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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.

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up vote 12 down vote accepted

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

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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
@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

$$\sqrt[n]{n}-1=\exp\left(\frac{\log n}n\right)-1\sim\frac{\log n}n.$$

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Can you give me a hint how I can obtain this approximation? – Huy Oct 17 '11 at 18:16
When $n\to\infty$, $\log(n)/n\to0$. When $x\to0$, $\exp(x)=1+x+o(x)$ hence $\exp(x)-1\sim x$. – Did Oct 17 '11 at 19:09
This is a nice and short proof. I like it. – xpaul Apr 20 '13 at 1:53

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 $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$

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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

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

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Multiply through by the square root to get $n^{(n+2)/2n} - n^{1/2}$. What happens to the exponent on the first term as $n$ becomes large? You need to fill in some details, but this might give you an alternate approach.

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This approach looks by far the easiest, but am I actually allowed to first evaluate $\lim_{n \to \infty} (n+2)/2n$ and then look at $n^{1/2}$ seperately? – Huy Oct 17 '11 at 18:49
I don't get it. Just because the exponent approaches $1/2$ it doesn't mean that the difference $n^{(n+2)/2n} - n^{1/2}$ should approach $0$. You need to control how fast the base (i.e., $n$ in this case) grows relative to how fast the exponent grows/shrinks. (And @Huy, intuitive and easy it may be, but this answer does not look fully rigorous to me. In fact, most of the work done by the other answers is to show that despite the growth of the base, the difference still tends to $0$.) – Srivatsan Oct 17 '11 at 19:43
@Srivatsan - The answer was never meant to be rigorous, just to provide an alternative view that might make the result seem plausible. I acknowledged that details need to be addressed. I should have been more explicit that I was providing an idea and not a solution. – Chris Leary Oct 19 '11 at 14:42
@Chris I am sorry, but I find the answer is a bit too vague. (I did not downvote though.) – Srivatsan Oct 19 '11 at 15:05
@Srivatsan I accept your criticism and agree with it. No problem. – Chris Leary Oct 19 '11 at 15:15

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