I know there is a lot of questions and resources around for limits of hypergeometric functions, but I was not able to find something useful for my problem.
I have the following function:
$$ f_n(z) = {}_2F_1(1,-n;1+n;z) $$
where $z$ is a real number in the interval $[-1,1]$, $n$ is an integer, and I want to compute the limit $f(z) = \lim _{n\to\infty}f_n(z)$. The limit seems to be well defined numerically, but I'm really struggling to find the asymptotic behaviour. I read this paper which seems to discuss similar stuff to what I have (in particular, I think case B is exactly my limit) but I don't grasp how to to apply it to my case.
If the limit cannot be explicitly evaluated, I would be happy with some series expansion for high $n$, specially at the values $z=0$ and $z=\pm 1$. From Gauss hypergeometric theorem I can evaluate $f(1)=1/2$ -but that's all I got.
I also took a look to book tables and other answers, trying several integral representations andthen taking the limit, but without success. Any help is welcomed, thank you so much!
UPDATE: using DLMF 15.2.7, as suggested, I get
$$f_n(-z) \sim \sqrt2 \left( \frac{1+z}{2\sqrt z} \right) ^n \left[ \sqrt n U\left (\frac{1}{2}, -\alpha \sqrt n \right) - \frac{\sqrt 2}{z-1} U\left(-\frac{1}{2}, -\alpha \sqrt n \right)\right]$$
being $U(k,x)$ the parabolic cylinder function and $\alpha$ given by 15.2.8 as $\alpha = \sqrt { \left(-2\log (1 - (z-1)^2/(z+1)^2) \right)}$. I tried to use the expansion DLMF 12.9.1 for the PC functions at first order,
$$U(k,x) \sim e^{-x^2/4} x^{-k-1/2}$$.
Then the expression above is simplified to
$$f_n(-z) \sim \sqrt2 \left( \frac{1+z}{2\sqrt z} \right) ^n e^{-\alpha^2 n /4} \left( \frac{\sqrt 2}{1-z} - \frac{1}{\alpha} \right) $$.
Now the only terms that depends on $n$ is the first one. However, this term goes just to $0$ or to $+\infty$ depending on the value of the fraction. Moreover, a clear divergence can be seen at $z\to 0$, in contradiction with my previous finding (and not consistent with my numerical experiments). I read that the approximation 15.12.7 just works for phase of $z$ less than $\pi$, making my approximation not useful for real $z$. Is this true? What can I do to circumvent this problem?
$$f_n(z) = {}_2F_1(1,-n;1+n;z)$$
Around $z=0$, we have $$f_n(z)=\sum_{k=0}^\infty \frac{n! \,(k-n-1)!}{(-n-1)! \,(k+n)!} z^k$$
Around $z=-1$, the first term of the expansion is $$f_n(z)=\frac{1}{2} \left(1+\sqrt{\pi }\,\frac{ \Gamma (n+1)}{\Gamma \left(n+\frac{1}{2}\right)}\right)-\frac{n}{2} (z+1)+O\left((z+1)^2\right)$$