Thread: radius of convergence of the power series

1. radius of convergence of the power series

Let $\displaystyle \displaystyle \sum_{n=0}^{\infty} a_n z^n$ be a power series with complex coefficients $\displaystyle a_n$ such that $\displaystyle \displaystyle \sum_{n=0}^{\infty} \mid a_n \mid$converges but $\displaystyle \displaystyle \sum_{n=o}^{\infty} n \mid a_n \mid$ 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 $\displaystyle R \geq 1$, 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.

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

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

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

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

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

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

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

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