Given that $a_n > 0$, prove: $\liminf \left(\frac{a_{n+1}}{a_n}\right)\leq \liminf\; \sqrt[n]{a_n}\leq\limsup\left(\frac{a_{n+1}}{a_n}\right)$
Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$