Proof check $(\log\log n) /(\log n) $ approaches zero

Question

Proof : If $|a| < 1$ then $(na^n)$ is a null sequence therefore if $b>1$ then ${n\over (b)^n}$ is a null sequence.There is always an $m$ such that for every $n > m$ $${n\over (b)^n} <\epsilon' = {\epsilon\over b}$$ Now if g is the characteristic of $\log\log n$ then $g\le \log\log n\lt g+1$. Therefore $${\log\log n\over \log n} < {g+1\over (b)^g}= b{g+1\over (b)^ {g+1}} $$ So if we choose $n>b^{b^m}$ we would have $g+1 > m$ and so $|{\log\log n\over \log n}| < \epsilon$. (Side note: for the logarithm the base b > 1 that is being used in here and the theorem).

Answer 1

An other way :

$$\lim_{n\to\infty }\frac{\ln(\ln(n))}{\ln n}\underset{m=\ln n}{=}\lim_{m\to\infty }\frac{\ln m}{m}\underset{\text{(Hospital)}}{=}\lim_{m\to\infty }\frac{1}{m}$$

Then if $N=\lfloor\frac{1}{\varepsilon}\rfloor+1$, you have that $$\left|\frac{\ln(\ln(n))}{\ln(n)}\right|<\varepsilon$$ if $n\geq N$ what prove the claim.