How to simplify the sum over 1/i? With the recurrence relation:
$$ T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log(n)}$$

The "sum for all levels" in the recurrence tree is:
     $$ \sum_{i=0}^{\log n -1} \frac{n}{\log n - i} =  \sum_{i=1}^{\log n} \frac{n}{i} = n \sum_{i=1}^{\log n} \frac{1}{i}$$

Inside the analysis of the recurrence, $\sum 1/i$ appears and then bounded by $\Theta(\log\log n)$. Why is this?
$\sum_{i=1}^k\frac{1}{i}$ is the $k$th harmonic number, which is $\Theta(\log k)$.