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Re: Sum of squares of binomial coefficients
Posted:
Oct 11, 2012 5:55 PM


In message <111020121212107442%edgar@math.ohiostate.edu.invalid> Jérôme Collet <Jerome.Collet@laposte.net> wrote:
> I need to compute the sum : > \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } > I know, because I used Stirling formula, Taylorpolynomials, and > ignored some problems on the borders, that this sum should be close to > \sqrt{2\pi m}. > The convergence is very fast, error is less than .5% if m>7. > Nevertheless, I do not know how to prove it correctly.
This statement worries me. The expression you give \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } is certainly larger than \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr}) } which evaluates to \binom{2m}{m}, unless I am mistaken. And that grows much faster than \sqrt{2\pi m}.
 Gavin Wraith (gavin@wra1th.plus.com) Home page: http://www.wra1th.plus.com/



