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

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

$ \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?

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    I looked Stirling's formula but cannot understand the connection,if there is one. – 2017-01-06
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    Have you tried any numerical experiments? The easiest could be using $m=10$? – 2017-01-06
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    Perhaps I'm missing something, but it would seem to me that the ceiling version would be much *greater* than the log in question, since it would *include* that log (rounded up), *plus* the log of all smaller numbers! So I would expect the floor version to be closer. – 2017-01-06
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    Why have you enclosed all your text in dollar signs? – 2017-01-06

1 Answers 1

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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.

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    I don't see where this answers the "asymptotically" part. As $n$ approaches infinity, of course $m \not\gt n!$. – 2017-01-06
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    You're right. I totally skipped the "asymptotically" part. – 2017-01-07