A follow-up question to this one basically Understanding a simplification in a theorem. If you want to see the original paper, see page 6/24 there.
We are given that $2 < n \leq M \leq n^2$. Then, the following inequality holds
$\log ( 1 + \dfrac{M}{n-2}) \geq \log n + O(1)$.
The paper goes on saying that now $\Omega (n^2)$ lower bound is obtained. How is this so? According to my understanding to actually show that, I would have to find a positive constant $c_1$, so that
$\log n + O(1) \geq c_1n^2$.
If I then divide this by $n^2$, I'll obtain
$\dfrac{\log n}{n^2} + \dfrac{O(1)}{n^2} \geq c_1$.
But as $n \to \infty$, the terms on the left side tend to 0. Therefore it is not bounded by $\Omega (n^2)$. I know I must be misunderstanding or neglecting something, what is it?