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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.

Martin Sleziak
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Huy
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4 Answers4

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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 $$ \frac{n(n-1)(n-2)}{6} > \frac{n^3}{8} \iff \left( 1-\frac1n \right) \left( 1- \frac2n \right) \geq \frac34. $$

Srivatsan
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  • Can you briefly explain why the second last term is greater than $\frac{n^3 x_n^3}{8}$ (for large enough n)? – Huy Oct 17 '11 at 18:28
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    @Huy Well, at that step, I wanted to lower bound $(n-1)(n-2)/6$ by $c n^2$ for some constant $c$. If $n$ is very large, $n-1$ and $n-2$ are essentially like $n$ each, so that quantity is essentially like $n^2/6$, but it is certainly a bit less. So you can pick a constant any $c < 6$. I chose $c=1/8$ for convenience (notice that $c$ will be a perfect cube, which is nice for us). To prove it, consider the ratio $$ \frac{1}{6} \frac{n(n-1)(n-2)}{n^3} = \frac{1}{6} (1-\frac{1}{n}) (1-\frac{2}{n}). $$ If $n \geq 20$, then this is at least $(1/6) \cdot (19/20) \cdot (18/20) > 1/8$ (verify!) – Srivatsan Oct 17 '11 at 18:38
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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 $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$

Community
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Aryabhata
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    Actually, the same proof can be generalized further to give a tight (up to the constant factor) bound of $n^{1/n}-1 = O((\log n) / n)$, just by setting $k = \log n$. I find that very interesting. :) – Srivatsan Oct 17 '11 at 21:01
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Use the fact that, when $n\to\infty$, $$\sqrt[n]{n}-1=\exp\left(\frac{\log n}n\right)-1\sim\frac{\log n}n$$

Did
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Write $$|\sqrt[n]n-1|=\left|\exp\left(\frac{\ln n}n\right)-1\right|=\int_0^{\frac{\ln n}n}e^tdt\leq \frac{\ln n}n\exp\left(\frac{\ln n}n\right).$$

Davide Giraudo
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