Relation between root test and radius of convergence

Question

We are able to use calculus to compute the radius of convergence of a power series. Now I want to look at a analytical way to that procedure. Say, we have a power series $\sum a_nx^n$. What if we should know that $\frac{a_{n+1}}{a_n} \to z$. What does that mean for the radius of convergence of $\sum a_nx^n$. Of course, the radius of convergence should be $\frac{1}{z}$. But how can I proof that when $\frac{a_{n+1}}{a_n} \to z$, the radius of convergence has to be $\frac{1}{z}$.

We could use the inequality $ \limsup a_n^{1/n} \leq \limsup \frac{a_{n+1}}{a_n}$ to find $\limsup a_n^{1/n}$ and compute the radius of convergence. But how can I show that $\frac{a_{n+1}}{a_n} \to z \Rightarrow \limsup a_n^{1/n} = z $?

Answer 1

This answer is for an earlier version of the question.

$\lim \sup a_n^{1/n}=z$ does not imply that $\frac {a_{n+1}} {a_n} \to 1$. For example, if $a_n=2n$ for $n$ even and $n$ for $n$ odd then $\lim \sup a_n^{1/n}=1$ but $\frac {a_{n+1}} {a_n}$ does not have a limit. Note that ratio test only gives a sufficient condition for convergence and this condition is not necessary.

Answer 2

We suppose that in $\sum a_nx^n$ all $a_n$ are $ \ne 0$ and that $\frac{a_{n+1}}{a_n} \to z$.

For $x \ne 0$ let $b_n(x):=a_nx^n$. Then we have

$\frac{|b_{n+1}(x)|}{|b_n(x)|}=|x| |\frac{a_{n+1}}{a_n}| \to |x||z|.$

Case 1: $z=0$. Then $\lim_{n \to \infty}\frac{|b_{n+1}(x)|}{|b_n(x)|}=0<1$ and the ratio-test says: $\sum a_nx^n$ converges absolutely for all $x$.

Case 2: $z \ne 0$. Then $\lim_{n \to \infty}\frac{|b_{n+1}(x)|}{|b_n(x)|}<1 \iff |x|<\frac{1}{|z|}$.

The ratio-test gives: $\sum a_nx^n$ converges absolutely for all $x$ with $|x|<\frac{1}{|z|}$ and $\sum a_nx^n$ is divergent for all $x$ with $|x|>\frac{1}{|z|}$ .

Conclusion: the radius of convergence is $\frac{1}{|z|}$ with $"\frac{1}{0}= \infty".$