Sufficient conditions for $\lim_{N \to \infty} \frac{{_2 F_1} (a,b;c+N;z)}{{_2 F_1} (a,b;c+N+1;z)}=1$

Question

As stated in the title, under what conditions on $a,b,c$ and $z$ the following holds?

$$\lim_{N \to \infty} \frac{{_2 F_1} (a,b;c+N;z)}{{_2 F_1} (a,b;c+N+1;z)}=1$$

I assume we are dealing with analytic continuation of the hypergeometric series, so any complex parameters are allowed, provided the analytic continuation exists.

If the answer is complicated, I would appreciate some references. I'm searching for information about asymptotics for the hypergeometric function, but I haven't yet found the answer to this question.

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An additional question is, under what conditions the following infinite product converges?

$${_2 F_1} (a,b;c;z)= \prod_{n=0}^\infty \frac{{_2 F_1} (a,b;c+n;z)}{{_2 F_1} (a,b;c+n+1;z)}$$

I think you understand why am I asking this. I'm curious if we can compute any hypergeometric function is this way, if we find a simple way to compute the ratios.

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Denoting:

$$F_n={_2 F_1} (a,b,c+n,z)$$

We have:

$$z (c+n-a) (c+n-b) F_{n+1}= \\ =(c+n) \left[\left(z\left(2 (c+n)-1-a-b \right)-(c+n-1)\right) F_n+(c+n-1) (1-z) F_{n-1} \right]$$

This should help in answering the question, I think.

Answer 1

I have some idea that might help you. In the book by Abramowitz and Stengun, there is a definition of the Hypergeometric function in terms of Gamma functions (15.1.1): \begin{align*} _2F_1(a,b;c;z)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(c)}\sum_{n=0}^\infty\frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)}\frac{z^n}{n!}. \end{align*}

For the Gamma function, we have the identity $\Gamma(z+1)=z\Gamma(z)$. This should help you in rewriting the denominator in terms of the numerator, and then compute the limit.

Hope this helps

Answer 2

The conditions can be found in DLMF $15.12$. If the listed conditions are met then $$ F(a,b;c;z)=1+\sum_{s=1}^{m-1}\frac{(a)_s(b)_s}{{\left(c\right)_{s}}s!}z^{s}+\mathcal O\left(c^{-m}\right), $$ such that for large $N$ $$ \frac{{_2 F_1} (a,b;c+N;z)}{{_2 F_1} (a,b;c+N+1;z)}\sim\frac{1+\mathcal O\left((c+N)^{-1}\right)}{1+\mathcal O\left((c+N+1)^{-1}\right)}, $$ which converges to $1$ in the limit.