The Laurent expansion of the digamma function at a negative integer

The Laurent expansion of the digamma function at a negative integer

To find the Laurent series of $\psi(z)$ at $z= 0$, I would first find the
Taylor series of $\psi(z+1)$ at $z=0$ and then use the functional equation
of the digamma function.
$$\displaystyle\psi(z + 1) = \frac{1}{z} + \psi(z) = \psi(1) + \psi'(1)x +
\mathcal{O}(z^{2}) = -\gamma + \zeta(2) + \mathcal{O}(z^{2}) = -\gamma +
\frac{\pi^{2}}{6}z + \mathcal{O}(z^{2})$$
$$ \implies \psi(z) = -\frac{1}{z} - \gamma + \frac{\pi^{2}}{6} z +
\mathcal{O}(z^{2}) $$
But I'm having trobule finding the Laurent series of $\psi(z)$ at a
negative integer $z=-n$.
It should be $ \displaystyle\psi(z) = - \frac{1}{z+n} + \psi(n+1) + O(z+n)
= -\frac{1}{z+n} + (H_{n} -\gamma) + O(z+n) $ (which would also be true
for $n=0$).