Proof of limit inequality

Question

Prove that for any sequence $\{x_n\}$ of positive real numbers $$\lim\text{sup}\sqrt[n]{x_n}\leq \lim\text{sup}\frac{x_{n+1}}{x_n}.$$

My attempt:

Let $A = \lim\text{sup}\frac{x_{n+1}}{x_n}$. Suppose $A<\infty$ and choose $\epsilon >0$, then $\exists$ an integer $N$ so that $N\le n \implies \frac{x_{n+1}}{x_n}\le \epsilon$. But I do not know how to proceed in finishing the proof?

Answer 1

Hint: if $x_{n+1}/x_n \le B$ for $n \ge N$, then $x_n \le x_N B^{n-N}$ for $n \ge N$. What does that tell you about $x_n^{1/n}$?