# Showing that $\underset{n \rightarrow \infty}{\lim} |a_n| / |a_{n+1}| = R$ implies that the radius of convergence of $\sum a_n z^n$ is also $R$

Hypothesis: Suppose that $\underset{n \rightarrow \infty}{\lim} |a_n| / |a_{n+1}| = R$.

Goal: Show that $\sum a_n z^n$ has radius of convergence $R$.

Attempt:

1. The radius of convergence of $\sum a_n z^n$ can be expressed as

$$\frac{1}{\limsup |a_n|^{1/n}}$$

2. From this we can derive

$$\frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_{n+1} \right|^{1/n}} = \left( \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_{n+1} \right|} \right)^{1/n} = \left( \frac{\underset{n \rightarrow \infty}{\limsup}|a_{n}|}{\underset{n \rightarrow \infty}{\limsup} \left|a_{n+1} \right|/|a_{n}|} \right)^{1/n}$$

so that then

$$\frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} = \left( R \frac{1}{\underset{n \rightarrow \infty}{\limsup} |a_{n}|} \right)^{1/n}$$

But at this point I'm not sure what to make of this subresult.

• You can't move the $n$-th root across the $\limsup$. The $n$ is bound in $\limsup\limits_{n\to\infty}$. – Daniel Fischer Mar 4 '14 at 20:51

Let $z\in\mathbb C$.
$$\frac{|a_{n+1} z^{n+1}|}{|a_nz^n|} \to \frac{|z|}{R}$$ so when $|z|<R$ there is convergence, and when $|z|>R$ there is divergence.
Hence, $R$ is the radius.