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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?

LAU
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4 Answers4

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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.

Reiner Martin
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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 $$\lim_{n\to \infty}\ln(a_n^{1/n})=\lim_{n\to \infty}\frac{\ln(a_n)}{n}\stackrel{\text{SC}}{=}\lim_{n\to \infty}\frac{\ln(a_{n+1})-\ln(a_n)}{(n+1)-n}=\lim_{n\to \infty}\ln\left(\frac{a_{n+1}}{a_n}\right)=\ln(L)$$ that is $a_n^{1/n}\to L$.

Robert Z
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  • Nice answer. Seems to hold for any $L \in [0, \infty]$. – Caleb Stanford Jun 30 '17 at 18:40
  • I am interesing to know how to say $\lim_{n\to \infty}\frac{\ln(a_n)}{n}=\lim_{n\to \infty}\frac{\ln(a_{n+1})-\ln(a_n)}{(n+1)-n}$. – Messi Lio Oct 28 '22 at 18:28
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    That is Stolz-Cesaro Theorem:https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem – Robert Z Oct 28 '22 at 18:29
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    @MessiLio: Recall that $\liminf_n\frac{|a_{n+1}|}{|a_n|}\leq\liminf_n\sqrt[n]{|a_n|}\leq\limsup_n\sqrt[n]{|a_n|}\leq\limsup_n\frac{|a_{n+1}}{|a_n|}$. – Mittens Oct 28 '22 at 19:10
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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.

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

farruhota
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