Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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
1  
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

1 Answer 1

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.

share|improve this answer
    
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
1  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.