# radius of convergence of the power series

• April 1st 2011, 05:27 PM
tsang
radius of convergence of the power series

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.
• April 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.
• April 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.