Find the residue at $z=-2$ for $g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$ Find the residue at $z=-2$ for
$$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$
I know that:
$$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$
Let $z \to -1 - z$ to get: 
$$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$
therefore we divide by the other part to get:
$$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$
I have to somehow get the coefficient of $\frac{1}{z+2}$ because I want to evaluate the residue of 
$g(z)$ at $z=-2$
The problem is I cant ever get a factor of $\frac{1}{z+2}$ what should I do?

$\psi(z)$ is regular over $\Re(z)>0$, hence $z=-2$ is a triple pole for $g(z)$. This gives:
$$\operatorname{Res}\left(g(z),z=-2\right)= \frac{1}{2}\left.\frac{d^2}{dz^2} (z+2)^3 g(z)\right|_{z=-2}=\frac{1}{2}\left.\frac{d^2}{dz^2} \frac{\psi(-z)}{z+1}\right|_{z=-2}\tag{1}$$
so:
$$\operatorname{Res}\left(g(z),z=-2\right)=\left.\frac{\psi(-z)}{(1+z)^3}+\frac{ \psi'(-z)}{(1+z)^2}+\frac{\psi''(-z)}{2(1+z)}\right|_{z=-2} = -\psi(2)+\psi'(2)-\frac{\psi''(2)}{2}$$
leading to:
$$\begin{eqnarray*}\operatorname{Res}\left(g(z),z=-2\right)&=& -(1-\gamma)+\left(1-\zeta(2)\right)-\frac{\psi''(2)}{2}\\&=&\gamma-\zeta(2)+\sum_{n\geq 0}\frac{1}{(n+2)^3}\\&=&\color{red}{\gamma-\zeta(2)+\zeta(3)-1}.\tag{2}\end{eqnarray*}$$
