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

Solution 1:

Here a simple counterexample: let $a_n=1$ if $n$ is even, and $a_n=2$ if $n$ is odd. Clearly $\lim\limits_{n \mapsto \infty}a_n^{\frac{1}{n}} = 1,$ while $a_{n+1}/a_n$ oscillates between $2$ and $1/2,$ i.e., does not converge.

Solution 2:

No, that is false (see Reiner Martin's answer). On the other hand, the converse is true by Stolz-Cesaro Theorem: if $a_{n+1}/a_n\to L$ then $$\lim_{n\to \infty}\ln(a_n^{1/n})=\lim_{n\to \infty}\frac{\ln(a_n)}{n}=\lim_{n\to \infty}\frac{\ln(a_{n+1})-\ln(a_n)}{(n+1)-n}=\lim_{n\to \infty}\ln\left(\frac{a_{n+1}}{a_n}\right)=\ln(L)$$ that is $a_n^{1/n}\to L$.

Solution 3:

Let $r_n = \frac{a_{n + 1}}{a_n}$ and $s_n = a_n^{1/n}$. Then the rule is

$$ \liminf r_n \le \liminf s_n \le \limsup s_n \le \limsup r_n. $$

So if both limits exist then they are equal but $\lim s_n$ might exist where $\lim r_n$ might not.