Prove limit of $n^{th}$ root test Given that $a_n>0, \forall{n}, \ \lim_{n \to \infty}\ a_n =a >0$, prove that$$ \ \lim_{n \to \infty}\ \sqrt[n]{a_n} =1 $$
This what I did: Let $a_n=\frac{1}{n}$ then $$\lim_{n \to \infty}\ \sqrt[n]{a_n} =  \lim_{n \to \infty} \left(\frac{1}{n}\right)^\frac{1}{n} =\lim_{n \to \infty} \frac{1}{n^\frac{1}{n}} =1 $$
But I don't know whether this right. Anyone can suggest an alternative way please.
 A: To prove it by that way is not enough. $a_n=\frac{1}{n}$ is but a special case, even if you show it works on $\frac{1}{n}$, the other possible $a_n$ is still undefined.
Since $\lim \limits_{n \to \infty} a_n=a>0$,
We take $\epsilon_0=\frac{1}{2} a>0$, then there is $N_0>0$, when $n>N_0$,
$$\frac{1}{2} a<a_n<\frac{3}{2}a.$$
Thus,
$$\sqrt[n]{\frac{1}{2} a}<\sqrt[n]{a_n}<\sqrt[n]{\frac{3}{2}a}.$$
(You can see the choice of $\epsilon_0$ is quite arbitrary.)
By squeeze theorem you can get the result.
P. S. Proposition: For any $c>0$, $\lim \limits_{n \to \infty} \sqrt[n] c=1$.
Proof: For $\epsilon>0$, assume that $c>1$.
$$\begin{align}
|\sqrt[n] c-1|<\epsilon &\Leftrightarrow \sqrt[n] c < 1+\epsilon \\
&\Leftrightarrow \frac{1}{n} \ln c < \ln (1+\epsilon) \\
&\Leftrightarrow n>\frac{\ln c}{\ln (1+\epsilon)}
\end{align}$$
Hence, take $N=\lceil \frac{\ln c}{\ln (1+\epsilon)} \rceil$, when $n>N$, $|\sqrt[n] c-1|<\epsilon$.
For $0<c<1$, the trick is similar. 
If $\epsilon >1$, $|\sqrt[n] c-1|<\epsilon$ always holds.
If $0<\epsilon<1$, notice that $\ln c<0$, $\ln (1-\epsilon)<0$.
$$\begin{align}
|\sqrt[n] c-1|<\epsilon &\Leftrightarrow 1-\epsilon < \sqrt[n] c \\
&\Leftrightarrow  \ln (1-\epsilon) < \frac{1}{n} \ln c\\
&\Leftrightarrow n>\frac{\ln c}{\ln (1-\epsilon)}
\end{align}$$
Hence, take $N=\lceil \frac{\ln c}{\ln (1-\epsilon)} \rceil$, when $n>N$, $|\sqrt[n] c-1|<\epsilon$.
From above we know 
$$\lim \limits_{n \to \infty} \sqrt[n] c=1.$$
A: Suppose $a>1$. Let $b_n=\sqrt[n]{a_n}-1$. If we prove $b_n \rightarrow 0$, we are done.
We know that:
$$1+nb_n \leq (1+b_n)^n=a_n.$$
This implies:
$$b_n \leq \frac{ a_n-1}{n}.$$
Since we are supposing $a>1$, we also have that (at least for sufficiently large $n$):
$$0 \leq b_n.$$
Combining both inequalities and using the squeeze theorem yields the result.
In the case that $a<1$, apply the previous result with the sequence $\sqrt[n]{\frac{1}{a_n}}$.
