I am trying to proof $f\in o(g)$
Let be $r,p\in \mathbb{R}$ with $p>0$
We have $f(n)=ln^r (n)$ and $g(n)=n^p$
I have already proofed that $ln(n)\in o(n)$ via l'Hospital $\lim\limits_{n\to \infty}\frac{\frac{1}{x}}{1}=0\Longrightarrow \lim\limits_{n\to \infty}\frac{ln(n)}{n}=0\Longleftrightarrow ln(n)\in o(n)$
I tried substituting n with ln(n) to recieve
$\forall c>0 \exists n_0 \forall n\geq n_0 : ln(ln(n))\leq cln(n)$
Then i tried splitting $c=\frac{a}{b}$ to get $\forall \frac{a}{b}>0\exists n_0 \forall n\geq n_0 : ln(ln(n)) \leq \frac{a}{b}ln(n)$
With this information i said, that $\frac{a}{b}*\frac{b}{a}*b*ln(ln(n))=b*ln(ln(n))\leq \frac{a}{b}*\frac{b}{a}*\frac{a}{b}*b*ln(n)=\frac{a}{b}*b*ln(n)=a*ln(n)$
to conclude $ln(n)^b=e^{\frac{a}{b}*\frac{b}{a}*b*ln(ln(n))}\leq e^{\frac{b}{a}*b*ln(n)}=e^{b*ln(n)}=n^a$
At this point i realized a huge problem. I assumed $\frac{a}{b}$ to be positive. But i need it to work for r and p, where only p is known to be postive, which in the other hand means, that r can be negative, what would make $\frac{p}{r}$ negative.
It means that my approach is propably not working a