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? 
 A: 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.
A: I'm not convinced by your proof; even if it's true, it's certainly not 'by definition'.
This problem a bit easier to think about if you take logarithms. The statement is equivalent to
$$\lim_{n \to \infty} a'_n = 0 \quad \Rightarrow \quad \lim_{n \to \infty} \frac{1}{n} a'_n = 0$$
where $a'_n=\log a_n$.
Hopefully you can convince yourself that this is true, prove the above statement, and then take exponentials and deduce (from continuity of $\exp$) the result you seek.
A: You need to be careful about the implications and how you phrase things. You use a result, this is not "by definition" and the result you quote also proves just one implication, not an equivalence. You could state things like this:

We know that if $\displaystyle\lim_{n \to \infty}\frac{b_{n+1}}{b_n}=L$ and $b_n > 0$ for all sufficiently large $n$, then
  $\displaystyle\lim_{n \to \infty}\sqrt[n] b_n=L$. 
Since  $\displaystyle\lim_{n \to \infty}a_n=1$, we have $\displaystyle\lim_{n \to \infty}\frac{a_{n+1}}{a_n}=1$ and  $a_n >0$ for all sufficiently  large $n$,  and therefore $\displaystyle\lim_{n \to \infty}\sqrt[n] a_n=1$. 

Commentary: This assumes you are free to use the result you quoted so that the "We know" is justified. Otherwise use an approach from other answers, or prove the result you use too.  
A: Observe that: $|\sqrt[n]{a_n} - 1| = \dfrac{|a_n-1|}{|(\sqrt[n]{a_n})^{n-1} + (\sqrt[n]{a_n})^{n-2}+\cdots +\sqrt[n]{a_n}+1|} < |a_n-1|$. This implies the answer.
