# Given that $a_n > 0$, prove: $\liminf \left(\dfrac{a_{n+1}}{a_n}\right)\leq \liminf\; \sqrt[n]{a_n}\leq\limsup\left(\dfrac{a_{n+1}}{a_n}\right)$

Given that $a_n > 0$, I need to prove: $\liminf \left(\dfrac{a_{n+1}}{a_n}\right)\;\leq \;\liminf\; \sqrt[\Large n]{a_n}\;\leq\;\limsup\left(\dfrac{a_{n+1}}{a_n}\right)$.

I am really confused about how to use the definitions to prove $\liminf \left(\dfrac{a_{n+1}}{ a_n}\right)\;\leq\;\liminf \;\sqrt[\Large n](a_n)$.

As most of you might have guessed the motivation for this comes from the ratio test and the root test for convergence of series.

Any help is much appreciated.

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0% accept rate? Hmmm, that won't appeal to many people to help you, as it seems to indicate you don't like the answers you get here. – DonAntonio Nov 27 '12 at 18:53
Hi, I am just new to this place but I find the answers really helpful, how do I show that? – UH1 Nov 27 '12 at 19:01
@UH1 see this thread – Norbert Nov 27 '12 at 20:10

In fact: If $a_n>0$, then
$$\liminf \left(\frac{a_{n+1}}{a_n}\right)\quad\leq\quad\liminf \sqrt[n]{a_n}\quad\leq \quad \limsup \sqrt[n]{a_n} \quad \leq\quad \limsup\left(\frac{a_{n+1}}{a_n}\right)$$
You need to use the definitions of the "$\liminf$" and "$\limsup$" of a sequence of numbers. (If you are confused about the definitions, you should edit your post to clarify what you find confusing.)
And you can use the fact that if $p>0$, then $$\lim_{n\to \infty}\sqrt[\large n]{p} =1.$$