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.

Let $z$ denote a complex number and $\{\alpha_n\}$ be a sequence in $\ell^2$. Would you help me to prove that series $\sum_{n=0}^{\infty} \alpha_n z^n$ has radius of convergence greater than or equal to $1$. I have proved special cases when $\alpha_n \neq 0$ by the ratio test, but can't do the same for the general one.

Thanks.

share|improve this question
    
Your should probably rewrite your question, as it is now it makes no sense at all, as the radius of convergence depends on $\alpha_n$. For instance, if $\alpha_n=\alpha \in \ell^2$ for every $n$, then the radius of convergence if $1$, and if $\alpha_n=\frac{1}{n!}\alpha$ for some $\alpha \in \ell^2$, the radius of convergence is infinite. –  Mercy Sep 24 '12 at 13:07

1 Answer 1

up vote 2 down vote accepted

As $\{\alpha_n\}\in\ell^2$, it's a bounded sequence, hence for $\,|z|\leq 1\,$ the sequence $\{\alpha_n z^n\}$ is bounded.

share|improve this answer
    
I said greater than one –  beginner Sep 24 '12 at 12:54
    
So it's necessarily greater or equal than $1$. –  Davide Giraudo Sep 24 '12 at 12:55
    
It just said that the series has radius of convergence equal 1 only –  beginner Sep 24 '12 at 13:01
    
Why? The radius of convergence of the power series $\sum_n a_nz^n$ is $\sup\{R,\{a_nR^n\}\mbox{ is bounded}\}\in\Bbb R_{\geq 0}\cup\{+\infty\}$, so what I said gives that this supremum is $\geq 1$. –  Davide Giraudo Sep 24 '12 at 13:08

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.