Prove that inserting $n$ sorted values in to an AVL using AVL insertion is $\Theta\left (n \log \left ( n \right ) \right )$
The upper bound is easy. Because a tree with $n$ nodes has height $\Theta(\log n)$, when inserting the sequence of elements each tree has height at most $O(\log n)$, so the total complexity is $O(n\log n)$.

You're right with the lower bound. After inserting $n$ elements the tree has height $\Theta(\log \frac{n}{2})=\Theta(\log n - \log 2)=\Theta(\log n)$. So we can split the sum into two and bound each part by its lowest element:
\begin{align}
\sum_{i=1}^n \Theta(\log i)
&amp;= \sum_{i=1}^{n/2} \Theta(\log i) + \sum_{i=n/2+1}^n \Theta(\log i) \\
&amp;= \frac{n}{2} \Omega(\log 1) + \frac{n}{2} \Omega(\log\frac{n+1}{2})\\
&amp;= n\,\Omega(\log n) \\
&amp;= \Omega(n \log n)
\end{align}