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!

  • $\begingroup$ The title is wrong. It is a lower bound, nor an upper bound. $\endgroup$ Jan 31 '19 at 23:46
  • $\begingroup$ 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^+.$ $\endgroup$ 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.

  • $\begingroup$ This is also what I thought. Thank you, for confirming that what I said is correct ! $\endgroup$ Jan 31 '19 at 21:42

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