Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$ 
Possible Duplicate:
$\lim_{n \to +\infty} n^{\frac{1}{n}} $ 

I know that 
$$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$
and I can imagine that $n$ grows linearly while $n$th root compresses it exponentially and therefore the result is $1$, but how do I calculate it?

Claim: For each $a>1$, there exists $N$ such that $n<a^n$ for all $n>N$.
Proof: Write $a=1+b$.  By the binomial theorem, $a^n=(1+b)^n\geq \frac{1}{2}n(n-1)b^2$ when $n\geq 2$.  Thus $\frac{a^n}{n} \geq \frac{1}{2}(n-1)b^2$.  It follows that if $N$ is at least $2/b^2+1$, then when $n>N$, $n<a^n$.

As a consequence, for each $a>1$, there exists $N$ such that $1\leq n^{1/n}<a$ for all $n>N$, and this implies that $\lim\limits_{n\to\infty}n^{1/n}=1$.
