# what is the summation from i=0 to log(n) [closed]

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, ShaileshFeb 13 '17 at 0:15

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

$$\sum_{i = 0}^{\ln(n)} C = C \sum_{i = 0}^{\ln(n)} 1 = C\cdot (\ln(n)+1) = C\ln(n) + C$$
$$\lim_{N\to \ln(n)} \sum_{i = 0}^{N} C = C\cdot (N+1) = C\ln(n) + C$$
• there are $\log(n) + 1$ indices when starting from zero. So it's $C(\log(n)+1)$. The sum is $\sum_{j=0}^{\log(n)-1} C = C\log(n)$. – user335907 Feb 12 '17 at 18:56