TITLE: Limit of the sequence $\lim_{n\rightarrow\infty}\sqrt[n]n$
QUESTION [11 upvotes]: 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?

REPLY [5 votes]: Let's see a very elementary proof. Without loss of generality we proceed replacing $n$ by $2^n$ and get that:
$$ 1\leq\lim_{n\rightarrow\infty} n^{\frac{1}{n}}=\lim_{n\rightarrow\infty} {2^n}^{\frac{1}{{2}^{n}}}=\lim_{n\rightarrow\infty} {2}^{\frac{n}{{2}^{n}}}\leq\lim_{n\rightarrow\infty} {2}^{\frac{n}{\dbinom{n}{2}}}=2^0=1$$
By Squeeze Theorem the proof is complete.