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