# proving radius of convergence for power series

Let $$\sum a_n z^n$$ be a power series. Proof that if $$lim _ {n \rightarrow \infty} |\dfrac{a_{n+1}}{a_n}|$$ exists, then it is equal to $$\dfrac{1}{R}$$, where $$R$$ is the radius of convergence.

So, using the ratio test, I get that if $$(1) \qquad |z| \; lim _ {n \rightarrow \infty} |\dfrac{a_{n+1}}{a_n}| \; < 1$$ (assuming it exists), then the series is absolutely convergent.

Now, I know the series is centered at $$0$$, and by definition if $$z$$ is contained an open ball of radius $$R$$, then the series is convergent. So, if we define $$|\dfrac{a_{n+1}}{a_n}|$$ to be $$R$$, then (1) is true for $$|z| < R$$. Which says exactly that $$R$$ is the radius of convergence.

My question: This seems to be a handwavy way of getting to the conclusion. Is there a better way of stating this formally?

Since your series is $$\displaystyle\sum_{n=0}^\infty a_nz^n$$, the natural approach is to study the limit$$\lim_{n\to\infty}\left\lvert\frac{a_{n+1}z^{n+1}}{a_nz^n}\right\rvert,$$which is equal to$$\lvert z\rvert.\lim_{n\to\infty}\left\lvert\frac{a_{n+1}}{a_n}\right\rvert=\frac{\lvert z\rvert}R.$$So, yes, the series converges absolutely if $$\lvert z\rvert and diverges if $$\lvert z\rvert>R$$, which implies that the radius of convergence is $$R$$ indeed.