$\frac{1}{n}=O(\frac{1}{\ln n})$ or $\frac{1}{n}=o(\frac{1}{\ln n})$?

Question

$\frac{1}{n}=O(\frac{1}{\ln n})$ or $\frac{1}{n}=o(\frac{1}{\ln n})$?

I know that $n\geq\ln n, (n> 0)$, with which $\frac{1}{n}\leq \frac{1}{\ln n}$ and so $\frac{1}{n}=O(\frac{1}{\ln n})$.

To know if $\frac{1}{n}=o(\frac{1}{\ln n})$, I need to know if $\lim_{n\to \infty}\frac{\ln n}{n}=0$, but how can I prove that this limit does not exist or does exist? Thank you very much.

Answer 1

Without L'Hopital or anything fancy, only elementary manipulations of what you know already.

If you know $x\geq \ln x$ for all $x\geq 1$, then you know $$ \sqrt{x} \geq \ln \sqrt{x} $$ for all $x\geq 1$ as well. Therefore, you know $$ \sqrt{x} \geq \frac{1}{2}\ln x $$ for all $x\geq 1$, and rearranging you get $$ \frac{\ln x}{\sqrt{x}} \leq 2\,. $$ It follows that, for $x\geq 1$, $$ 0\leq \frac{\ln x}{x} \leq \frac{2}{\sqrt{x}} $$ and this is in particular true for integers: $$ 0\leq \frac{\ln n}{n} \leq \frac{2}{\sqrt{n}} \xrightarrow[n\to\infty]{} 0 $$

Answer 2

You have a typical $\frac{\infty}{\infty}$ indeterminacy. Consider the continuous version of your ratio where you replace $n$ by $x\in\mathrm{R}$ and use the L'Hospital rule to conclude that $\lim_{x\rightarrow\infty}\frac{\ln x}{x}=0$.