Timeline for Simple approximation to a sum involving Stirling numbers?
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2013 at 8:07 | comment | added | user1748601 | Thank you. Your answer is correct. To finish it off, define $n=xy$. Taking $\lim_{n\rightarrow\infty}{(\frac{y-1}{y})^x}=e^{-\frac{x}{y}}$. | |
| Sep 8, 2013 at 8:01 | vote | accept | user1748601 | ||
| Sep 8, 2013 at 3:12 | history | answered | Lucia | CC BY-SA 3.0 |