# Inequality involving $\limsup$ and $\liminf$

This may have been asked before, however I was unable to find any duplicate. This comes from pg. 52 of "Mathematical Analysis: An Introduction" by Browder. Problem 14:

If $(a_n)$ is a sequence in $\mathbb R$ and $a_n > 0$ for every $n$. Then show: $$\liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$$

The middle inequality is clear. However I am having a hard time showing the ones on the left and right. (It seems like the approach should be similar for each). This is homework, so it'd be great if someone could give me a hint to get started on at least one of the inequalities.

Thanks.

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It may sound somewhat unhelpful, but the best technique is usually to ditch the tricks (at least at first) and go one step at a time with the definitions for every term. After a while that you have become comfortable with the definitions, it gets a lot easier and a lot clearer how "tricks" and other techniques work. –  Asaf Karagila Oct 2 '11 at 23:56
This is also given as Problem 2.4.26 in the book Kaczor, Nowak: Problems in Mathematical Analysis. The problem is stated on p.46 and a solution is given on p.205. –  Martin Sleziak Jul 23 '12 at 5:55

Suppose $$r := \lim \sup \frac{a_{n+1}}{a_n}.$$ (If the above expression is $\infty$, then there is nothing to prove. So assume $0 \leq r < \infty$.) Fix any $\epsilon > 0$. This means that there exists $N$ such that for $n \geq N$, we have $$\frac{a_{n+1}}{a_n} \leq r + \epsilon.$$ From this, can you deduce that for $n \geq N$, we have $$\frac{a_n}{a_N} \leq (r+\epsilon)^{n-N}?$$ Rearranging a bit, $$a_n \leq (r+\epsilon)^{n} \left( \frac{a_N}{(r+\epsilon)^N} \right),$$ so that $$a_n^{1/n} \leq (r+\epsilon) \left( \frac{a_N}{(r+\epsilon)^N} \right)^{1/n}.$$ Can you take it from here?