Question: > 
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> [$\lim\_{n \to +\infty} n^{\frac{1}{n}} $](https://math.stackexchange.com/questions/115822/lim-n-to-infty-n-frac1n) 
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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?

Answer: 
Perhaps one of the most elementary ways to prove it: since $\,n\geq 1\,\,\forall n\in\mathbb{N}\,$ , we can put$$\sqrt[n]{n}=1+c\_n\,,\,c\_n\geq0\Longrightarrow n=(1+c\_n)^n\geq \frac{n(n-1)}{2}c\_n^2$$using the binomial expansion ,so that$$0<c\_n\leq\sqrt\frac{2}{n-1}$$and now just apply the squeeze theorem and get $\,c\_n\to 0\,$, which is precisely what we need.
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 $n>N$, $n<a^n$.

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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 $\lim\limits\_{n\to\infty}n^{1/n}=1$.