How can I solve this comparsion between sums?

Question

$Suppose\;m\;intergal\;and\;m\ge2$

$Which \;of\; these \;sums\; is\; asymptotically\; closer \;to\; the\; value\; log_mn!?$

$ \sum_{k=1}^n\lfloor\;log_m k\;\rfloor$

$Or$

$\sum_{k=1}^n\lceil\;log_m k\;\rceil$

Has anything to do with Stirling's formula?

Answer 1

Assume that $m$ is greater than $n!$, and $n\geq 2$. Thus, $\log_m{n!}<1$, but for $1< k\leq n$ $$\lfloor\log_m k\rfloor=0\\\lceil\log_m k\rceil=1$$ which results in $$\sum_{k=1}^n \lfloor\log_m k\rfloor=0\\\sum_{k=1}^n \lceil\log_m k\rceil=n-1\\$$ Obviously for such cases the floor function provides better approximation, though nothing but zero. However, there may be cases with the opposite result.