I have to show that
$\lim_{n\to\infty} \sqrt{n}(\sqrt[n]{n}-1)=0$
We proofed that
$\lim_{n\to\infty} \sqrt[n]{n}=1$
My problem is, that I do not know how to solve that.
That $\lim_{n\to\infty}(\sqrt[n]{n}-1)=0$ is clear. But $\lim_{n\to\infty} \sqrt{n}$ is not bounded.
So I can not simply calculate:
$\lim_{n\to\infty}\sqrt{n}\cdot \lim_{n\to\infty}(\sqrt[n]{n}-1)$
right?
I would be thankfull for every hint. :-)
for $n>1$ we have that $\sqrt[n] {n}=1+δ_n$. Then $δ_n>0$ and we have that $n=(1+δ_n)^n\geq \frac { n(n-1)(n-2)δ_n^3}{6}$(Newton's binomial)$<=>δ_n^3\leq \frac {6}{(n-1)(n-2)}<=> n^{3/2}δ_n^3\leq \frac {n^{3/2}6}{(n-1)(n-2)}$
($\frac {n^{3/2}6}{(n-1)(n-2)}\to 0$)and thus $ n^{3/2}δ_n^3\to 0$ for $n\to \infty$. This means that $n^{3/2}(\sqrt[n] {n}-1)^3\to 0$ or $(\sqrt n (\sqrt[n] {n}-1))^3\to 0$ or $\sqrt n (\sqrt[n] {n}-1)\to 0$ for $n\to \infty$