## Why is $\sum_{i=0}^n\sqrt{i}\log_2^2i \geq \Omega(n\sqrt{n}\log_2n)$?

Where $$\Omega(f)$$ denotes the set of lower bounds, why is $$\sum_{i=0}^n\sqrt{i}\log_2^2i \geq \Omega(n\sqrt{n}\log_2n)$$?

1. How can the function on the left be compared to a whole set? I thought usually a function is an element of the set, i.e. $$g\in\Omega(f)$$ or it is not, i.e. $$g\notin\Omega(f)$$.
2. If it would say $$\sum_{i=0}^n\sqrt{i}\log_2^2i \in \Omega(n\sqrt{n}\log_2n)$$ instead, I still would not understand why it is true. How do you evaluate the limit of the left side?