Can you get the following asymptotic expression for an hypergeometric function? 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?
 A: 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, $I_N$ 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.
