Timeline for aproximate sum involving binomial coefficients
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 16, 2016 at 9:06 | vote | accept | user3209698 | ||
| Mar 15, 2016 at 20:48 | vote | accept | user3209698 | ||
| Mar 16, 2016 at 9:06 | |||||
| Mar 15, 2016 at 18:21 | comment | added | Max Alekseyev | @user3209698: It's possible (under replacement of $\alpha$ with $\log \alpha$ in the integral). You know better what kind of result you need to get. A particular approach may also depend on actual values of $\alpha,\beta$. | |
| Mar 15, 2016 at 15:07 | comment | added | user3209698 | Do you think it would be a good approach to the nice exact result you got? | |
| Mar 15, 2016 at 14:59 | comment | added | user3209698 | Thank you for the answer, I found it so useful. At the end I think that the series you get before defining the Eularian numbers can be sumed up by converting the sum into an integral:\begin{equation}\sum_{k=1}^\infty k^{j-1}/\alpha^k\to \int_{1}^{\infty}dx\;x^{j-1} e^{-\alpha x}\end{equation} | |
| Mar 15, 2016 at 12:59 | history | edited | Max Alekseyev | CC BY-SA 3.0 |
substitution of s=0
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| Mar 15, 2016 at 4:22 | history | answered | Max Alekseyev | CC BY-SA 3.0 |