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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.