Does $\lim_{n \to \infty} |a_n|^{\frac{1}{n}}=l$ imply $\lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}=l$?

Question

Consider a series $\sum_{n \geq 0} a_n$. How can I show the following?

$$\lim_{n \to \infty} |a_n|^{\frac{1}{n}}=l \implies \lim_{n \to \infty} \frac{|a_{n+1}|}{|a_n|}=l$$

Answer 1

You can't, because it's not true. For example, take $$a_n=\cases{4^n&if $n$ is even\cr2\times4^n\quad&if $n$ is odd.\cr}$$ Then $$a_n^{1/n}=\cases{4&if $n$ is even\cr4\times2^{1/n}\quad&if $n$ is odd\cr}$$ which has limit $l=4$. But $$\frac{a_{n+1}}{a_n}=\cases{8&if $n$ is even\cr2\quad&if $n$ is odd\cr}$$ which has no limit.

Answer 2

Take $a_n=0$

$$\lim_{n\to +\infty} |a_n|^{\frac{1}{n}}=0$$

but

$\lim_{n\to +\infty}\frac{|a_{n+1}|}{|a_n|}$ is indeterminate.