Let $n$ be a positive integer. Is the function $F\_n=\frac{\Gamma^{(n)}}{\Gamma^{2n}}$ entire on $\mathbb C$? If yes, is $F\_n(0)$ a ramarkable value?
Following <https://en.wikipedia.org/wiki/Digamma_function#Infinite_product_representation>, it is the case for $n=1$.



higher derivatives of gamma function
Yes, it is entire: $\Gamma$ has no zeros and only simple poles. At a pole of $\Gamma$, $\Gamma^{(n)}$ has a pole of order $n+1$, while the denominator has a pole
of order $2n$. Therefore the ratio has no poles. We also have $F\_n(0)=0$ when $n\geq 2$ which is of course a remarkable value.


