Prove ∑pq=α(qα)(pq)(−1)q(−q)pqα=p!α!.\sum_{q=\alpha}^p \binom{q}{\alpha} \binom{p}{q}\frac{(-1)^q(-q)^p}{q^\alpha}=\frac{p!}{\alpha!}.

How to prove

pq=α(qα)(pq)(1)q(q)pqα=p!α!

for 1αp?

EDIT: This is a result that I derived after playing around with the (given) fact that

pq=1qj=1qp2(1+h/q)j1pi=1,iq1qi=pq=1[(1+h/q)q1](1)pqqpq!(pq)!=pq=1hqq!

and grouping the coefficients of each hα in both sides.

I have tried some values of α and p and it works, so the formula seems to be true. I just don’t know how to proceed to prove this result.

Answer

Suppose we seek to verify that

(1)ppq=r(pq)(qr)(1)qqpr=p!r!.

We use the integral representation

(qr)=(qqr)=12πi|z|=ϵ(1+z)qzqr+1dz

which is zero when q<r (pole vanishes) so we may extend q
back to zero.

We also use the integral

qpr=(pr)!2πi|w|=γexp(qw)wpr+1dw.

We thus obtain for the sum

(1)p(pr)!2πi|w|=γ1wpr+1×12πi|z|=ϵzr1pq=0(pq)(1)q(1+z)qzqexp(qw)dzdw=(1)p(pr)!2πi|w|=γ1wpr+1×12πi|z|=ϵzr1(11+zzexp(w))pdzdw=(1)p(pr)!2πi|w|=γ1wpr+1×12πi|z|=ϵ1zpr+1(exp(w)+z(1exp(w)))pdzdw=(pr)!2πi|w|=γ1wpr+1×12πi|z|=ϵ1zpr+1(exp(w)+z(exp(w)1))pdzdw.

We extract the residue on the inner integral to obtain

\frac{(p- r)!}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}}
{p\choose p- r} \exp( r w)
(\exp(w)-1)^{p- r}
\; dw
\\ = \frac{p!}{ r!}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{p- r+1}}
\exp( r w)
(\exp(w)-1)^{p- r}
\; dw.

It remains to compute
[w^{p- r}]\exp( r w)
(\exp(w)-1)^{p- r}.

Observe that \exp(w)-1 starts at w so (\exp(w)-1)^{p- r}
starts at w^{p- r} and hence only the constant coefficient from
\exp( r w) contributes, the value being one, which finally
yields

\frac{p!}{ r!}.

Attribution
Source : Link , Question Author : Jeze Ken , Answer Author : Marko Riedel

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