Is there any approximation with an "elementary function" for the following generalized hypergeometric function, especially for very large values of $n$ and $0 < p < 1$? I mean, without using binomials or other special functions.
$_3F_2 \left(1, 1, 1-n; 2, 2; \frac{p}{p-1}\right)$
For simplicity write $z=p/(p-1)$ and assume $p<1/2$ for having $-1<z<0$ and $1-z=\frac{1}{1-p}.$ Then $$F_n(z)=\sum_{k=0}^{\infty}\frac{(1-n)_k}{k!}\frac{z^k}{(k+1)^2}.$$ Then $$[z(zF_n(z))']'=\sum_{k=0}^{\infty}\frac{(1-n)_k}{k!}z^k=(1-z)^{n-1}.$$ This implies that $$z(zF_n(z))'=\frac{1}{n}(1-(1-z)^n)$$$$ zF_n(z)=\frac{1}{n}\int_0^z\frac{1-(1-t)^n}{t}dt=\frac{1}{n}\int_{1-z}^1\frac{1-u^n}{1-u}du=\frac{1}{n}\int_{1-z}^1\sum_{k=0}^{n-1}u^kdu$$ Therefore $$F_n(z)=\frac{1}{nz}\sum_{k=1}^{n}\frac{1}{k}(1-(1-z)^k)=\frac{1-p}{np}\sum_{k=1}^{n}\frac{1}{k}\left(\frac{1}{(1-p)^k}-1\right)$$
Now the approximation, using the following lemma: if $R>1$ then $$\sum_{k=1}^n\frac{R^k}{k}\sim_{n\to \infty}\frac{R^n}{n}\times \frac{R}{R-1}.$$Proof: $$\sum_{k=1}^n\frac{R^k}{k}=\frac{R^n}{n}\times \sum_{k=0}^{n-1}\frac{n}{n-k}\frac{1}{R^k}\leq \frac{R^n}{n}\times \sum_{k=0}^{n-1}(k+1)\frac{1}{R^k}$$ Hence by dominated convergence $$\sum_{k=0}^{n-1}\frac{n}{n-k}\frac{1}{R^k}\to_{n\to \infty}\frac{R}{R-1}.$$ We conclude by applying the lemma to $R=1/(1-p).$ We can ignore the term $\frac{1}{n}\sum_1^n\frac{1}{k}\sim \frac{\log n}{n}\to 0$ and write$$F_n(\frac{p}{p-1})\sim \frac{1-p}{p^2n^2}\times \frac{1}{(1-p)^n}.$$
Finally, to expell the hypothesis $p<1/2$, it has been done in order to garantee the convergence of the series $F_n(z)$. But the calculation has shown that $F_n(z)$ is actually a polynomial and all above holds for all $z$, in particular for all $0<p<1.$