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

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?

• Well, the limit of $\ln$ depends on where the interior function approaches. As the interior function approaches infinity, the exponent approaches zero and the term approaches 1. The $\ln$ of that expression therefore approaches zero. Think about what L'Hopital's rule would tell us. The denominator of the original expression approaches infinity faster than the numerator, so the expression goes to zero. This is equivalent to showing that the interior function's exponent goes to zero at the same rate as $n$ goes to infinity; that is to say the limit of $n^{\frac{1}{n}}$ is 1. This is enough. – KR136 Feb 6 '16 at 3:54

## 3 Answers

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

Hence,

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

• Definitely the symplest proof. – Thomas Andrews Feb 6 '16 at 5:04

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 $\ln(x)$

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.)