How to prove this limit: $\lim\limits_{n \to \infty} \frac1{\sqrt[n] n}=1 $? This is what I am trying to prove:
$$\lim_{n \rightarrow \infty} \frac1{\sqrt[n] n}=1 $$
How do I go about this?
 A: $$\lim_{n\to\infty}\frac 1{\sqrt[n]n}=\lim_{n\to\infty}n^{-1/n}=\lim_{n\to\infty}e^{-(\ln n)/n}=e^{-\lim_{n\to\infty}(\ln n)/n}=e^0=1$$
Because e.g. by L'Hôpital's rule $\lim_{n\to\infty}\frac{\ln n}n=\lim_{n\to\infty}\frac1n=0$.
A: A direct proof without L'Hopital.
We will show that $n^{1/n}\to 1$. Specifically, let $x_n=n^{1/n}-1$. Then we will show that $x_n\to 0$. 
Now $(1+x_n)^n = n$.
And $$(1+x_n)^n\geq 1+nx_n + \frac{n(n-1)}{2} x_n^2$$
So you need:
$$nx_n + \frac{n(n-1)}{2}x_n^2 \leq n-1$$
Dividing by $n(n-1)/2$ we need:
$$x_n^2 + \frac{2}{n-1}x_n \leq \frac{2}{n}$$
Completing the square:
$$\left(x_n+\frac{1}{n-1}\right)^2 \leq \frac{2}{n} + \frac{1}{(n-1)^2}$$
That means that $$x_n \leq \sqrt{\frac{2}{n}+\frac{1}{(n-1)^2}}-\frac{1}{n-1}$$
But clearly, $\sqrt{\frac{2}{n}+\frac{1}{(n-1)^2}}-\frac{1}{n-1}\to 0$, so, since $x_n>0$, we have $x_n\to 0$.
A: From Ross (Elementary Analysis, Theory of Calculus):
12.3 Corollary.
If $\lim \left|\frac{s_{n+1}}{s_n}\right|$ exists and it is equal to $L$, then $\lim \left|s_n\right|^{\frac{1}{n}}$ exists and it is equal to $L$.
A: Hint: let $y = n^{-1/n}$, then work with $\log y = -\dfrac{1}{n}\log n$ as $n \to \infty$.
A: Here are the steps
$$\lim_{n \to \infty} \frac1{\sqrt[n] n}= \lim_{n \to \infty} \frac{1}{n^{\frac1n}}= \frac{1}{\lim\limits_{n \to \infty} n^{\frac1n}} $$
$$ =\frac{1}{\lim\limits_{n \to \infty} \exp\left(\ln n^{\frac1n}\right)} =\frac{1}{\lim\limits_{n \to \infty} \exp\left(\frac{\ln n}{n}\right)} $$
$$
=\frac{1}{\exp\left( \lim\limits_{n \to \infty} \frac{\ln n}{n}\right)} =\frac{1}{\exp\left( \lim\limits_{n \to \infty} \frac{\frac{d}{dn}[\ln n]}{\frac{d}{dn}[n]}\right)}
$$
$$
= \frac{1}{\exp\left( \lim\limits_{n \to \infty} \frac{1}{n}\right)} = \frac{1}{\exp\left(0\right)} =1$$
