Let $\Gamma$ be the pointwise limit on $]0, +\infty[$ of the sequence $\left(\Gamma_n\right)_{n \geqslant 1}$ where
$$\Gamma_n(x) = \frac{1}{x} e^{-\gamma x} \prod_{k=1}^{n} \frac{e^{x k^{-1}}}{1 + x k^{-1}}.$$
Show that $\lim_{x \rightarrow +\infty} (\ln(\Gamma))''(x) = 0$.