calculus - How to prove $lim_{n to infty}a_n=1 rightarrow lim_{n to infty}sqrt[n] a_n=1$

Let $a_n\geq0$

Prove/disprove: $$\lim_{n \to \infty}a_n=1 \rightarrow \lim_{n \to \infty}\sqrt[n] a_n=1$$

Proof: By definition a sequence $\displaystyle\lim_{n \to \infty}\sqrt[n] b_n=L$ iff $\displaystyle\lim_{n \to \infty}\frac{b_{n+1}}{b_n}=L$ since $\displaystyle\lim_{n \to \infty}a_n=1$ $\displaystyle\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=1$ and therefore $\displaystyle\lim_{n \to \infty}\sqrt[n] a_n=1$

Am I right?

Answer

I don't see where does the fact you use come from (certainly not from the definition). From the link you provided it seems at least the "if" part is true, so I guess you can prove it like that.

But there's also a quick and easy way to see it:
$$\begin{align*}1\le\sqrt[n]x\le x&\text{ if }x\ge1\\1\ge\sqrt[n]x\ge x&\text{ if }x\le1\end{align*}$$
Therefore $\sqrt[n]{a_n}\to1$ because all its members are closer to $1$ than in the original sequence.