2
$\begingroup$

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.

  • 0
    I don't know whether it is part of your arsenal, but for this problem the *Root Test* is marginally easier to apply.2011-11-11

1 Answers 1

5

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.

  • 1
    @Joh$n$: 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).2011-11-11