radius of convergence of the power series

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Apr 1st 2011, 05:27 PM
tsang

radius of convergence of the power series

Hi, can anyone please help me with this question?

Let $\displaystyle \sum_{n=0}^{\infty} a_n z^n$ be a power series with complex coefficients $a_n$ such that $\displaystyle \sum_{n=0}^{\infty} \mid a_n \mid$ converges but $\displaystyle \sum_{n=o}^{\infty} n \mid a_n \mid$ diverges. Prove that the radius of convergence of the power series is 1.

I had some hint that I need to try to prove it by contradiction, first prove $R \geq 1$ , then prove R can't be great than 1 by contradiction. But I'm not sure how to do it. Can anyone please help me? Thanks a lot.
Apr 1st 2011, 06:02 PM
hatsoff

Recall that the radius of convergence is the same for $|a_n z^n|$ as it is for $a_nz^n$ . Then that $R\geq 1$ follows immediately from the hypotheses. Now suppose towards a contradiction that $R>1$ , and choose $m\in(1,R)$ . Then $\sum |a_n m^n|$ converges.

Notice that $\lim_{n\to\infty}\frac{(\ln n)/\ln m}{n}=\lim_{n\to\infty}\frac{1}{n\ln m}=0$ , so we can choose $N$ such that $n>\ln n/\ln m$ and therefore $m^n>n$ for all $n\geq N$ .

It follows that $\sum_{n=N}^\infty |a_n m^n|\geq\sum_{n=N}^\infty n|a_n|=\infty$ , a contradiction.
Apr 3rd 2011, 02:22 AM
tsang

Quote:

Originally Posted by hatsoff

Recall that the radius of convergence is the same for $|a_n z^n|$ as it is for $a_nz^n$ . Then that $R\geq 1$ follows immediately from the hypotheses. Now suppose towards a contradiction that $R>1$ , and choose $m\in(1,R)$ . Then $\sum |a_n m^n|$ converges.

Notice that $\lim_{n\to\infty}\frac{(\ln n)/\ln m}{n}=\lim_{n\to\infty}\frac{1}{n\ln m}=0$ , so we can choose $N$ such that $n>\ln n/\ln m$ and therefore $m^n>n$ for all $n\geq N$

It follows that $\sum_{n=N}^\infty |a_n m^n|\geq\sum_{n=N}^\infty n|a_n|=\infty$ , a contradiction.

Hi, thanks a lot for your reply. However, I just wonder, how you get that $R \geq 1$ ? Because the question gives series of $|a_n|$ converges, how do I connect that with $|a_n z^n|$ ? Thanks.