0
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This is also what I thought. Thank you, for confirming that what I said is correct ! $\endgroup$ Jan 31 '19 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.