Given that $\lim_{z\rightarrow0}az-\frac{\ln z}{z!}=\gamma$, find the value of $a$ that gives the quickest convergence

Question

You could prove that $$\lim_{z\rightarrow0}az-\frac{\ln (z!)}{z}=\gamma$$For any $a\in\mathbb{R}$. Where $z!$ is the factorial extended to all real numbers (except negative integers). Which value gives the quickest convergence, and why? I think it is $\dfrac{\pi^2}{12}$, but I don't know how to prove it. I think it is because of the fact that $$\frac{\pi^2z}{12}-\frac{\ln (z!)}{z}>\gamma$$when $z>-1$, while all other values of $a$ are greater only when $z>0$, which makes this function "flatter" near $\gamma$. Is this a valid argument? And if so (or not), is there another proof?

Answer 1

You can use Taylor polynomials: $$ \log (z!) = \log \Gamma (z + 1) = - \gamma z + \frac{{\pi ^2 }}{{12}}z^2 + \mathcal{O}(z^3 ) $$ as $z\to 0$. See here (take $z+1$ in place of $z$).