Let $\{a_n\}$ be a sequence of positive numbers. Prove that if $$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n} = L$$ then $$\lim_{n\rightarrow\infty}a_n^{1/n}=L.$$
The first condition means that for any $\epsilon$, then exists $N$ such that for all $n\geq N$, we have $$L-\epsilon < \dfrac{a_{n+1}}{a_n} < L+\epsilon.$$ This means $$(L-\epsilon)^na_N<a_{N+n}<(L+\epsilon)^na_N$$ for all $n\geq 0$. Then $$(L-\epsilon)^{\frac{n}{N+n}}a_N<a_{N+n}^{\frac{1}{N+n}}<(L+\epsilon)^{\frac{n}{N+n}}a_N$$
How can I finish from here?