
Re: Sum of squares of binomial coefficients
Posted:
Oct 12, 2012 9:25 AM


In message <111020121555505579%edgar@math.ohiostate.edu.invalid> Gavin Wraith <gavin@wra1th.plus.com> wrote:
> 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}.
Erratum: the penultimate line should read
"which evaluates to 4^m\binom{2m}{m}, unless I am mistaken."
This is the number of ways of choosing from a 2melement set an melement subset and, independently a subset, call it A, of any size. If we suppose that A has r+s elements, r of which are in the melement set, you can see why. In any case, the expression given by Jerome has to exceed this number but be less than its square.
 Gavin Wraith (gavin@wra1th.plus.com) Home page: http://www.wra1th.plus.com/

