Finding the coefficients of $h(z)$ laurent series Consider:
$$h(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$
Find the coefficients $a_n$ of the Laurent Series of $h(z)$ centered at $z=-2$
I got this from the approach here: Infinite sum complex analysis
You see the complex-analytic approach. My question is, why does @MarkoRiedel
Only use $\psi(-z)$ part of the function $h(z)$ in finding the coeffcients. As you see, his coefficient integral is:
$$\frac{1}{2\pi i} \cdot \int_{|z+2| = \epsilon} \frac{\psi(-z)}{(z+2)^{n+1}} dz $$
He does not consider $$\frac{1}{(z+1)(z+2)^3} \space \text{of} \space h(z) \space \text{just the} \space \psi(-z) \space \text{digamma.}$$ 
How and why?
Thanks!
 A: We can write the series for $\psi(-z)$ at $z=-2$ as
$$
\begin{align}
\psi(-z)
&=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}-\frac1{k+2-(z+2)}\right)\\
&=-\gamma+\sum_{k=0}^\infty\left(\frac1{k+1}+\frac1{(z+2)-(k+2)}\right)\\
&=1-\gamma+\sum_{k=1}^\infty(1-\zeta(k+1))(z+2)^k\tag{1}
\end{align}
$$
where we have used that
$$
\begin{align}
\frac1{n!}\frac{\mathrm{d}^n}{\mathrm{d}z^n}\sum_{k=0}^\infty\frac1{(z+2)-(k+2)}
&\stackrel{\hphantom{z=-2}}{=}\sum_{k=0}^\infty\frac{(-1)^n}{((z+2)-(k+2))^{n+1}}\\
&\stackrel{z=-2}{=}1-\zeta(n+1)\tag{2}
\end{align}
$$
Therefore, we get
$$
\begin{align}
\frac{\psi(-z)}{((z+2)-1)(z+2)^3}
&=-\left[\sum_{k=0}^\infty(z+2)^{k-3}\right]\left[1-\gamma+\sum_{k=1}^\infty(1-\zeta(k+1))(z+2)^k\right]\\
&=\bbox[5px,border:2px solid #C00000]{\sum_{n=0}^\infty\left[-(1-\gamma)+\sum_{k=1}^n(\zeta(k+1)-1)\right](z+2)^{n-3}}\tag{3}
\end{align}
$$

I see that this result is used to compute a sum that is far more simply computed using partial fractions.
$$
\begin{align}
\sum_{n=1}^\infty\frac1{(n+1)(n+2)^3}
&=\sum_{n=1}^\infty\left(\color{#C00000}{\frac1{n+1}-\frac1{n+2}}\color{#00A000}{-\frac1{(n+2)^2}}\color{#0000FF}{-\frac1{(n+2)^3}}\right)\\
&=\color{#C00000}{\frac12}\color{#00A000}{-\left(\zeta(2)-1-\frac14\right)}\color{#0000FF}{-\left(\zeta(3)-1-\frac18\right)}\\
&=\frac{23}8-\frac{\pi^2}6-\zeta(3)\tag{4}
\end{align}
$$
