Given that $a_n > 0$, prove: $\liminf \left(\frac{a_{n+1}}{a_n}\right)\leq \liminf\; \sqrt[n]{a_n}\leq\limsup\left(\frac{a_{n+1}}{a_n}\right)$ 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.
 A: 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 $$\lim_{n\to \infty}\sqrt[\large n]{p} =1.$$
A: 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.
