# Relation between root test and radius of convergence

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 $$\frac{a_{n+1}}{a_n} \to z \Rightarrow \limsup a_n^{1/n} = z$$?

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.

• You are right, I will change the last sentence. Excuse me – Paul Jan 29 at 11:53

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 $$"\frac{1}{0}= \infty".$$