Question: > 
> **Possible Duplicate:**  
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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.
One could use the fact that for a sequence of positive terms, if $\lim\limits\_{n\rightarrow\infty}{a\_{n+1}\over a\_n}$ exists, then so does $\lim\limits\_{n\rightarrow\infty} \root n\of {a\_n}$ and the two limits are equal. A proof of this general fact can be found in [these notes](http://web.archive.org/web/20130728110356/http://math.uga.edu/~pete/243series4.pdf) of Pete L. Clark. This result can also be found in many analysis texts; e.g., baby Rudin.

That your sequence has limit $1$ is easily shown using the above fact. A detailed proof that your sequence has limit $1$, based on the proof of the above fact, can be found in this [thread](https://math.stackexchange.com/questions/115822).