Limit of hypergeometric function in proof of Gauss's summation formula

72 Views Asked by At

I've been studying special functions from Askey,Roy and in proof of Gauss's summation formula (theorem 2.2.2) it states that $\lim_{n\to \infty}\frac{\Gamma (c+n-a)\Gamma(c+n-b)}{\Gamma (c+n) \Gamma(c+n-a-b)}\ _2F_1(a,b;c+n;1)=1$. Only thing we know about $a,b,c$ is that $Re(c-a-b)>0$

I dont understand how is that correct. I have tryed using $lim_{n\to\infty}\frac{\Gamma (n+a)}{\Gamma(n)n^a}=1$ but I dont know what to do with hypergeometric function.