# Missing conclusion in proof convergence radius

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!

• What is the question, anyway?? – DonAntonio May 12 '17 at 8:44
• 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. – jacmeird May 12 '17 at 8:46
• @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? – DonAntonio May 12 '17 at 8:47
• Yeah, I know, but why I can assume that the limit is $L$? – jacmeird May 12 '17 at 8:49
• @j Because you say that limit exists!! You wrote "Let the limit exist..." ... and I still don't know was ist die Frage... – DonAntonio May 12 '17 at 8:50

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 $|z-z_0|>l$.