Timeline for Limit of $\frac{1}{n}\sum_{r=1}^n\frac{n\ (\mathrm{mod}\ r)}{r}$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| S Jul 25, 2017 at 16:45 | history | suggested | jeq | CC BY-SA 3.0 |
Replaced double-backslash-brace pair with single-backslash-brace pair, to fix rendering.
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| Jul 25, 2017 at 16:42 | review | Suggested edits | |||
| S Jul 25, 2017 at 16:45 | |||||
| Apr 18, 2012 at 16:11 | comment | added | Emil Jeřábek | @Anonymous: Yes, I guess I should have mentioned in the answer that this estimate of $\sum_{r\le n}\lfloor n/r\rfloor$ is a classical fact. | |
| Apr 17, 2012 at 17:59 | vote | accept | unknown | ||
| Apr 17, 2012 at 16:32 | comment | added | Anonymous | To add to Emil's nice answer: The determination of the OP's limit is implicit in Dirichlet's estimate for the average order of the divisor function, namely $\sum_{k=1}^{n} d(k) = n \log{n} + (2\gamma-1)n + O(\sqrt{n})$. (See section 18.2 in Hardy and Wright, for example.) To see the connection, notice that the sum in question is exactly the number of ordered pairs of natural numbers $(r,s)$ with $rs \leq n$ (and now compare with Emil's answer above). | |
| Apr 17, 2012 at 15:45 | comment | added | Emil Jeřábek | You are right, hopefully it is more clear now. | |
| Apr 17, 2012 at 15:43 | history | edited | Emil Jeřábek | CC BY-SA 3.0 |
improve formatting
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| Apr 17, 2012 at 15:32 | comment | added | Gerhard Paseman | Fir clarity, I would change the indexing rs <= n to set notation that said something like the set of pairs of positive integers (r,s) with r,s <= n. For me that would remove the ambiguity of whether r or s or both are the index, as well as whether negative r and negative s were allowed. Gerhard "Nice Derivation, By The Way" Paseman, 2012.04.17 | |
| Apr 17, 2012 at 15:17 | history | answered | Emil Jeřábek | CC BY-SA 3.0 |