Geometric series? Common ratio of $\sum_{n=0}^\infty 2^{-n}z^{n^2}$ 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 $\frac{1}{2}z^{something?}$
Thanks.

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.
