How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$? I want to show that $$\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n}-1) = 0$$ and my assistant teacher gave me the hint to find a proper estimate for $\sqrt[n]{n}-1$ in order to do this. I know how one shows that $\lim_{n \to \infty} \sqrt[n]{n} = 1$, to do this we can write $\sqrt[n]{n} = 1+x_n$, raise both sides to the n-th power and then use the binomial theorem (or to be more specific: the term to the second power). However, I don't see how this or any other trivial term (i.e. the first or the n-th) could be used here. 
What estimate am I supposed to find or is there even a simpler way to show this limit?
Thanks for any answers in advance.
 A: Write 
$$|\sqrt[n]n-1|=\left|\exp\left(\frac{\ln n}n\right)-1\right|=\int_0^{\frac{\ln n}n}e^tdt\leq \frac{\ln n}n\exp\left(\frac{\ln n}n\right).$$
A: The OP's attempt can be pushed to get a complete proof. 
$$
n = (1+x_n)^n \geq 1 + nx_n + \frac{n(n-1)}{2} x_n^2 + \frac{n(n-1)(n-2)}{6} x_n^3 > \frac{n(n-1)(n-2) x_n^3}{6}  > \frac{n^3 x_n^3}{8},
$$
provided $n$ is "large enough" 1. Therefore, (again, for large enough $n$,) $x_n < 2 n^{-2/3}$, 
and hence $\sqrt{n} x_n < 2n^{-1/6}$. Thus $\sqrt{n} x_n$ approaches $0$ by the sandwich (squeeze) theorem. 

1In fact, you should be able to show that for all $n \geq 12$, we have
$$
\frac{n(n-1)(n-2)}{6} > \frac{n^3}{8} \iff \left( 1-\frac1n \right) \left( 1- \frac2n \right) \geq \frac34. 
$$
A: Use the fact that, when $n\to\infty$, $$\sqrt[n]{n}-1=\exp\left(\frac{\log n}n\right)-1\sim\frac{\log n}n$$
A: An elementary proof using $\text{AM} \ge \text{GM}$:
We have that, for sufficiently large $n$, 
$$ \frac{1 + 1 + \dots + 1 + n^{1/3} + n^{1/3} + n^{1/3}}{n} \ge n^{1/n}$$
using $\text{AM} \ge \text{GM}$ on $n-3$ copies of $1$ and three copies of $n^{1/3}$.
i.e we get the estimate
$$ 1 - \frac{3}{n} + \frac{3}{n^{2/3}} \ge n^{1/n}$$
This proof can be generalized to show that
$$n^{(k-1)/k} (n^{1/n} - 1) \to 0$$
for any positive integer $k$.
A variant of: Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$
