Let $q = \frac{a}{a-1}, a \in (0,1)$. I am trying to figure the asymptotics of $$ { }_3 F_2\left(1,1,1-n ; 2,2 ; q\right)$$
when $n \rightarrow \infty$.
In particular, I am interested in the behavior of \begin{equation} \lim _{n \rightarrow \infty} { }_3 F_2\left(1,1,1-n ; 2,2 ; q\right) \cdot b_{n}, \end{equation} where $\left(b_{i}\right)_{i \in \mathbb{N}}$ is null sequence that convergences at least linearly or faster.
It would seem that the series could be reduced by using $$ f_{n}(x) = {}_{3}F_{2}(1-n, \, 1, \, 1; \, 2 \, 2; \, x) = \sum_{k=0}^{\infty} \binom{n-1}{k} \, \frac{(-x)^{k}}{(k+1)^2}. $$ When $n=0$ this is $$ f_{0}(x) = \frac{\text{Li}_{2}(x)}{x}, $$ where $\text{Li}_{2}(x)$ is the dilogarithm function, and $f_{n}(x)$ is a finite polynomial for $n > 0$. If one considers a differential equation for $f_{n}(x)$ then $$ x^2 \, f_{n}^{''} + 3 \, x \, f_{n}^{'} + f_{n} = (1-x)^{n-1} $$ which has the solution $$ f_{n}(x) = \frac{c_{0}}{x} + \frac{c_{1} \, \ln(x)}{x} + \frac{(1-x)^{n+1}}{n (n+1) \, x} \, {}_{2}F_{1}(1, n+1; \, n+2; \, 1-x). $$
For a limit depending upon $n$ leads to some form of \begin{align} \lim_{n \to \infty} \, b_{n} \, f_{n}(x) &= \lim_{n \to \infty} \, b_{n} \, \sum_{k=0}^{\infty} \binom{n-1}{k} \, \frac{(-x)^{k}}{(k+1)^2} \\ &= \sum_{k=0}^{\infty} \frac{(-x)^{k}}{(k+1)^2} \, \lim_{n \to \infty} \binom{n-1}{k} \, b_{n} \\ &= \sum_{k=0}^{\infty} \phi_{k} \, \frac{(-x)^k}{(k+1)^2}, \end{align} where $\phi_{k}$ represents the evaluated limit if there is any dependency upon $k$.