Asymptotic expansion of hypergeometric function ${}_3F_2$ for large parameters I encountered the following hypergeometric function in my research: $${}_3F_2(2,1+n,1+n;1,2+n;z)$$ where $0<z<1$.  I'm interested in its behavior for large $n$.  Semilog plot suggests exponential increase in $n$, however, I'm having trouble deriving the expression for the asymptotic expansion.  There are lots of references to 1969 volume by Yudell Luke.  I scanned it to no avail for a result that fits the above formula.  I am a mere computer scientist, and am unfamiliar with literature on hypergeometric functions (which seems quite extensive).  I would appreciate any help.
 A: Below is a closed-form evaluation in terms of simple functions.  Let $y=z/(z-1).$  Then
$${}_3F_2(2,n,n;1,n+1;z)=n(-z)^{-n}\Big((n-1)\big(\log(1-y)+\sum_{k=1}^{n-1}\frac{y^k}{k} \big) + y^n \Big)$$
I proved this by using Pochhammer symbol properties, which gets me to a linear combination of ${}_2F_1.$  Then I used a linear transformation to get from the argument $z$ to $y.$  That series can be manipulated to give the logarithm and a finite sum.  Variable $y$ is always negative, but if it is small, the sum will converge rapidly. If you put this on a computer, watch out for large negative $y,$ which occurs for $z$ close to 1.  In the finite sum, I'd probably add terms pairwise. If the $z\sim 1$ case is your most important case, then it's probably worth thinking about this some more.
Added: The sum for $z\sim 1$ can be done by an expansion found in 'Asymptotic Expansions Pertaining to the Logarithmic Series and Related Trigonometric Sums,' G. Fikioris & P. Andrianesis, J. Class. Analysis vol 7 #2, (2015) 113-127.
$$ \sum_{k=1}^{n-1}\frac{y^k}{k} \sim y^n\sum_{k=0}^\infty \frac{A_k(y)}{(y-1)^{k+1}} \frac{1}{n^{k+1}} \, ,n \to \infty $$
where the $A_k(y)$ are the Eulerian polynomials and begin with $A_0(y)=1$ and $A_1(y) = y.$
