Radius convergence of a power series...

Question

"Suppose that $\sum_{n=0}^{+\infty}a_nx^n$ has convergence radius $R$, $R>0, \text{or }R=+\infty$. Proof that the convergence radius of $\sum_{n=0}^{+\infty}na_nx^{n-1}$ is also $R$."

This seems to be a very fundamental result, but I don't know how to solve it... Has anybody a hint of how to do it? Or maybe a link for some site that does it... As I said, it looks to be a very fundamental result...

Answer 1

Hint: $\lim_{n \to \infty}\sqrt[n]{n}=1 $. The Cauchy-Hadamard theorem will do the rest.

Answer 2

I done it. It's simpler than I thought. Using the convergence radius formula for the series $\sum_{n=0}^{+\infty}b_nx^{n-1}$, $$R=\lim_{n\to+\infty}\left|\frac{b_n}{b_{n+1}}\right|$$ for $b_n=na_n$, the result is straightfoward.