# 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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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

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

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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I am really stuck, if possible can you show me how to do the first inequality where liminf (a_(n+1) / a_n)<= liminf (a_n)^(1/n) –  UH1 Nov 27 '12 at 23:42
I can help with the last inequality. Give me a little time, I've got to grab some dinner, etc. –  amWhy Nov 27 '12 at 23:47
actually i have figured something out, the manipulation of definitions can be a little tricky sometimes. Thanks for the help! –  UH1 Nov 28 '12 at 0:24
You're welcome! Congrats! –  amWhy Nov 28 '12 at 0:26

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? –  Chou Apr 2 at 6:15