I'm having trouble understanding why this identity holds:
$$\sum_{k=0}^{(\log n) - 1} \frac{n}{\log (n - k)} + \theta(1) = \sum_{k=1}^{\log n} \frac{n}{k}+ \theta(1) $$
Any pointers to a proof would be very appreciated.
I'm having trouble understanding why this identity holds:
$$\sum_{k=0}^{(\log n) - 1} \frac{n}{\log (n - k)} + \theta(1) = \sum_{k=1}^{\log n} \frac{n}{k}+ \theta(1) $$
Any pointers to a proof would be very appreciated.