I am trying to simplify an expression and find a closed form for $$\sum_{m=0}^l \binom{s-m}{s-l} \binom{s-1+m}{s-1}x^m$$
How could I get rid of this summation?
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Sign up to join this communityI am trying to simplify an expression and find a closed form for $$\sum_{m=0}^l \binom{s-m}{s-l} \binom{s-1+m}{s-1}x^m$$
How could I get rid of this summation?
You may argue as GH from MO from your other post.
Therefore, the sum on your LHS equals to
Unfortunately, this has no closed form. How can we be sure? To this end, denote your sum by $$f(\ell):=\sum_{m=0}^\ell\binom{s+m-1}{s-1}\binom{s-m}{s-\ell}x^m.$$ As I explained the WZ-method in the other post, the procedure generates a recurrence. However, this time it is a three-term relation $$(\ell+2)f(\ell+2)+(-sx-\ell x-s+2\ell-x+2)f(\ell+1)+(x-1)(s-\ell)f(\ell)=0$$ which reveals that $f(s)$ can not have a closed form.
If you're not interested in the sum, then formulate this as a contour integral. Let $\gamma$ be a closed path (oriented positive) around $z=0$, and apply Cauchy's Integral Formula: $$f(\ell)=\frac1{2\pi i}\int_{\gamma}\frac{dz} {z^{\ell+1}(1-xz)^s(1-z)^{s+1-\ell}}.$$ On a positive note, we can derive a generating function for the sequence $f(\ell)$: $$\sum_{\ell=0}^{\infty}f(\ell)y^{\ell}=\left(\frac{(1+y)^2}{1+y-xy}\right)^s.$$ To see this, start by interchanging summations to proceed as follows: \begin{align} \sum_{\ell\geq0}f(\ell)y^{\ell} &=\sum_{m\geq0}\binom{s+m-1}mx^m\sum_{\ell=m}^s\binom{s-m}{\ell-m}y^{\ell} \\ &=\sum_{m\geq0}\binom{s+m-1}mx^my^m(1+y)^{s-m} \\ &=(1+y)^s\sum_{m=0}^{\infty}\binom{s+m-1}m\left(\frac{yx}{1+y}\right)^m \\ &=(1+y)^s\left(1-\frac{yx}{1+y}\right)^{-s} \\ &=\left(\frac{(1+y)^2}{1+y-xy}\right)^s. \end{align}
Mathematica says: $$\binom{s}{s-l} \, _2F_1(-l,s;-s;x).$$
(without hypergeometrics for special values of $x,$ like $x=1:$
$$\frac{(-1)^l (l-2 s-1)!}{l! (-2 s-1)!}$$