$$\sum_{k=1}^{n}\left \lfloor \log _{m}k \right \rfloor$$ $$\sum_{k=1}^{n}\left \lceil \log_{m}k\right \rceil$$ I found myself stuck trying to solve these two summations but i can't make any progress. Any ideas?
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If we try $m = 10$ and then calculating the first and second one is easy. The first one $\sum_{k=1}^{100} (\lfloor \log_{10}(k) \rfloor) = 90$. I think to generalize it we have to work with intervals. If $m$ is an integer, we should look at powers of $m$. – user19405892 Jan 28 '16 at 21:02
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I forgot to mention that m is indeed an integer and also $$m\geq 2$$...My mistake – George Giannopoulos Jan 28 '16 at 21:07
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The key to the first problem is to observe that $\left\lfloor\log_mk\right\rfloor=\ell$ if and only if $\ell\le\log_mk<\ell+1$, which is true if and only if $m^\ell\le k<m^{\ell+1}$. There are $m^{\ell+1}-m^\ell=m^\ell(m-1)$ integers $k$ in this range, and each of them contributes $\ell$ to the sum, so if $m^{\ell+1}\le n$, this interval contributes $\ell(m-1)m^\ell$ to the total.
Let $L=\lfloor\log_mn\rfloor-1$; for $\ell=1,\ldots,L$ we have $m^{\ell+1}\le n$, so these intervals contribute a total of
$$(m-1)\sum_{\ell=1}^L\ell m^\ell\tag{1}$$
to the sum. This answer shows one way to derive a closed form for $(1)$. It then only remains to add the contribution of
$$\sum_{k=m^{L+1}}^n\lfloor\log_mk\rfloor=\left(n-m^{L+1}+1\right)\lfloor\log_mn\rfloor\;.$$
The second problem can be done similarly.
Brian M. Scott
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so i have to add the result of (1) to $$\left(n-m^{L+1}+1\right)\lfloor\log_mn\rfloor;.$$? – George Giannopoulos Jan 28 '16 at 22:09
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@Παναγιώτης: Yes, that’s correct. The idea is to divide the sum up into blocks on which the floor is constant; $(1)$ covers the full blocks, and the other term covers the last, possibly incomplete block. – Brian M. Scott Jan 28 '16 at 22:12
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