TITLE: Does $\lim_{n \to \infty}a_n^{1/n} = 1$ imply $\lim_{n \to \infty} \frac{a_{n + 1}}{a_n} = 1$?
QUESTION [13 upvotes]: Supose I have a sequence $\{a_n\}$ of positive real numbers such that $\lim\limits_{n \to \infty}a_n^{1/n} = 1$. Is it true that $\lim\limits_{n \to \infty} \frac{a_{n + 1}}{a_n} = 1$ or depends of the sequence that a choose?

REPLY [5 votes]: Even if one exists, the other may not. Consider: $a_n=\frac{3+(-1)^n}{2^{n}}$
$$\lim_\limits{n\to\infty} \sqrt[n]{a_n}=\frac{1}{2}.$$
$$\lim_\limits{n\to\infty} \frac{a_{n+1}}{a_n}=\begin{cases} 1, n =\ odd \\ \frac{1}{4}, n= \ even \end{cases}.$$