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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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    See [Inequality involving $\limsup$ and $\liminf$](http://math.stackexchange.com/questions/69386/inequality-involving-limsup-and-liminf) and other posts linked there.2015-06-12

2 Answers 2

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In fact: If $a_n>0$, then

$\liminf \left(\frac{a_{n+1}}{a_n}\right)\quad\leq\quad\liminf \sqrt[n]{a_n}\quad\leq \quad \limsup \sqrt[n]{a_n} \quad \leq\quad \limsup\left(\frac{a_{n+1}}{a_n}\right)$

Hints:

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 $\lim_{n\to \infty}\sqrt[\large n]{p} =1.$

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    You're welcome! Congrats!2012-11-28
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The proof can go as follows. Set $\ell =\liminf_{n\to\infty}\frac{a_{n+1}}{a_n}$

Choose $\alpha <\ell$. By the definition of $\liminf$, there must exist an $N$ such that, for each $n\geq N$, we have that $\alpha <\frac{a_{n+1}}{a_n}$ That is, for $k\geq 0$, we have $ a_{N+k}>\alpha\cdot a_{N+k-1}$

This gives that $a_{N+k}>\alpha^k \cdot a_{N}$

In paricular $n-N\geq 0$ when $n=N+1,\dots$; so

$a_{n}>\alpha^{n-N} \cdot a_{N}$

Now, taking the $n$-th root gives that for $n\geq N+1 $ $\root n \of {{a_n}} > \alpha \cdot{\left( {\frac{{{a_N}}}{{{\alpha ^N}}}} \right)^{1/n}}$

and taking $\liminf\limits_{n\to\infty}$ gives

$\liminf\limits_{n\to\infty} \root n \of {{a_n}} \geq \alpha $

This means that for each $\alpha <\ell$,

$\liminf\limits_{n\to\infty} \root n \of {{a_n}} \geq \alpha $

which is saying that $\ell \leq \liminf\limits_{n\to\infty} \root n \of {{a_n}}$

The proof for $\limsup$ is completely analogous.

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    The restriction $\alpha \neq 0$ seems to be omitted?2015-04-02