Find the asymptotic expansion of
$$ {}_2F_1\left(2v,v+n,2v+n,1-\tau/n \right), \quad n \to \infty $$
where $v,\tau > 0$.
My first attempt was to consider a fix value $z$ and then look at the limit $z\to 1^{-}$. Using the integral representation here, one has
\begin{align*} {}_2F_1\left(2v,v+n,2v+n,z \right) &= \frac{\Gamma(2v+n)}{\Gamma(v+n)\Gamma(v)} \int_0^1 t^{v+n}(1-t)^{v-1}(1-zt)^{-2v} dt \\ & = \frac{\Gamma(2v+n)}{\Gamma(v+n)\Gamma(v)} \int_0^{\infty} e^{-nu}q(u)du \end{align*}
with the variable substitution $u=-\ln t$ and $q(u) = e^{-(v+1)u}(1-e^{-u})^{v-1}(1-ze^{-u})^{-2v}$. Then, by Watson's lemma, one may evaluate and e.g. the two first terms of the integral becomes
$$ \int_0^1 t^{v+n}(1-t)^{v-1}(1-zt)^{-2v} dt \sim \frac{\Gamma(v)(1-z)^{-2v}}{n^v} - (1-z)^{-2v-1}(z(v-1)+3v+1)\frac{\Gamma(v+1)}{2n^{v+1}} $$
Now, the issue is that from here, the expression works fine when $z$ is smaller than $1$ but in the limit $z\to 1^{-}$, the expansion fails to be useful when I compare with numerical results.
Suggestions on how to handle this are much appreciated!
Assume $\operatorname{Re} v > 0$, $|\arg \tau|<\pi$ and $n\gg |\tau|$. By $(15.8.1)$, $$ F(2v,v + n,2v + n,1 - \tau /n) = \left( {\frac{n}{\tau }} \right)^{2v} F(2v,v,2v + n,1 - n/\tau ). $$ Then, by $(15.6.1)$, \begin{align*} \left( {\frac{n}{\tau }} \right)^{2v} F(2v,v,2v + n,1 - n/\tau ) & = \left( {\frac{n}{\tau }} \right)^{2v} \frac{{\Gamma (2v + n)}}{{\Gamma (v)\Gamma (v + n)}}\int_0^1 {\frac{{t^{v - 1} (1 - t)^{v + n - 1} }}{{(1 - (1 - n/\tau )t)^{2v } }}{\rm d}t} \\ & = \left( {\frac{n}{\tau }} \right)^{2v} \frac{{\Gamma (2v + n)}}{{\Gamma (v)\Gamma (v + n)}}\frac{1}{{n^v }}\int_0^n {\frac{{s^{v - 1} (1 - s/n)^{v + n - 1} }}{{(1 - s/n + s/\tau )^{2v } }}{\rm d}s} . \end{align*} By the dominated convergence theorem, $(13.4.4)$ and $(13.2.40)$, \begin{align*} \int_0^n {\frac{{s^{v - 1} (1 - s/n)^{v + n - 1} }}{{(1 - s/n + s/\tau )^{2v } }}{\rm d}s} \to \int_0^{ + \infty } {\frac{{s^{v - 1} {\rm e}^{ - s} }}{{(1 + s/\tau )^{2v } }}{\rm d}s} & =\tau ^v \Gamma (v )\,U(v,1 - v,\tau )\\ &= \tau ^{2v} \Gamma (v )\, U(2v,v + 1,\tau ) \end{align*} as $n\to+\infty$, where $U$ is the Kummer function. By $(5.11.12)$, $$ \frac{{\Gamma (2v + n)}}{{\Gamma (v + n)}} \sim n^v $$ as $n\to+\infty$. Accordingly, $$ F(2v,v + n,2v + n,1 - \tau /n) \sim n^{2v} U(2v, v+1,\tau ) $$ as $n\to+\infty$.