Asymptotically closer to $\log_mn!$

Question

Let $m∈ \Bbb Z $ and $m \ge 2 $. Analyze two sums $$\sum_{k=1}^n \lfloor \log_mk \rfloor $$ and $$\sum_{k=1}^n \lceil \log_mk \rceil $$ Which one is asymptotically closer to $\log_mn!$ I know that we can start solving this by writing $\log_mn = l + θ , 0 \le θ \lt 1$. Now I don't know how to move on calculating the sums as $l,m,θ$. It's also the same exercise as 9.47 from Concrete Mathematics.

Answer 1

Let $t=\log_{m}{k}$ and $\{t\}$ denote fractional part of t. $t$ is closer to $\lfloor{t}\rfloor$ if $$\{t\}<0.5 \implies m^n\le k < m^{n+0.5}$$ and $\lceil{t}\rceil$ otherwise : $$\{t\}>0.5 \implies m^{n+0.5} < k < m^{n+1}$$ For each $n$, number of terms in $\sum\log_m k$ with between $n$ and $n+0.5$ = $m^{n+0.5}-m^{n}$, and between $n+0.5$ and $n+1$ = $m^{n+1}-m^{n+0.5} > m^{n+0.5}-m^{n}$. Hence more terms are closer to their ceiling then floor. Subsequently, sum is closer to sum of ceilings. This is true for any convexcurve.