Mathematics Asked on December 8, 2021
How to prove the following identity?
$$prod_{nge0}^{ }frac{left(n+aright)left(n+bright)}{left(n+cright)left(n+dright)}tag{$a,b,c,d in mathbb R$}=frac{Gamma(c)Gamma(d) }{ Gamma(a)Gamma(b)}$$
For $a+b=c+d$.
The product can be rewritten as :
$$prod_{nge0}^{ }frac{left(n+aright)left(n+bright)}{left(n+cright)left(n+dright)}=prod_{nge0}^{ }frac{n^{2}+left(a+bright)n+ab}{n^{2}+left(c+dright)n+cd}$$$$=prod_{nge0}^{ }frac{n^{2}+left(a+bright)n+ab}{n^{2}+left(a+bright)n+cd}=prod_{nge0}^{ }left(1+frac{ab-cd}{n^{2}+left(a+bright)n+cd}right)$$
But I think this is useless.
Or we can say:
$$prod_{nge0}^{ }frac{left(n+aright)left(n+bright)}{left(n+cright)left(n+dright)}$$$$=lim_{N to infty} frac{ab}{cd}cdotfrac{left(a+1right)left(b+1right)}{left(c+1right)left(d+1right)}cdot…cdotfrac{left(a+Nright)left(b+Nright)}{left(c+Nright)left(d+Nright)}$$$$=lim_{N to infty}frac{left(a+Nright)!left(b+Nright)!}{left(c+Nright)!left(d+Nright)!}cdotfrac{left(c-1right)!left(d-1right)!}{left(a-1right)!left(b-1right)!}$$$$=frac{ Gamma(c)Gamma(d)}{ Gamma(a)Gamma(b)}lim_{N to infty}frac{left(a+Nright)!left(b+Nright)!}{left(c+Nright)!left(d+Nright)!}$$
How to finish?
You can finish the proof by using the asymptotics $$ mathop {lim }limits_{N to + infty } frac{{(N + a)!}}{{N^{N + a + 1/2} e^{ - N} sqrt {2pi } }} = 1, $$ which can be derived from Stirling's formula ($a$ is an arbitrary fixed complex number).
Answered by Gary on December 8, 2021
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