Prove that $\lim_{n \to \infty} {\sqrt[n]{n}} = 1$ using the inequality that $(1+x)^n\geq 1 + nx + \frac{n(n-1)}{2}x^2 $ $(1+x)^n\geq 1 + nx + \frac{n(n-1)}{2}x^2 $ holds  for all  $ n \in \mathbb{N} $ and  $x \ge  0$
I proved that $ a \geq b \Leftrightarrow a^n \geq b^n $. 
Then I plug $x_{n} $  into the inequality. 
But I don't know what to do next. Please help me
 A: We have $(1+x)^n\geq 1 + nx + \frac{n(n-1)}{2} x^2\geq 1$.
So $(1+x)\geq \sqrt[n]{1 + nx + \frac{n(n-1)}{2} x^2} \geq \sqrt[n]{1} =1$.
It would be great if what is in the middle would equal $n$ (that's how far the hint goes, the rest is the full answer:

$$n=1 + nx + \frac{n(n-1)}{2} x^2$$
$$0=(1-n) + nx + \frac{n(n-1)}{2} x^2$$
Which has two roots, but since we want $x\geq 0$, we take the positive root, namely:
$$x_0=\frac{-n+\sqrt{n^2+2n(n-1)^2}}{n(n-1)}$$
Then letting $x=x_0$ in the original inequality we get:
$$(1+x_0)\geq \sqrt[n]{n}\geq1$$
And low taking the limit as $n\to\infty$, $(1+x_0)\to 1$, therefore
$$\lim_{n\to\infty}\sqrt[n]{n}=1$$

(I wanted to make it appear and disappear, but I don't know how, I'm sorry!)
A: An alternative way is to exploit the AM-GM inequality. Since for any $n\geq 2$ we have
$$ n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right),\qquad \sqrt[n]{n}=\text{GM}\left(1,1+\frac{1}{1},\ldots,1+\frac{1}{n-1}\right)$$
we also have
$$ \sqrt[n]{n}\leq \text{AM}\left(1,1+\frac{1}{1},\ldots,1+\frac{1}{n-1}\right) = 1+\frac{H_{n-1}}{n}\leq 1+\frac{\log n}{n} $$
and $\lim_{n\to +\infty}\sqrt[n]{n}=1$ follows by squeezing.
