Interesting growing behaviour of hypergeometric function Doing some computations and plottings I've found out that the function
$$_2F_1(2s-1,s-\tfrac{1}{2};s;-1)$$
behaves for large real $s$ like $4^{-s}$. More precisely: It seems that
$$_2F_1(2s-1,s-\tfrac{1}{2};s;-1)4^s$$
grows, but very very slowly. No exponential growth or decay at all. If you modify the $4$ only slightly this of course changes and you get exponential groth or decay. So my question is to explain this phenomenon. Is
$$\lim_{s\to \infty}\ _2F_1(2s-1,s-\tfrac{1}{2};s;-1)4^s = \infty?$$
Is there an easy to detemine for every $x>0$ the respective $b(x)>0$ such that you have
$$_2F_1(2s-1,s-\tfrac{1}{2};s;-x) \sim b^{-s}?$$
What is
$$\lim_{s\to \infty}\  _2F_1(2s-1,s-\tfrac{1}{2};s;-x) b^{s} $$
then?
 A: $$f(s)=\, _2F_1\left(2s-1,s-\frac{1}{2};s;-1\right)=\, _2F_1\left(s-\frac{1}{2},2 s-1;s;-1\right)$$
Even for small values of $s$, we have a nice logarithmic behaviour
$$\log[ f(s)] \sim a -b \,s$$ Using the data computed for $1 \leq n \leq 100$, we have, with $R^2=0.9999982$,
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\
 a & 1.640373 & 0.021344 & \{1.598011,1.682736\} \\
 b & 1.379308 & 0.000367 & \{1.380037,1.378580\} \\
\end{array}$$ and, as you noticed, $b$ is quite close to $\log(4)=1.38629$.
Pushing the numerical analysis much further, $b$ is closer and closer to $(\log(4)-\epsilon)$. This is normal since
$$\, _2F_1\left(2s,s;s;-1\right)=4^{-s}$$
What is interesting is that, if $\epsilon=0$, $\log[ f(s)]$ is an increasing function going to infinity while, if $\epsilon=10^{-4}$,  $\log[ f(s)]$   goes through a maximum value.
What is interesting is that
$$4^s \, _2F_1(s-1,2 s-1;s;-1)=2+2 \sqrt{\pi }\frac{ \Gamma (s)}{\Gamma \left(s-\frac{1}{2}\right)}$$
