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

Question

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.

Answer 1

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: $$ \sqrt[n]{n} \geq 1+\frac{\log n}{n}.$$

Answer 2

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.

Answer 3

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.