# How do I simplify $O\left({n^2}/{\log{\frac{n(n+1)}{2}}}\right)$

I'm not very certain about how to deal with asymptotics when they are in the denominator. For $$O\left(\frac{n^2}{\log{\frac{n(n+1)}{2}}}\right)$$, my intuition tells me that it should be treated in a similar way as little o. Would that be correct?

Also, is this in any way comparable with $$O(n)$$ or $$O(n^2)$$?

$$E = \frac{cn^2}{\log \frac{n(n+1)}{2}}$$

where $$c$$ is some constant. The simple upper bound for $$E$$ is

$$E\le c n^2$$

which implies that $$\mathcal{O}(n^2)$$. For a better bound

$$E = \frac{cn^2}{\log \frac{n(n+1)}{2}} = \frac{cn^2}{ 2 \log n + \log n - \log 2 }$$

Now it is an easy verification that $$E$$ is $$\mathcal{O}(\frac{n^2}{\log n})$$. In other words $$E$$ is $$o(n^2)$$(weaker statement as compare to the previous one).

• You can't express $\log(n^2 + n)$ as $\log n^2 + \log n$ that would be equal to $3\log n = \log n^3$. But surely you can ignore the smaller summand. That is, the final result is correct. – Albjenow Jan 9 at 13:09