# How to prove $\lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0$.

I am trying to prove that $\lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0$. My attempt is as follows.

Since for each $n\in \mathbb{N}$, $n>0$ by arithmetic geometric inequality $$\sqrt[n]{n}=\sqrt[n]{n.1...1}=\le \dfrac{n+1+...+1}{n}=\dfrac{n+n-1}{n}=2-\dfrac{1}{n}$$.

Hence $$0<\sqrt[n]{n}-1\le 1-\dfrac{1}{n}<1.$$

So $\lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0$.

Is this correct ? If not how to show that $\lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0$ ? Please help me. Thanks.

• For every $n\geqslant1$, $n<(3/2)^n$ hence $1\leqslant\sqrt[n]{n}<3/2$ hence the $n$th term of the sequence is between $0$ and $1/2^n$. (Note also that $\sqrt[n]{n}-1<1$ does not imply $(\sqrt[n]{n}-1)^n\to0$, for example $(1-1/n)^n\to1/e\ne0$.) – Did Jun 12 '15 at 17:18
• @Did please could you help me to prove $\lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0$? – Aven Jun 12 '15 at 17:23
• @Did argued that $(n^{1/n} - 1)^n \leq (3/2 - 1)^n \leq (1/2)^n$, which approaches 0 as $n$ goes to infinity. (This is a correct argument, while your original attempt was not.) More strongly, one may note that $log_2(n^{1/n}) = (1/n)\log_2 n$, which approaches 0, so that $n^{1/n} \rightarrow 1$. This also implies what you want to show. – Andy Drucker Jun 12 '15 at 17:27
• @Aven "please could you help me to prove..." I just did. – Did Jun 12 '15 at 17:30
• – Martin Sleziak Jan 12 '20 at 19:05

The inequality $$\sqrt[n]{n} \leq \frac{1+\ldots+1+n}{n}= 2-\frac{1}{n}$$ just gives: $$\limsup_{n\to +\infty}\left(\sqrt[n]{n}-1\right)^n \leq\frac{1}{e}$$ but if you consider the slightly improved inequality: $$\sqrt[n]{n}\leq\frac{1+\ldots+1+\sqrt{n}+\sqrt{n}}{n}=1-\frac{2}{n}+\frac{2}{\sqrt{n}}$$ you easily get: $$\lim_{n\to +\infty}\left(\sqrt[n]{n}-1\right)^n = 0$$ as wanted.

For a sharp inequality, you need to write $n$ as a product of $n$ numbers, quite close one each other: that is not difficult to achieve. Since: $$n=1\cdot\prod_{k=1}^{n-1}\frac{k+1}{k}$$ by the AM-GM inequality it follows that: $$\sqrt[n]{n}\leq 1+\frac{H_{n-1}}{n}$$ that is essentially optimal since $e^x\geq 1+x$ implies: $$\sqrt[n]{n} \geq 1+\frac{\log n}{n}.$$

• Thanks for this incredible explanation! $\lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0$ because of the rule: "the limit of the products is the product of the limits" which in this case would be $0*0*0*...*0$ n times? – GniruT Nov 20 '15 at 12:08

Not exactly rigorous, but you can make it rigorous without too much effort, I think: It suffices to show that

$$\lim_{n \to \infty} \sqrt[n]{n} < 2$$

But this is trivial, since $2^n \gg n$. Hence the limit follows. (In fact, $\lim_{n \to \infty} \sqrt[n]{n} = 1$.)

ETA: If the $2$ makes you nervous, you can substitute any value $q, 1 < q < 2$. @Did chose $q = 3/2$ in the comments.

We can also approach this using brute force via application of L'Hospital's Rule.

$$\lim_{n\to \infty}(n^{1/n}-1)^n=\exp{\lim_{n\to \infty}n\log(n^{1/n}-1)}$$

Now, let's look at the interior limit.

$$\lim_{n\to \infty}n\log(n^{1/n}-1)=\lim_{n\to \infty}\frac{\log(n^{1/n}-1)}{n^{-1}}=-\lim_{n\to \infty}n^2\frac{1-n^{-2}\log n}{n^{1/n}-1}=-\infty$$

Inasmuch as $\lim_{x\to -\infty}e^x=0$, we obtain the expected result.