
Re: Sum of squares of binomial coefficients
Posted:
Oct 12, 2012 9:27 AM


Le 11/10/2012 23:55, Gavin Wraith a écrit :
> 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}.
I made a stupid mistake, I forgot a normalizing term, which is \binom{2m}{m}^2 Thank for your attention.
So my question is \sum_{r,s}{ (\binom{r+s}{r} \binom{2mrs}{mr})^2 } seems to be equivalent to \sqrt{2\pi m} \binom{2m}{m}^2 How can I prove it ?

