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I'm trying to teach myself complex analysis, and I've been working on this idea.

Suppose $\lim_{n\to\infty}|a_n|/|a_{n+1}|=R$, I would like to know why $\sum a_nz^n$ also has $R$ as its radius of convergence.

I believe I want to show $$ \limsup \sqrt[n]{|a_n|}=\frac{1}{\lim_{n\to\infty}|a_n|/|a_{n+1}|}=\lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|}. $$

I notice that $|a_n|$ can be written as $|a_n|=\frac{|a_n|}{|a_{n-1}|}\frac{|a_{n-1}|}{|a_{n-2}|}\cdots\frac{|a_{k+1}|}{|a_{k}|}|a_k|$. I hoped this would be helpful since it is a product of ratios of the absolute values of successive terms.

I didn't know how to proceed after these observations. How can I finish this? Thanks!

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2 Answers 2

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It might be easier to show it more directly. Assume $|z|<R$. Choose $\epsilon>0$ so that $|z|<R-\epsilon$. Then find $N$ so that $|\frac{|a_n|}{|a_{n+1}|}-R|<\epsilon$ for $n\geq N$. In particular, $|\frac{a_n}{a_{n+1}}|>R-\epsilon$.

Then for $n\geq N$, $$\left|\frac{a_N}{a_n}\right| = \left|\frac{a_N}{a_{N+1}}\frac{a_{N+1}}{a_{N+2}} ...\frac{a_{n-1}}{a_n}\right| > (R-\epsilon)^{n-N}$$

So $|a_n|< \frac{|a_N|}{(R-\epsilon)^{n-N}}$, and therefore:

$$|a_nz^n| < |a_Nz^N|\cdot \left |\frac{z}{R-\epsilon}\right |^{n-N}$$

for $n\geq N$. But $|\frac{z}{R-\epsilon}|< 1$, so $\sum |a_nz^n|$ converges because it is bounded above by a geometric series with quotient less than $1$ for $n\geq N$. Therefore $\sum a_nz^n$ is absolutely convergent.

Similarly, you can show that if $|z|>R+\epsilon$, then there exists an $M$ such that for $n\geq M$, $|a_nz^n|>|a_Mz^M|\cdot |\frac{z}{R+\epsilon}|^{n-M}$. Therefore $|a_nz^n|$ does not converge to zero, and hence $\sum a_nz^n$ cannot converge.

(You'll probably need to put some condition to make the limit make sense, such as $a_n$ is non-zero, or non-zero for all but a finite number of values. If you allow finitely many zero values, then choose $M,N$ so that no zero values occur after them.)

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Thanks Thomas Andrews, I was able to follow this. Please excuse me if I am wrong, but are the exponents supposed to be $n-N$ and $n-M$ instead? –  Dedede Jan 31 '12 at 0:28
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Yes, you are correct. Will fix when I get back to a computer - too painful to edit LaTex on my iPad. –  Thomas Andrews Jan 31 '12 at 0:55

Instead of trying to estimate the individual $a_i$s, what I would do is focus on the ratio between successive terms in the power series:

Suppose $\lim \frac{|a_n|}{|a_{n+1}|} = R$. Then for $|z|>R$ it is pretty clear that $\sum a_n z^n$ cannot converge, because from a certain point, its successive terms will increase in magnitude by a ratio as close to $\frac{|z|}{R}>1$ as you like.

On the other hand, consider $|z|<R$. Choose $K$ strictly between $\frac{|z|}{R}$ and $1$. Then from a certain point in $\sum a_nz^n$, the ratio between successive terms will be at most $K$. But then we can show $\sum a_n z^n$ converges by majorizing with the geometric series $\sum c K^n$ for some appropriate $c$ (which I assume we already know converges for $|K|<1$).

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Thank you, this makes the result seem more intuitive. –  Dedede Jan 31 '12 at 0:29

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