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Consider a power series $\sum\limits_{n=0}^\infty c_n z^n$, for which the limit $\lim\limits_{n\to\infty}\left| \frac{c_n}{c_{n+1}} \right|=R$ holds. I want to prove that this limit implies that the radius of convergence of the series is also $R$. But I'm struggling with this, because I do not understand the relationship between the ratio of the coefficients of a sequence and the radius of convergence of the respective infinite series. Namely, we are dealing with a radius of convergence, which implies that $\exists$ an open ball $B(0, R)$ such that the series converges whenever $z\in B(0, R)$. But how does this relate to the coefficients? I also understand that the limit above looks like the ratio test, but I'm having difficulty extracting anything useful from this fact for my proof.

So, how do I formalize what exactly I need to show, in such a way that would allow me to understand how to build the proof around that? I also need to understand the relationship between $c_n$'s and the set $B(0, R)$. Would appreciate your help!

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    If the limit of the ratio test is finite, then that means at large n the $c_n$s behave like a geometric series. What's the condition on a geometric series converging? – 2017-02-23
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    Consider $\sum \left| \frac{c_n}{c_{n+1}} \right|=\sum x^n=\frac{1}{1-x}$ for some $x\in \mathbb{R}$. But then, shouldn't it be true that $\{x^n\}$ must converge to $0$ in order for the preceding expression to hold true? How do I arrive at $\frac{1}{1-R}$? Also, how do we ascertain that such an $x$ exists? – 2017-02-23

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Here's the key, if you were to apply the ratio test to $c_nz^n$, you would get the following: $$ \lim_{n\rightarrow\infty}\left|\frac{c_nz^n}{c_{n+1}z^{n+1}}\right|=\frac{1}{|z|}\lim_{n\rightarrow\infty}\left|\frac{c_n}{c_{n+1}}\right|=\frac{1}{|z|}R. $$ By the ratio test (in the way you set it up), we know that this converges absolutely if $$ \frac{1}{|z|}R>1 $$ In other words, if $$ R>|z|. $$

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    I've already arrived at this, but then how do I prove that $R\le|z|$ to arrive at the equality? – 2017-02-23
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    @sequence You don't. $R \le |z|$ is a closed ball, not an open one. – 2017-02-23
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    @eyeballfrog Oh, I see. Just the fact that $R>|z|$ is sufficient for the proof. Thank you very much. I was over-complicating this. – 2017-02-23