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If the limit of $\lim\limits_{n\to\infty} \left\vert\frac{c_n}{c_{n+1}}\right\vert$ with $c_n\neq 0$ exisits, then $R:=\lim\limits_{n\to\infty} \left\vert\frac{c_n}{c_{n+1}}\right\vert$ is the convergence radius of power series $\sum\limits_{n=0}^{\infty} c_n(z-z_0)^n$

I know that these proof is really common but I only found proofs by using some formulars and consequences of the ratio test. So, these things I've noted:


Let $\lim\limits_{n\to\infty} \left\vert\frac{c_n}{c_{n+1}}\right\vert$ exist and let $f(z)=\sum\limits_{n=0}^{\infty} c_n(z-z_0)^n$ a given power series. Then one can say that $$\lim\limits_{n\to\infty} \left\vert\frac{c_n(z-z_0)^n}{c_{n+1}(z-z_0)^{n+1}}\right\vert=\frac{1}{\vert z-z_0\vert}\lim\limits_{n\to\infty}\left\vert\frac{c_n}{c_{n+1}}\right\vert=(*)$$


The first fraction is a fact of the ratio test, isn't it? Unfortunately I don't know another starting because we didn't introduce the ratio test in the lecture yet.

Furthermore I'm not sure how to go on after the (*). I know that the series converges if $r<\vert z-z_0\vert$ and that $R=\sup\{r\geq 0:\exists z_0\in\mathbb{C}$ with $r=\vert z-z_0\vert\}$.


Any hints? Thank you so much!

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  • $\begingroup$ What is the question, anyway?? $\endgroup$ – DonAntonio May 12 '17 at 8:44
  • $\begingroup$ Oh sorry, it's not clearly enough. I don't know how to go on after (*). And where the first fraction comes from. I saw this in many proofs. $\endgroup$ – jacmeird May 12 '17 at 8:46
  • $\begingroup$ @ja "To go"... where ? If the limit of $\;\left|\frac{c_n}{c_{n+1}}\right|\;$ is $\;R\;$ , then after (*) you have $\;\frac{R}{|z-z_0|}\;$ ... what else? $\endgroup$ – DonAntonio May 12 '17 at 8:47
  • $\begingroup$ Yeah, I know, but why I can assume that the limit is $L$? $\endgroup$ – jacmeird May 12 '17 at 8:49
  • $\begingroup$ @j Because you say that limit exists!! You wrote "Let the limit exist..." ... and I still don't know was ist die Frage... $\endgroup$ – DonAntonio May 12 '17 at 8:50
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The ratio test for the series $\sum_{n=0}^{\infty}a_n$ tells you that

if the limit $\lim\limits_{n\to\infty}\dfrac{|a_{n+1}|}{|a_n|}$ exists and is less than $1$, then the series converges.

Let's apply this to the power series $$ \sum_{n=0}^{\infty} c_n(z-z_0)^n $$ You want to compute the limit $$ \lim_{n\to\infty}\frac{|c_{n+1}(z-z_0)^{n+1}|}{|c_n(z-z_0)^n|}= \lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}|z-z_0| $$ Since we're assuming the limit $$ \lim_{n\to\infty}\frac{|c_{n}|}{|c_{n+1}|}=l $$ exists, we have three cases.

Case 1: $l=\infty$

In this case we can conclude that $$ \lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}|z-z_0|=0<1 $$ so the series converges for every $z$.

Case 2: $l$ is finite and $l>0$

In this case we can conclude that $$ \lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}|z-z_0|=\frac{|z-z_0|}{l} $$ This is less than $1$ if and only if $|z-z_0|<l$.

Case 2: $l=0$

In this case we can conclude that $$ \lim_{n\to\infty}\frac{|c_{n+1}|}{|c_n|}|z-z_0|= \begin{cases} 0 & \text{if $z=z_0$} \\[4px] \infty & \text{if $z\ne z_0$} \end{cases} $$ Thus the series converges only if $z=z_0$.

Another known consequence of the ratio test is that

if the limit $\lim\limits_{n\to\infty}\dfrac{|a_{n+1}|}{|a_n|}$ exists and is greater than $1$, then the series does not converge.

This is only relevant for case 2, where we can say that the power series

  • converges if $|z-z_0|<l$;
  • does not converge if $|z-z_0|>l$.
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