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Inequality involving $\limsup$ and $\liminf$
limit of $\frac{a_{n+1}}{a_n}$

Show that $\limsup|s_n|^{1\over n}\le \limsup|{s_{n+1}\over s_n}|$ and similarly $\liminf|s_n|^{1\over n}\ge \liminf|{s_{n+1}\over s_n}|$. I have no idea where to start. I tried to show the inequality through subsequence but still don't quite get where to start. Any explaination how to link to the concept of related topic would be appreciated.

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marked as duplicate by Martin Sleziak, Gerry Myerson, BenjaLim, J. M., t.b. Jul 25 '12 at 4:03

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.

This inequality appeared in this answer and this answer. See also here. – Martin Sleziak Jul 23 '12 at 5:34
@MartinSleziak: How did you find those answers? Good memory or good search technique (or both, of course)? I have had limited success (read basically none) with searches for common terms such as $\limsup$. – copper.hat Jul 23 '12 at 6:05
@copper.hat When I first saw this inequality at MSE, it was new to me and it seemed interesting, so I noted down the link. Later I added a few other links. Perhaps the next time searching will be easier, since I've added tags limsup+inequality to that question. (There are probably a few other questions which would deserve these tags, too.) – Martin Sleziak Jul 23 '12 at 6:32
Anyway, if I wanted to search for similar questions, this Google search seems promising: limsup frac Some tips and tricks for searching MSE can be found at meta in this question and other question tagged search. – Martin Sleziak Jul 23 '12 at 6:37
@MartinSleziak: Much appreciated! – copper.hat Jul 23 '12 at 6:44

1 Answer 1

Here’s a start for you.

For convenience we may suppose that the $s_n$ are non-negative and omit the absolute value signs. Suppose that $$\limsup_ns_n^{1\over n}>\limsup_n\,{s_{n+1}\over s_n}\;.$$ Let $$b=\frac12\left(\limsup_ns_n^{1\over n}+\limsup_n\,{s_{n+1}\over s_n}\right)\;;$$ then there is an $n_0\in\Bbb N$ such that $$\frac{s_{n+1}}{s_n}<b$$ for all $n\ge n_0$. Thus, for all $n>n_0$ we have $s_n<s_{n_0}b^{n-n_0}$. Let $a=s_{n_0}b^{-n_0}$, so that $s_n<ab^n$ for $n>n_0$. Then $s_n^{1/n}<a^{1/n}b$ for $n>n_0$.

  1. What is $\lim\limits_{n\to\infty}a^{1/n}$?

  2. How does $b$ compare with $\limsup\limits_n\,s_n^{1/n}$?

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