Show the power series $\sum_{n=1}^\infty na_nx^{n-1}$ diverges when $|x|>R$.

Question

Suppose that the power series $$\sum_{n=1}^\infty a_nx^n$$ has radius of convergence $R$. Then the power series $$\sum_{n=1}^\infty na_nx^{n-1}$$ diverges if $|x|>R$.

Attempt: I can show that $\sum_{n=1}^\infty |na_nx^{n-1}|$ diverges by comparison test, but how do I show that $\sum_{n=1}^\infty na_nx^{n-1}$ diverges also?

Answer 1

Showing that it diverges for $|x|>R$ is equivalent to showing that the radius of convergence $R'$ of the second power series is $\leq R$.

But $\sum_{n=1}^{+\infty} |na_{n}x^{n-1}| = \frac{1}{x}\sum_{n=1}^{+\infty} |na_{n}x^{n}| \geq \frac{1}{x}\sum_{n=1}^{+\infty} |a_{n}x^{n}|$ shows just that.

P.S.: in fact the other direction ($R' \geq R$) is more difficult to prove

Answer 2

Given that $\sum_{n=1}^\infty a_nx^n$ has a radius of convergence $R$, then we have from the Root Test that

$$R=\frac{1}{\limsup_{n\to \infty}\sqrt[n]{|a_n|}}$$

Inasmuch as $\lim_{n\to \infty}\sqrt[n]{n}=1$, then the radius of convergence of the series $\sum_{n=1}^\infty na_nx^{n-1}$ is

$$\frac{1}{\limsup_{n\to \infty}\sqrt[n]{|na_n|}}=\frac{1}{\limsup_{n\to \infty}\sqrt[n]{|a_n|}}=R$$

as was to be shown!