Prove that $a^{\frac{1}{n}}$ converges to 1 as $n$ goes to $\infty$ for $a>0$ I am having trouble proving that the sequence $a^{\frac{1}{n}}$ converges to $1$ as $n \to \infty$ for $a>0$. 
My attempt at a proof:
I want to show that for every $\epsilon>0$ there exists $N \in \mathbb{N}:n \geq N$ implies $$|a^{\frac{1}{n}}-1|<\epsilon$$. So let $\epsilon>0$ be given and by the Archimedean Property, we can choose a $n$ such that $\frac{1}{n} <\epsilon$. Then $$|a^{\frac{1}{n}}-1|<|a^{\epsilon} -1|<\epsilon$$ since $\epsilon$ can be made small. I think am getting lost in this problem since I am not sure how to deal with $a^{\frac{1}{n}}$. It is obvious that for large n, $a^{\frac{1}{n}}$ will approach 1 and so $$|a^{\frac{1}{n}}-1|=0$$, which is less than $\epsilon$. Any help will be appreciated. 
 A: If $a = 1$ then $a^{1/n} = 1$. If the result holds for all $a > 1$, then it holds for $0 < a < 1$ because of the inequalities $0 < 1- a^{1/n}  < (1/a)^{1/n} - 1$ for $0 < a < 1$. So it suffices to consider $a > 1$. Let $\delta_n = a^{1/n} - 1$ and use the binomial theorem to show that $a > \frac{n(n-1)}{2}\delta_n^2$. Then $\delta_n < \sqrt{\frac{2}{n(n-1)}a}$. Now argue that $\delta_n \to 0$.
A: 
CASE $1$:  $0<a<1$

If $0<a<1$, then let $a^{1/n}=1-b_n$, where $0<b_n<1$.  We can write
$$a=(1-b_n)^n=\frac{1}{\left(1+\frac{b_n}{1-b_n}\right)^n} \tag 1$$
where we note that $0<\frac{b_n}{1-b_n}$.
Using Bernoulli's Inequality in $(1)$ reveals
$$a\le \frac{1}{1+n\frac{b_n}{1-b_n}} \tag 2$$
whereupon solving for $b_n$ in $(2)$ yields
$$0<b_n\le \frac{1-a}{1+(n-1)a} \tag 3$$
Finally, applying the squeeze theorem to $(3)$, we find that $\lim_{n\to \infty}b_n=0$ from which we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}a^{1/n}=1}$$

as was to be shown.


CASE $2$:  $1<a$

If $1<a$, then let $a^{1/n}=1+b_n$, where $b_n>0$.  Then we have from Bernoulli's Inequality
$$\begin{align}
a&=\left(1+b_n\right)^n\\\\
&\ge 1+nb_n
\end{align}$$
whereupon solving for $b_n$ yields
$$0<b_n\le \frac{a-1}{n} \tag4$$
Applying the squeeze theorem to $(4)$, we see that $\lim_{n\to \infty}b_n=0$ from which we obtain the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}a^{1/n}=1}$$

as expected!
