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.

How would I find the region of convergence of the series of $\frac{1}{n^3}(\frac{z+1}{z-1})^n$. I thought about rewriting $\frac{z+1}{z-1}$ as $\frac{2}{z-1}+1$ but I don't think that helps.

Thanks

share|improve this question
2  
Consider the linear fractional transformation $\frac{z+1}{z-1}$ as mapping some regions onto the open disks centered at the origin, and work backwards from $\sum \frac{x^n}{n^3}$ to find the region of convergence for variable $z$. –  hardmath Jul 21 at 11:26
    
That's not a power series, so it doesn't have a radius of convergence. –  Thomas Andrews Jul 21 at 11:31
    
@Thomas: I read "region" of convergence in the problem statement. –  Yves Daoust Jul 21 at 11:40
    
Yeah, I either misread it or it was quickly edited. Still, it is not a power series :) @YvesDaoust –  Thomas Andrews Jul 21 at 11:40

1 Answer 1

up vote 4 down vote accepted

Let $w = \frac{z+1}{z-1}$. Then you have a power series in $w$, centered at $0$. Find its radius of convergence, call that $R$. Then find which $z$ correspond to $\lvert w\rvert < R$. The map $z \mapsto \frac{z+1}{z-1}$ can be explicitly inverted.

share|improve this answer
    
There's a pretty obvious way to express $|z-(-1)|<|z-1|$, without inverting that transformation... –  Thomas Andrews Jul 21 at 11:37
    
@Daniel Fischer I get that the series converges when $|\frac{z+1}{z-1}|<1$. How do I now check on the boundary of that ? Thanks –  user137090 Jul 21 at 11:38
    
@user137090 $\sum \frac{1}{n^3}$ converges, so it converges everywhere on the boundary. –  Thomas Andrews Jul 21 at 11:39
    
@Daniel Fischer I'm confused. If I make the transformation, can I treat it as a power series? –  user137090 Jul 21 at 11:49
    
@ThomasAndrews Yes, but what if $R\neq 1$? (whistles) –  Daniel Fischer Jul 21 at 11:50

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.