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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?

  • 0
    So the "trick" is $a^b=e^{b*\ln{a}}$2012-06-05

6 Answers 6

25

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$0and now just apply the squeeze theorem and get $\,c_n\to 0\,$, which is precisely what we need.

  • 1
    @experimentX: Specifically, $c_n$ by definition is $\sqrt[n]{n}-1$. It is a case of introducing notation to make the exposition easier.2012-06-06
6

Claim: For each $a>1$, there exists $N$ such that $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.


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

  • 1
    @DonAntonio: Inspired by your post I cleaned mine up a little (but didn't essentially change the method).2012-06-05
5

$\lim_{n\rightarrow\infty}\sqrt[n]n$ $=\lim_{n\rightarrow\infty}e^{\frac{\ln(n)}{n}}$ and as we know that $\lim_{n\rightarrow\infty}\frac{\ln(n)}{n} = 0$ {apply l'Hospital's rule}
So $\lim_{n\rightarrow\infty}\sqrt[n]n=1$

5

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 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.

  • 0
    See also [here](http://math.stackexchange.com/questions/28476/finding-the-limit-of-frac-n-sqrtnn/28487#28487) and [here](http://math.stackexchange.com/a/76800/) for the result on relation between $\frac{a_{n+1}}{a_n}$ and $\sqrt[n]{a_n}$.2012-06-05
4

Let$y=n^{\frac{1}{n}}$ $\log y=\log( n^{\frac{1}{n}})$ $\Rightarrow \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{\log n}{n}$ We have $\left(\frac{\infty}{\infty}\right)$ form so we apply L'hospital rule, we get, $ \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{\frac{1}{n}}{1}$ $\therefore \lim_{n\to\infty}\log y=\lim_{n\to\infty}\frac{1}{n}=0$ Hence $\lim_{n\to\infty} y=e^{0}=1.$ It follows that, $\lim_{n\to\infty}n^{\frac{1}{n}}=1.$

4

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.

  • 0
    @experimentX: you're welcome!2012-06-05