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I am trying to find the common ratio of $\sum_{n=0}^\infty 2^{-n}z^{n^2}$. Writing it out in full I got: $\frac{1}{2^1}+(z^1)^1+\frac{1}{2^2}+(z^2)^2+\frac{1}{2^3}+(z^3)^3+...$

So the common ratio is $\frac{1}{2}z^{something?}$


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Some of your pluses should be timeses. $\sum a_n$ is geometric if and only if $\frac{a_{n+1}}{a_n}$ does not depend on $n$. – Jonas Meyer Nov 11 '11 at 3:33
If you are trying to use the ratio test, then you should figure out when $\limsup_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$. Even if a limit does not exist, the $\limsup$ of a bounded sequence always exists and is finite. – JavaMan Nov 11 '11 at 3:49
I don't know whether it is part of your arsenal, but for this problem the Root Test is marginally easier to apply. – André Nicolas Nov 11 '11 at 8:45
up vote 5 down vote accepted

Let us be systematic about how we find the ratio of consecutive terms. The terms are given by $ a_n = \frac{z^{n^2} }{2^n} $ so the ratio is $$ \frac{a_{n+1} }{a_n } = \frac{ z^{(n+1)^2} }{2^{n+1} } \cdot \frac{2^n}{z^{n^2} }= \frac{ z^{(n+1)^2-n^2} }{2} = \frac{z^{2n+1 }}{2}. $$

Note that this value changes with each $n$ so it is not "common", and this series is not a geometric series.

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Thanks for your help. If I was asked to find the radius of convergence of this series, but since its not geometric and not "common", can I still do the following? $$ \frac{|z|^{2n+1}}{2} \to 0\text{ as }n\to\infty\text{ if } |z|<1 \text{ and }\to\infty\text{ if } |z|>1. $$ If not, how would I find the radius of convergence? – John Southall Nov 11 '11 at 3:44
@John: yes, you can do that. You're using the ratio test, which is related to geometric series, but works for lots of series that aren't geometric (like yours). – Dan Drake Nov 11 '11 at 3:55

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