When proving $\liminf(a_n)>L$, proving $\liminf(a_n)>a$ with $a<L$ is enough?

Question

Let $\epsilon>0$, can I prove $\liminf(a_n)>L$ by proving $\liminf(a_n)>1-\epsilon$?

I'm proving that $$\liminf \sqrt[n]{a_n}\geq \liminf \frac{a_{n+1}}{a_n}$$

To do so I took $\liminf \frac{a_{n+1}}{a_n}=L$ and than with $a=L-\epsilon \quad$ I proved that $\liminf \sqrt[n]{a_n}\geq a$. Is it enough to say that $\liminf \sqrt[n]{a_n}\geq \liminf \frac{a_{n+1}}{a_n}$?

Answer 1

You proved that, for every $\epsilon>0$ $$\liminf\sqrt[n]{a_n}\geq L-\epsilon$$ We can then take the supremum on the right-hand side: $$\liminf\sqrt[n]{a_n}\geq\sup_{\epsilon>0}L-\epsilon=L$$