Timeline for Finite sum of reciprocal factorials
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 21, 2019 at 15:02 | vote | accept | slomil | ||
| Mar 20, 2019 at 23:44 | answer | added | Robert Israel | timeline score: 0 | |
| Mar 20, 2019 at 23:33 | comment | added | Daniel Schepler | It looks like a convolution of $\frac{1}{k}$ with $\frac{1}{k!}$ - so its generating function would probably be something like $e^x \cdot (-\ln(1-x))$. | |
| Mar 20, 2019 at 23:24 | comment | added | Sidharth Ghoshal | Maybe try sticking an $x^k$ to it each term. Then it becomes some polynomial. If you differentiate this polynomial it takes on the form. $$ \frac{1}{(n-1)!} + \frac{1}{(n-2)!}x + ...$$ but it’s not clear if this direction is fruitful. | |
| Mar 20, 2019 at 23:23 | comment | added | Arthur | Are you certain that it's $k$ in the denominator and not $k!$? (That's a factorial and a question mark, not an interrobang.) | |
| Mar 20, 2019 at 23:23 | history | edited | Sidharth Ghoshal |
Better tags
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| Mar 20, 2019 at 23:22 | comment | added | PrincessEev | Hm. With the multiplication by $n!$, each summand has the form $$\binom n k \cdot (k-1)!$$ I wonder if that means anything. | |
| Mar 20, 2019 at 23:21 | history | edited | slomil | CC BY-SA 4.0 |
added 3 characters in body
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| Mar 20, 2019 at 23:20 | review | First posts | |||
| Mar 20, 2019 at 23:27 | |||||
| Mar 20, 2019 at 23:15 | history | asked | slomil | CC BY-SA 4.0 |