \lim_{n \to \infty} \vert \dfrac{2^{-(n+1)}z^{(n+1)^2}}{2^{-n}z^{n^2}} \vert

I know how to simplify this further. However, i have been told that you cannot use the ratio test to find the radius of convergence for this series, for some reason that was explained to me but i didn't really understand why. If anyone could explain why the ratio test cannot be used here clearly, that would be great.

Reply 1

little_wizard123

I'm not too sure either (sorry, been a while since I did all this). What was the reason, can you remember?

Reply 2

DFranklin

Unless I'm missing something, you can use the ratio test in the way that you are using it. However, there's another ratio test people sometimes use:

Given a series

\sum_0^\infty a_n z^n

, if

\lim |a_n / a_{n+1}| = L

, (i.e. the limit exists and equals L) then the radius of convergence is also L.

You can't use this ratio test here (because you have z^n^2, not z^n).