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?
Best Answer
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 $\lvert z\rvert<R$ and diverges if $\lvert z\rvert>R$, which implies that the radius of convergence is $R$ indeed.