I have been trying to solve the sequence
$T(n)= 4T(\frac{n}{2} )+\frac{n^2}{\log n},$
$T(1)=1$
after calculations, I came to this
$4^k Τ(\frac{n}{2^k})+ \sum_{i=0}^{k-1} \frac{n^2}{log\frac{n}{2^i}}$
The only thing I can imagine is the $4^k T(\frac{n}{2^k}) + \sum_{i=0}^{k-1} \frac{n^2}{log\frac{n}{2^i}}$ behaves the same as the harmonic series and therefore can be approximated as $\log(k)$, but I cant find a proof for that.
Because when I set $\log n=k,$ I end up with: $n^2 \sum_{i=0}^{k-1} \frac{1}{k-i}$ for which I can't find a "proof" for the relation between this and the harmonic series, since the harmonic series has a form of $\sum_{i=0}^{k} \frac{1}{i}$, and not $\sum_{i=0}^{k-1} \frac{1}{k-i}$.
Any suggestions on how I can solve the problem will be greatly appreciated! Thanks!
Case 2b of the master theorem immediately gives $T(n)=\Theta(n^2\log\log n)$.