Calculate the residue of:
$$f(z) = \frac{\psi(-z)}{(z+1)(z+2)^3} \space \text{at} \space z=-2$$
Where $\psi(z)$ is the digamma function, and $\zeta(z)$ is the Riemann-zeta function (below).
The answer given is: $\displaystyle \mathrm{Res}_{z=-2} f(z) = \frac{\pi^2}{6} +\zeta(3) - 3 +\gamma$
Using the Laurent Series. Obviously we must find, the coefficient of $\frac{1}{z+2}$ in the series expansion.
Using the series definition of $\psi(-z)$ zeta definition
$$\psi(-z) = 1-\gamma + \sum_{k=1}^{\infty} (1-\zeta(k+1)) (z+2)^k$$
Or
$$\psi(-z) = -\gamma + \sum_{n=0}^{\infty} \frac{z + 1}{(n+1)(z-n)}$$
How can I approach this?
Thanks!
