Is $(n^{1/n}-1)\in O(n^{-\frac12})$ as $n\to \infty$?

Question

Let $$x_n=n^{1/n}-1$$ An exercise asks me to prove that $\{x_n\}\to 0$ and that $x_n\in O(n^{-\frac12})$ as $n\to \infty$. Now, I could easily prove the first part. But to me $x_n\notin O(n^{-\frac12})$. This is because, as far as I understand, $x_n \in O(n^a)$ if $a>\frac1n$. Am I right?

Thank you for your help.

Answer 1

I can even prove a stronger statement, namely that $(n^{1/n}-1) \in o(n^{-1/2})$ because $$\lim_{n \to \infty}\dfrac{n^{1/n}-1}{n^{-1/2}}=0 $$ This limit can be calculated by using l'Hôpitals rule twice: $$\lim_{n \to \infty}\dfrac{n^{1/n}-1}{n^{-1/2}}=\dfrac{0}{0}=\lim_{n \to \infty}(-2)\dfrac{n^{(1/n)-2}(1-\ln n)}{n^{-3/2}}=-2\lim_{n \to \infty}\dfrac{n^{(1/n)}(1-\ln n)}{\sqrt{n}}=\dfrac{-\infty}{\infty}=$$ $$-4\lim_{n \to \infty}\dfrac{n^{(1/n)-2}(1-\ln n)^2-n^{(1/n)-1}}{n^{-1/2}}=-4\lim_{n \to \infty}n^{(1/n)}\lim_{n \to \infty}\left(\dfrac{(1-\ln n)^2}{n^{3/2}}-\dfrac{1}{n^{1/2}}\right)=$$ $$-4\cdot 1 \cdot 0 = 0$$

Answer 2

By a simple expansion, $$ \left(\frac{1}{x} \right)^x = 1-x \log x + \frac{1}{2} (\log x)^2 x^2 +O(x^3) $$ as $x \to 0$. Hence $$ n^{1/n} - 1 \approx -\frac{1}{n}\log \frac{1}{n} $$ as $n \to +\infty$. Now you have to analyze $$ \sqrt{n} \left( n^{1/n}-1 \right) \approx \sqrt{n} \left( - \frac{1}{n}\log \frac{1}{n} \right). $$ Can you draw any conclusion?

Answer 3

I write here a heuristic argument:

$$n^{\frac{1}{n}}-1=e^{\frac{\ln n}{n}}{-1}\approx\frac{\ln n}{n}<\frac{\sqrt{n}}{n}$$ for $n$ large enough.

Hope you can make it rigorous or come up with another way to prove!! :)