Given the radius of convergence of $\sum_{n=1}^{\infty} c_{n} z^{n}$ is $R$ ($0<R<\infty$). Find the radius of convergence of $\sum_{n=1}^{\infty} b_n z^n =\sum_{n=1}^{\infty} n^{k} c_{n} z^{n}$.
Solution
$$R_{1}=\lim _{n \rightarrow \infty}\left|\frac{b_{n}}{b_{n+1}}\right|=\lim _{n \rightarrow \infty}\left|\frac{n^{k} c_{n}}{(n+1)^{k} c_{n+1}}\right|=\lim _{n \rightarrow \infty}\left|\frac{n}{n+1}\right|^{k} \cdot \lim _{n \rightarrow \infty}\left|\frac{c_{n}}{c_{n+1}}\right|=R.$$
Thus, the radius of convergence of $\sum_{n=1}^{\infty} b_n z^n$ is also $R$.
So this is the solution of a lecturer and I'm just curious that how can we apply the ratio test in this case? We do not know if the $\lim _{n \rightarrow \infty}\left|\frac{c_{n}}{c_{n+1}}\right|$ exists or not. I think we have to calculate it by using root test, not the ratio test. Am I right? Thank you.
Best Answer
You are right. It may happen that some $c_n$'s are $0$ and it also may happen, even if every $c_n$ is not $0$, that the limit $\lim_{n\to\infty}\left|\frac{c_n}{c_{n+1}}\right|$ doesn't exist (but if it exists, then, yes, it must be $R$).
The statement is correct though.