Supposing $u = \cfrac{(n^2+n)}{2n^2+n}$ and v = $\cfrac{\log_e n}{n}$ I've tried to do this question by finding out that $\lim_{x\to \infty}(\cfrac{u}{v})$ is undefined.
As the big $O$ notation need the $|\cfrac{u}{v}| < K$, where $K$ is some constant for all sufficiently large $n$. Doest finding out that the limit of it is undefined answer the question?
Thank you :)
That limit is not undefined (in case $x=n$). In fact as $n\to+\infty$, $$\frac{u(n)}{v(n)}=\frac{n^2+n}{2n^2+n}\cdot \frac{n}{\ln n}=\frac{n+1}{2n+1}\cdot \frac{n}{\ln n}\sim \frac{n}{2\ln n}\to +\infty$$ So your statement is FALSE.