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?
 A: Try this, let $n> e^{e^N}$. Then you have that
$${1\over\log\log n}< {1\over N}$$
So for any $\epsilon >0$, Choose $N>{1\over\epsilon}$ and then for all $n > e^{e^N}$ we have
$${1\over\log\log n}<\epsilon$$
showing the limit condition.
A: We know that
$\lim_{n\to+\infty}\log(n)=+\infty$
thus
$$\lim_{n\to+\infty}\frac{1}{\log(\log(n))}=\lim_{N\to+\infty}\frac{1}{\log(N)}=0$$
A: No it is not alright since the inverse function reverse the inequalities. 
But you have 
$$\log(\log n)\xrightarrow[n\to\infty]{} +\infty$$
so 
$$\frac{1}{\log(\log n)}\xrightarrow[n\to\infty]{} 0.$$
A: 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
$\lim_{x \to \infty} {1 \over \log (\log x) } = 0$.
A: Let $n=2^{2^k}$ and note that $\ln \ln x$ is monotonic
