# upper bound on the radius of convergence of a power series

Let $$(a_n)$$ be a sequence of complex numbers. Now suppose that $$\sum na_n$$ converge absolutely. Prove that the radius of convergence of $$\sum_{n = 0}^{\infty} a_nx^n$$ is $$\geq 1$$.

I don't understand where I am going wrong. For me it just comes from the fact that $$\forall z, \mid z \mid \leq 1, \sum_{n = 0}^\infty \mid a_n \mid \mid z^n \mid \leq \sum_{n = 0}^\infty \mid a_n \mid \leq \sum_{n = 0}^\infty n \mid a_n \mid$$

Hence, for all $$z, \mid z \mid \leq 1$$ the series $$\sum a_nz^n$$ converges absolutely and hence converges. So for all $$z, \mid z \mid \leq 1,$$ the series $$\sum_{n = 0}^\infty a_nz^n$$ converges, thus the radius of convergence of $$x \mapsto \sum_{n = 0}^\infty a_nx^n$$ is $$1$$.

Where is the problem in what I said ?

In my book they are using the fact that $$\lim_{n \to \infty} n\mid a_n \mid \to 0$$ to prove that the radius of convergence is $$1$$. So I might be wrong somewhere, since what I said is quite trivial.

Thank you!

• The title is wrong. It is a lower bound, nor an upper bound. Jan 31 '19 at 23:46
• Look up and apply the Cauchy-Hadamard Radius Formula, using the fact that $|a_n|^{1/n}=\frac {|na_n|^{1/n}}{n^{1/n}}<\frac {1}{n^{1/n}}$ for all but finitely many $n\in \Bbb Z^+.$ Feb 1 '19 at 3:25

There is nothing wrong with your approach. Actually, the problem is silly. You would be able to get the some conclusion simply assuming that the series $$\sum_{n=0}^\infty a_n$$ converges.