Question about Growth Rates

Question

I see in some notes from my instructor in Algorithm course that $\Sigma_{i=0}^{log n} (n/2^i)$ has growth bigger than $\Sigma_{i=1}^{n} (i log i)$. i couldn't understand why? any tutorial or hint?

Answer 1

$\Sigma_{i=0}^{log n} (n/2^i) = n*\Sigma_{i=0}^{log n} (1/2^i)$ is equivalent to $2n$ when $n$ is huge, so this is $O(n)$.

For n big enough, $\Sigma_{i=1}^{n} (i log i) > \Sigma_{i=1}^{n} i = \frac{n(n+1)}{2}$ which is $O(n^2)$.

So unless I'm missing something, I think the statement is wrong.