Prove $\lim_{n \to \infty} \frac{\ln(n)}{n}=0$ without L'Hospital's Rule

Question

Prove the following without using L'Hospital's Rule, integration or Taylor Series: $$\lim_{n \to \infty} \frac{\ln(n)}{n}=0 $$

I began by rewriting the expression as: $$\lim_{n \to \infty}{\ln(n^{1/n})} $$

Since the text shows $$\lim_{n \to \infty}{n^{1/n} = 1} $$

I was wondering is the proof just as simple as stating:

$$\lim_{n \to \infty}{\ln(1) = 0} $$

or do I need to apply the squeeze theorem, use a $\varepsilon$-N proof, or etc?

Answer 1

Since $e^x> x$, we have $\ln x < x$ for all $x >0$.

Hence,

$$0 \leqslant \frac{\ln n}{n} = \frac{2 \ln \sqrt{n}}{n} < \frac{2 \sqrt{n}}{n} = \frac{2}{\sqrt{n}} \to 0$$

Answer 2

As long as you have proven that $\ln(x)$ is a continuous function and that $\lim n^{1/n}=1$, then the proof is as easy as:

$$ \lim_{n \to \infty} \frac{1}{n}\ln{n}=\lim_{n \to \infty} \ln(n^{1/n})=\ln\left(\lim_{n \to \infty}n^{1/n}\right)=\ln(1)=0$$

and note that the second equality is true by the continuity of $\ln(x)$

Answer 3

If you know that $e^x> 1+x$, then $e^x=(e^{x/2})^2 > (1+x/2)^2=1+x+\frac{x^2}{4}$.

Now let $x=\log n$, so $n=e^{\log n} > 1+\log(n)+\frac{\log^2(n)}{4}$.

So:

$$\frac{\log n}{n} < \frac{1}{1+\log(n)/4}$$

Now, if you know $\log(n)$ is increasing, you are done. [You don't even need to know that $\log(n)\to+\infty$, because if $\log(n)$ bounded above, then $\frac{\log n}{n}\to 0$ trivially.)