Timeline for Sum of 'the first k' binomial coefficients for fixed $N$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2021 at 23:33 | comment | added | Bjørn Kjos-Hanssen | @Make42 The proof by Joe Dohn doesn't show explicitly the use of $e$ but here it is: $$\sum_{i=0}^k \frac{k^i}{i!} (N/k)^i \leq \sum_{i=0}^k \frac{k^i}{i!} (N/k)^k =(N/k)^k \sum_{i=0}^k \frac{k^i}{i!} \leq (N/k)^k \sum_{i=0}^\infty \frac{k^i}{i!} =(N/k)^k e^k.$$ | |
| Sep 18, 2018 at 9:48 | review | Suggested edits | |||
| Sep 18, 2018 at 11:49 | |||||
| S Aug 14, 2018 at 9:02 | history | suggested | Manolito Pérez | CC BY-SA 4.0 |
I improved the formatting of the mathematical expressions
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| Aug 14, 2018 at 8:35 | review | Suggested edits | |||
| S Aug 14, 2018 at 9:02 | |||||
| Aug 5, 2017 at 23:22 | comment | added | Gerry Myerson | @Make, $e$ is the base of natural logarithms, $2.718281828459045\dots$. | |
| Aug 5, 2017 at 20:16 | comment | added | Make42 | Is $e$ the euler constant? | |
| Apr 13, 2017 at 18:26 | comment | added | Joe Dohn | For completeness, let me state the proof of that: $$\sum_{i=0}^k \binom{N}{i} \leq \sum_{i=0}^k \frac{N^i}{i!} = \sum_{i=0}^k \frac{k^i}{i!} (N/k)^i \leq e^k (N/k)^k $$ | |
| Sep 17, 2015 at 21:50 | comment | added | Thomas Dybdahl Ahle | Actually the entire sum turns out to be upper bounded by $(eN/k)^k$ | |
| Mar 5, 2010 at 20:43 | history | answered | Justin Melvin | CC BY-SA 2.5 |