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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.

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    Small hint: if you skip the floors/ceilings, the sum is precisely $\log_mn!$, so you are interested in the sum of differences $\log_mk-\lfloor\log_mk\rfloor$. – 2017-01-21
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    I would try to split the integers in $[1,n]$ into bins satisfying $m^{\ell} \leq k < m^{\ell+1}$ then if $k$ is in bin-$\ell$ we have $\lfloor\log_m(k)\rfloor = \ell$ and $\lceil\log_m(k)\rceil = \ell+1$. Then to estimate the sums one needs to count how many numbers are in each bin. – 2017-01-21
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    Exactly the same question asked two days ago: http://math.stackexchange.com/questions/2103384/concrete-mathematics-9-47?rq=1 This is just a plain copy of this question. When a question gets closed it means you should try to improve the question, not create a new account and post it again. – 2017-01-21

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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.

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    It's true that there are more terms that are closer to the ceiling, however you also need to weight this to the average *distance* from the ceiling/floor to be able to conclude. For example if we have the numbers $\{0.2,0.6,0.6\}$ then the floor is $\{0,0,0\}$ and the ceiling is $\{1,1,1\}$. The sum of the terms is $1$, the floor sum is $0$ and the ceiling sum is $3$. The floor sum is closer here even though more terms closer to the ceiling. – 2017-01-21