Rate of convergence of ${}_2F_1(an,-r,n+1,\frac{n+1}{an} )$ to $0$ Consider the following function:
\begin{align}
{}_2F_1(an,-r,n+1,\frac{n+1}{an} )
\end{align}
where $r \ge 2$ is a positive integer and $a \in (0,1)$ and ${}_2F_1$ is the hypergeometric function.
The function  ${}_2F_1(an,-r,n+1,\frac{n+1}{an} )$ goes to zero as $n \to \infty$.  I am interested in finding the rate at which this is happening.
By using Wolram-Alpha I found that
\begin{align}
\lim_{n \to \infty} n \cdot {}_2F_1(an,-2,n+1,\frac{n+1}{an} )&=\frac{1}{a}-1,\\
\lim_{n \to \infty} n^2 \cdot {}_2F_1(an,-3,n+1,\frac{n+1}{an} )&=-4+\frac{-2+6a}{a^2},\\
\lim_{n \to \infty} n^2 \cdot {}_2F_1(an,-4,n+1,\frac{n+1}{an} )&=\frac{3(1-a)^2}{a^2}\\
\end{align}
Basically, from play with Wolram-Alpha the behavior seems to be as follows:
\begin{align}
\begin{array}{cc}
\lim_{n \to \infty} n^{ \frac{r}{2}} \cdot {}_2F_1(an,-r,n+1,\frac{n+1}{an} )=c(r,a),  & \text{ if } r \text{ even }\\
\lim_{n \to \infty} n^{\frac{r+1}{2}} \cdot {}_2F_1(an,-r,n+1,\frac{n+1}{an} )=c(r,a),  & \text{ if } r \text{ odd }\\
\end{array} 
\end{align}
Note that I am not so much interested in the final limit as the rate.

The relation between contiguous hypergeometric functions
\begin{equation}
 (c-a)F\left(a-1,b;c;z\right)+\left(2a-c+(b-a)z\right)F\left(a,b;c;z\right)+a(z
-1)F\left(a+1,b;c;z\right)=0
\end{equation}
can be used to prove the proposed property for $n\to\infty$ by induction
\begin{equation}
{}_2F_1(an,-r,n+1,\frac{n+1}{an} )=
\begin{cases}
   \frac{c(r,a)}{n^{r/2}}+o(n^{-r/2})  & \text{ if } r \text{ even }\\                        \frac{c(r,a)}{n^{(r+1)/2}}+o(n^{-(r+1)/2})  & \text{ if } r \text{ odd }
\end{cases}
\end{equation}
Denoting
\begin{equation}
f_q={}_2F_1(-q,an;n+1;\frac{n+1}{an} )
\end{equation}
we suppose that for a given positive integer $p$
\begin{align}
 f_{2p}&=\frac{c(2p,a)}{n^p}+o(n^{-p})\\
 f_{2p+1}&=\frac{c(2p+1,a)}{n^{p+1}}+o(n^{-p-1})
\end{align}
(verified for $p=1$).
Then, with $a=-2p-1,b=an,c=n+1,z=(n+1)/(an)$, the contiguity relation reads
\begin{equation}
 (n + 2p + 2)f_{2p+2}-\frac{(2p+1)(2an-n-1)}{an}f_{2p+1}+\frac{(2p+1)(an-n-1)}{an}f_{2p}=0
\end{equation}
and thus
\begin{align}
 \frac{1}{2p+1}f_{2p+2}=\left( \frac{2a-1}{an}+o(1/n) \right)\left( \frac{c(2p+1,a)}{n^{p+1}}+o(n^{-p-1}) \right)-\left(\frac{a-1}{an} +o(1/n) \right)\left( \frac{c(2p,a)}{n^p}+o(n^{-p}) \right)
\end{align}
so
\begin{equation}
 f_{2p+2}=(2p+1)\left( \frac1a-1 \right)\frac{c(2p,a)}{n^{p+1}}+o(n^{-p-1})
\end{equation}
which is of the expected form with
\begin{equation}
 c(2p+2,a)=(2p+1)\left( \frac1a-1 \right)c(2p,a)
\end{equation}
The same procedure for $a=-2p-2$ gives
\begin{equation}
 f_{2p+3}=(2p+2)\frac{(1-a)c(2p+1,a)+(2a-1)c(p+2,a)}{an^{p+2}}+o(n^{-p-2})
\end{equation}
which corresponds to the expected form with:
\begin{equation}
 c(2p+3,a)=(2p+2)\frac{(1-a)c(2p+1,a)+(2a-1)c(2p+2,a)}{a}
\end{equation}
By induction, the proposed asymptotic behavior is thus verified.
Remark: a simple expression exists for the even parameters:
\begin{equation}
 c(2p,a)=(2p-1)!!\left( \frac{1-a}{a} \right)^p
\end{equation}
