Asymptotic expansion involving hypergeometric function ${}_2F_1$ For $n\in\mathbb Z$, define $p_n$ as
$$p_n =  e^{-2\pi \eta|n|} \begin{cases}
   \binom{-\alpha}{n} \: _2{F}{_1}(-\alpha, n+\alpha; n+1; e^{-4\pi \eta}), & n\geq 0, \\
   \binom{\alpha}{-n} \: _2{F}{_1}(-n-\alpha, \alpha; 1-n; e^{-4\pi \eta}), & n\leq 0.
  \end{cases},$$
where $\eta>0$ and $\alpha\in\mathbb R$ are constants, $_2F_1$ is the hypergeometric function, and $\binom{\cdot}{\cdot}$ is the (generalized) binomial coefficients.

Question: What is the asymptotic behavior of $|p_n|^2$, for each
$n\to\infty$ and $n\to-\infty$?

Remarks:

*

*I tried some numerical experiments. For $\eta=0.02$ and $\alpha=1.5$, the log-plot of $|p_n|^2$ is,


I think that $|p_n|^2$ decays as $\sim n^a e^{-b n}$ for some constants $a,b>0$, different for $n\to\infty$ and $n\to-\infty$.


*I know that $\sum_{n\in\mathbb Z} |p_n|^2=1$, because it comes from a certain probability distribution.

 A: For brevity, I've denoted ${}_2F_1$ simply as $F$.
It is simple to see that
$$F(a,b;c;z) \sim 1 +\frac{ab}{c}z$$
as $|c|\rightarrow \infty$. Then, using a relationship of the hypergeometric function gives for the first relation
$$F(-\alpha,n+\alpha;n+1;v) = (1-v)^{\alpha}F\left(-\alpha,1-\alpha;n+1;\frac{v}{v-1}\right) \sim (1-v)^{\alpha}\left(1+\frac{\alpha(\alpha-1)}{n+1}\frac{v}{v-1} \right)$$
where $v = \exp(-4\pi\eta)$. For the second, I changed $-n\rightarrow n$, then the function of interest is
$$F(n-\alpha;\alpha;1+n,v) = (1-v)^{-\alpha}  F\left(1+\alpha,\alpha;1-n;\frac{v}{v-1}\right) \sim (1-v)^{-\alpha} \left(1+\frac{\alpha(1+\alpha)}{1+n}\frac{v}{v-1} \right)$$
Note that normally, the convergence depends on if $\left|\left( \frac{v}{v-1}\right)\right| <1$ or not. In this case, this is true for all values of $v$ except when $v=1$, occuring at $\eta=0$. Close to this point, convergence is slower.
An example of the first approximation is given in the figure below.
For the binomial coefficients, you can write these in terms of Gamma functions, using the asymptotic expression for these and then piecing things together should give you the asymptotic expression for the whole.
For the first part, note that $$\binom{-\alpha}{n} = (-1)^n\binom{n+a-1}{n}=(-1)^n \frac{\Gamma(\alpha+n)}{\Gamma(n+1)\Gamma(\alpha)}$$
then,
$$ p_n^2 \sim e^{-4\pi\eta n} \left(\frac{\Gamma(\alpha+n)}{\Gamma(n+1)\Gamma(\alpha)} \right)^2 \left(1-e^{-4\pi\eta}\right)^{2\alpha} \sim e^{-4\pi\eta n} \left(\frac{(n+\alpha)^{n+\alpha-1/2}e^{1-\alpha}}{(n+1)^{n+1/2}\Gamma(\alpha)} \right)^2 \left(1-e^{-4\pi\eta}\right)^{2\alpha}$$
where I've only kept the first term of the hypergeometric function and used that $\Gamma(z) \sim \sqrt{2\pi}z^{z-1/2}e^{-z}$.
Small comment: the legend in the figure is wrong wrt the parameters in F, should be the first hypergeometric function without modifications.
