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Re: Sum of squares of binomial coefficients
Posted:
Oct 12, 2012 9:27 AM
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Le 11/10/2012 23:55, Gavin Wraith a écrit :
> In message <111020121212107442%edgar@math.ohio-state.edu.invalid>
> Jérôme Collet <Jerome.Collet@laposte.net> wrote:
> >
> > I need to compute the sum : \sum_{r,s}{ (\binom{r+s}{r}
> > \binom{2m-r-s}{m-r})^2 } I know, because I used Stirling formula,
> > Taylor-polynomials, 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{2m-r-s}{m-r})^2 } is certainly larger than
> \sum_{r,s}{ (\binom{r+s}{r} \binom{2m-r-s}{m-r}) } 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{2m-r-s}{m-r})^2 }
seems to be equivalent to
\sqrt{2\pi m} \binom{2m}{m}^2
How can I prove it ?
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