I am working on an analysis problem and I am looking for some insight.
Problem:
Let R be the radius of convergence of a power series $\sum c_nx^n$ with nonzero coefficients $c_n$. Prove that as n tends to infinity that
$$\liminf|\frac{c_{n}}{c_{n+1}}|\leq R\leq \limsup|\frac{c_{n}}{c_{n+1}}|$$
What I know so far:
$R=\frac{1}{\limsup [(|c_{n}|)^{\frac{1}{n}}]}$
$\limsup(a_{n}b_{n})=\lim(a_{n})\limsup(b_n)$
and by the ratio test
$R=\lim |\frac{c_{n}}{c_{n+1}}|$ provided that it exists.
Would it be accurate to state that $$\liminf|\frac{c_{n}}{c_{n+1}}|\leq \lim |\frac{c_{n}}{c_{n+1}}|\leq \limsup|\frac{c_{n}}{c_{n+1}}|?$$ I am not really sure what would be sufficient to prove this problem.
Best Answer
You cannot assume that any of limit $\frac {|c_n|} {|c_{n+1}|}$ exists. I will give a proof of the right hand inequality and leave the other one to you. Suppose $|x| > \lim \sup \frac {|c_n|} {|c_{n+1}|}$. Choose $t$ such that $|x| >t> \lim \sup \frac {|c_n|} {|c_{n+1}|}$. Then $t >\frac {|c_n|} {|c_{n+1}|}$. for $n$ sufficiently large, say for $n \geq n_0$. Thus $|c_{n+1}| >|c_n|/t$ for $n \geq n_0$. Iterating this we get $|c_{n+1}| >|c_{n_0}|/t^{n+1-n_0}$. Hence $\sum\limits_{k=n_0}^{\infty} |c_n||x|^{n} >\sum\limits_{k=n_0}^{\infty} |c_{n_0}| |\frac x t|^{n}|t|^{n_0}=\infty$. Hence the series $\sum c_nx^{n}$ is divergent whenever $|x| > \lim \sup \frac {|c_n|} {|c_{n+1}|}$. This implies that $R \leq \lim \sup \frac {|c_n|} {|c_{n+1}|}$.