Calculating $\Gamma'(n)$ for $n \in \mathbb{N}^{*}$

Question

I've been doing some math for fun. I wanted to find $\ \Gamma'(n)$ for $n \in \mathbb{N}^{*}$. I've shown that the sequence ( if noted $\left(a_n\right)_{n \in \mathbb{N}^{*}}$ ) is given by $$ \forall n \in \mathbb{N}^{*}, \ a_{n+1}=\left(n-1\right)!+na_n\text{ and }a_1=-\gamma $$ Is there a simple way to explicit $a_n$ for $n \in \mathbb{N}^{*}$ ? I guess yes but what i've tried doesnt work, i'm sure it is linked to the harmonic sum.

Answer 1

$\frac{df}{dx} = f\cdot\frac{d}{dx}\log f(x)$ leads to $$ \Gamma'(n) = \Gamma(n)\,\psi(n) = \color{red}{(n-1)!\left(H_{n-1}-\gamma\right)} $$ where $H_k=\sum_{m=1}^{k}\frac{1}{m}$ is the $k$-th harmonic number and $\gamma=\sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$ is the Euler-Mascheroni constant.