I conjecture that the following statement holds for large values of $N$
$$
{}\_3F\_1\left(-N+1,1,1;2;-\frac{1}{N}\right)\to\frac{1}{2}\bigg({}\_2F\_1(1,1;2;1-\frac{1}{N})+\log 2+\gamma\bigg)
$$
where $\gamma$ is the Euler-Mascheroni constant. PLugging in big values for $N$ in the above formula it looks like the conjecture holds, but I am unable to prove it. Anybody can do better?



Can you get the following asymptotic expression for an hypergeometric function?
As the OP remarked in a comment, the ultimate purpose of the question is to derive the asymptotic large-$N$ expansion of the integral


$$I\_N=\int\_0^1 dq\, \frac{q^2}{\log q}\left[\left(1-\frac{3}{N}\log q\right)^N-1\right]$$


According to equation (69) of this [reference](http://arxiv.org/abs/hep-ph/0505034), $I\_N$ should converge to
$$I\_N=\tfrac{1}{2}\left(\log N + \log 2 +\gamma\_E\right)+{\cal O}(N^{-1/2})\qquad\qquad(\*)$$ 


Noting the expansion
$$(1+a/N)^N=\exp\left[a-\tfrac{1}{2}a^2/N+{\cal O}(N^{-2})\right]$$ 
I first transform from $q$ to $x=\log q$ and then take the large-$N$ limit as follows:


$$I\_N=\int\_{-\infty}^0 dx\, \frac{e^{3x}}{x}\left[\left(1-\frac{3x}{N}\right)^N-1\right]=\int\_{-\infty}^0 dx\, \frac{e^{3x}}{x}\left[\exp\left(-3x-\tfrac{9}{2}x^2/N+{\cal O}(N^{-2})\right)-1\right]$$
$$\quad\rightarrow \int\_{-\infty}^0 dx\, \frac{1}{x}\left[\exp(-\tfrac{9}{2}x^2/N)-e^{3x}\right]$$
$$\quad=-\tfrac{1}{2}\left(\log N+\log 2+\gamma\_E\right).\qquad\qquad(\*\*)$$


This differs by a minus sign from the result $(\*)$ in the reference. I checked it numerically, and it's pretty clear that (\*\*) and not (\*) is the correct asymptotics.


