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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.


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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 King Sep 24 '12 at 13:07

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

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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

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