Find the residue of:
$$f(z) = \frac{(\psi(-z) + \gamma)}{(z+1)(z+2)^3} \space \text{at} \space z=n$$
Where $n$ is every positive integer because those $n$ are the poles of $f(z)$
This is a simple pole, however:
$$\lim_{z \to n} \frac{(z-n)\psi(-z)}{(z+1)(z+2)^3}$$
But this gives $0$?
Since: $$\Gamma(z+1)=e^{-\gamma z}\prod_{n\geq 1}\left(1+\frac{z}{n}\right)^{-1}e^{\frac{z}{n}}$$ we have: $$ \log\Gamma(z+1) = -\gamma z+\sum_{n=1}^{+\infty}\left(\frac{z}{n}-\log\left(1+\frac{z}{n}\right)\right)$$ and by differentiating both sides: $$ \gamma+\psi(z+1) = \sum_{n=1}^{+\infty}\left(\frac{1}{n}-\frac{1}{n+z}\right)\tag{1}$$ hence the LHS has simple poles with residue $-1$ in the points $z=-1,-2,\ldots$ Since $\psi(x+1)=\frac{1}{x}+\psi(x)$, we can write $(1)$ as: $$ \gamma+\psi(-z) = \frac{1}{z}+\sum_{n=1}^{+\infty}\left(\frac{1}{n}+\frac{1}{z-n}\right)\tag{2}$$ hence the LHS of $(2)$ has simple poles with residue $1$ for $z=0,1,2,\ldots$. Since $\frac{1}{(z+1)(z+2)^3}$ is holomorphic over $\Re(z)>-1$, we simply have: $$ \operatorname{Res}\left(f(z),z=n\right)=\frac{1}{(n+1)(n+2)^3}.\tag{3}$$
