Limit of product with unbounded sequence $\lim_{n\to\infty} \sqrt{n}(\sqrt[n]{n}-1)=0$

Question

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. :-)

See also How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$? and How to calculate the following limit: $\lim_{x\to+\infty} \sqrt{n}(\sqrt[n]{x}-1)$? — Martin Sleziak, Jan 14 '17 at 17:30
@AnneBauval Yes, the I posted the other link by mistake - I should have mentioned only the first one: How to prove $\lim_{n \to \infty} \sqrt{n}(\sqrt[n]{n} - 1) = 0$? — Martin Sleziak, Nov 10 '23 at 15:29

Haha · Answer 1 · 2013-11-09T16:08:43.940

2

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 $n\to \infty$

edited Nov 09 '13 at 16:08

answered Nov 09 '13 at 15:27

Haha

5,648

Unfortunately we do not know the L'Hospital criterion yet. :( – user105916 Nov 09 '13 at 15:30
@user105916 You want elementary calculus only? – Haha Nov 09 '13 at 15:39
I am studying for 3 weeks now. So, yes i think so. – user105916 Nov 09 '13 at 16:01
Thanks, but I dont understand that. I do not know Newton's binomial either. – user105916 Nov 09 '13 at 16:21
Or is Newtons binomial just $(x+y)^n=\sum_{k=0}^n \binom{n}{k}x^{n-k}y^k$ – user105916 Nov 09 '13 at 16:23
@user105916 Yes this is and if you take the 4th element of the series this is the $n(n-1)(n-2)/6$ and thus you have the inequality – Haha Nov 09 '13 at 16:33
@user105916 Just to inform you that if you are an undergratuate mathematician this must be the proper way to solve this problem with elementary calculus. I know that elementary calculus sometimes is more difficult than advanced – Haha Nov 09 '13 at 16:44
Thanks for your help. :-) And your nice words. – user105916 Nov 09 '13 at 18:29

Limit of product with unbounded sequence $\lim_{n\to\infty} \sqrt{n}(\sqrt[n]{n}-1)=0$

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