Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(a_n)$ be a non-zero real sequence with $\left (\frac{a_{n+1}}{a_n} \right)$ bounded. How might we prove that $$\liminf_{n\to\infty} \, \frac{|a_{n+1}|}{|a_n|} \leq \liminf_{n\to\infty} \, |a_n|^{\frac{1}{n}}$$ I've tried a number of approaches, including stuff with the AM-GM inequality, but haven't quite been able to get it out. Thanks for the help.

share|cite|improve this question

marked as duplicate by Martin Sleziak, Arctic Char, Claude Leibovici, Najib Idrissi, egreg Apr 17 '14 at 10:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Various proofs of this inequality can be found here:… and in the questions that are linked there among linked questions. – Martin Sleziak Apr 17 '14 at 8:23
up vote 1 down vote accepted

You may as well assume that $I = \liminf \frac{|a_{n+1}|}{|a_n|}$ is positive.

Let $0 < \epsilon < I$. There exists an index $N$ with the property that $k \ge N$ implies $|a_{k+1}| \ge (I-\epsilon) |a_k|$. By induction we deduce $|a_{N+k}| \ge (I-\epsilon)^k |a_N|$ for all $k \ge 0$. Writing $n = N + k$ we have $$ n \ge N \implies |a_n|^{1/n} \ge (I-\epsilon) \left( \frac{|a_N|}{(I-\epsilon)^N} \right)^{1/n}. $$ The fraction in parentheses is positive and independent of $n$. Let $n \to \infty$ to obtain $$ \liminf_{n \to \infty} |a_n|^{1/n} \ge I-\epsilon. $$ Now let $\epsilon \to 0^+$.

share|cite|improve this answer

Assuming $a_0 = 1$, $a_n > 0$ for all $n$.

Set $b_n = a_{n+1} / a_{n}$, then $a_n = \prod_{i=0}^{n-1}{b_i}$. So this is equivalent to the original expression:

$$\liminf_{n\to\infty} \, b_n \leq \liminf_{n\to\infty} \, (\prod_{i=0}^n b_i)^{\frac{1}{n}}$$

Take the log of both sides:

$$\liminf_{n\to\infty} \, c_n \leq \liminf_{n\to\infty} \, {\frac{1}{n}}(\sum_{i=0}^n c_i)$$ which is implied by "the average of a set of elements is greater than or equal to the smallest element in the set".

share|cite|improve this answer
Actually, I'm not sure this argument works. Before we pass to the limit, we're looking at $\inf_{k \ge n} \ln b_k$, which we want to be less or equal to $\inf_{k \ge n} \frac{1}{k} \sum_{i=0}^{k-1} \ln b_i$. But you have no guarantee that $any$ term in the sum from $i = 0$ to $k-1$ is greater than or equal to than your left-hand-side. – Joshua Ciappara Mar 25 '13 at 23:52
Well, you have some idea when $k > n$, but in the case $k = n$ you don't know about any of the terms from $i = 0$ to $k-1$ in comparison to $\inf_{k \ge n} \ln b_k$. – Joshua Ciappara Mar 26 '13 at 0:01
Good point. If we start with ${0,0,0,0,5,...}$, the average at $5$ is $1$, which is strictly less than the $5$. Can I fix it? Well, if instead of averaging from $0$ we average from $k$, then $inf \{c_k, c_{k+1},...,c_{d}\} \leq inf \{x_{k,k}, x_{k,{k+1}}, ..., x_{k,d} \}$ where $x_{a,b} = \frac{c_a + ... + c_b}{b-a+1}$. $x_{0,n} = \frac{x_{a,n}(n-a) + c_0 + ... + c_{a-1}}{n}$, which for small $a$ relative to $n$ differs only slightly (and differs less and less as $n$ gets large). I think that leads towards a fix, but it is messy. – Yakk Mar 26 '13 at 0:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.