Given $T(n) = \left( \sum_{i=1}^{\log n} T(\lfloor \frac {n}{k^i} \rfloor) \right) + n $ where $k \ge 3$, how do I prove that $T(n) = \theta(n) $?
I tried replacing $n$ with $2^m$ when $m = \log n$ bug it doesn't seem to help.
In general, how do I calculate the handle these kind of questions? I saw this: How to tackle a recurrence that contains the sum of all previous elements? but I'm struggling to implement it in the mentioned above case...
