Timeline for Estimating a sum involving binomial coefficients [refined]
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 6, 2014 at 11:10 | comment | added | Dima Pasechnik | Your sum is certainly hypergeometric too, but it is the ordinary, aka Gaussian, one, i.e. $ _2F_1(q-m, q-m; -m; -1)$, (if I got it right), and there is a lot of stuff known about them, e.g. explicit integral representations. Should be routine to compute an asymptotic using the latter. | |
| Feb 27, 2014 at 15:06 | comment | added | The Masked Avenger | In both comments I mean the sum I wrote. I am hoping the factor in front is (up to a small multiplicative adjustment) an estimate for your sum. | |
| Feb 27, 2014 at 10:15 | comment | added | W-t-P | You mean, an upper bound for the original sum, or for your sum (to be multiplied by $(m+1)!/((m-q)!(q!)^2)$? | |
| Feb 27, 2014 at 5:56 | comment | added | The Masked Avenger | Preliminary (and likely error-prone) fiddling gives $\frac{(2m-q)^q(4m)^{m-q}}{(2m+q)^m}$ as an upper bound for the sum. I suspect the lower bound won't be too different. | |
| Feb 27, 2014 at 5:40 | comment | added | The Masked Avenger | Taking another look, it seems many of the terms of the sum are less than 1, so it might be worth exploring. | |
| Feb 27, 2014 at 5:29 | history | answered | The Masked Avenger | CC BY-SA 3.0 |