I need to know how to get the summation of a constant (c) from i=0 to log(n) of a constant


closed as off-topic by TheGeekGreek, Namaste, kingW3, Daniel W. Farlow, Shailesh Feb 13 '17 at 0:15

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  • 4
    $\begingroup$ $\;\log n-1\;$ won't usually be a natural number, so what do you exactly mean? $\endgroup$ – DonAntonio Feb 12 '17 at 18:50
  • $\begingroup$ Maybe there's a ceiling or flooring operation on $\log(n) - 1$, for $n>0$? $\endgroup$ – rookie Feb 12 '17 at 18:52
  • $\begingroup$ Your question is not enough clear ! $\endgroup$ – Khosrotash Feb 12 '17 at 18:52
  • $\begingroup$ Welcome to MSE! It's important that posts are clearly written in TeX and explained so users can help as much as possible! $\endgroup$ – Test123 Feb 12 '17 at 18:53
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    $\begingroup$ @Alexandralopez That doesn't matter: the number's still not a natural one. $\endgroup$ – DonAntonio Feb 12 '17 at 18:56

$$\sum_{i = 0}^{\ln(n)} C = C \sum_{i = 0}^{\ln(n)} 1 = C\cdot (\ln(n)+1) = C\ln(n) + C $$

"More rigorously"

$$\lim_{N\to \ln(n)} \sum_{i = 0}^{N} C = C\cdot (N+1) = C\ln(n) + C$$

  • 1
    $\begingroup$ there are $\log(n) + 1$ indices when starting from zero. So it's $C(\log(n)+1)$. The sum is $\sum_{j=0}^{\log(n)-1} C = C\log(n)$. $\endgroup$ – user335907 Feb 12 '17 at 18:56
  • $\begingroup$ @james.nixon Whoops, thank you! $\endgroup$ – Von Neumann Feb 12 '17 at 18:56

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