Prove that $\lim_\limits{n\to\infty }\frac{1}{\log(\log(n))}=0$ How to prove this limit?
$$\lim_{n\to\infty }\frac{1}{\log(\log(n))}=0$$
I thought of something like 
$$0 \le \frac{1}{\log(\log(n))} \le \frac{1}{n}$$
Is it alright?

We have $\log x = \int_1^x {1 \over t} dt $, hence $\log$ is increasing. From the integral we see that $\log (n+1) \ge {1 \over 2} + {1 \over 3} + \cdots + { 1 \over n}$, and since the Harmonic series is divergent, we see that
$\lim_{x \to \infty} \log x = \infty$.
It follows that $\lim_{x \to \infty} \log (\log x) = \infty$ and so
$\lim_{x \to \infty} {1 \over \log (\log x) } = 0$.
