\( \forall \,\, n\geq 1 \)

$$\psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z)$$

$$\textit{proof}$$

Use the series representation of the digamma

$$\psi_{0}(z) = -\gamma-\frac{1}{z}+ \sum_{n=1}^\infty\frac{z}{n(n+z)}$$

This can be written as the following

$$\psi_{0}(z) = -\gamma + \sum_{k=0}^\infty\frac{1}{k+1}-\frac{1}{k+z}$$

By differentiating with respect to \( z \)

$$\psi_{1}(z) = \sum_{k=0}^\infty\frac{1}{(k+z)^2}$$

$$\psi_{2}(z) = -2\sum_{k=0}^\infty\frac{1}{(k+z)^3}$$

$$\psi_{3}(z) = 2 \cdot 3 \, \sum_{k=0}^\infty\frac{1}{(k+z)^4}$$

$$\psi_{4}(z) = -2 \cdot 3 \cdot 4 \,\sum_{k=0}^\infty\frac{1}{(k+z)^5}$$

Continue like that to obtain

$$\psi_{n}(z) = (-1)^{n+1}n!\,\sum_{k=0}^\infty\frac{1}{(k+z)^{n+1}}$$

We realize the RHS is just the Hurwitz zeta function

$$\psi_{n}(z) = (-1)^{n+1}n!\,\zeta(n+1,z)\,\,$$

The proof can be finished by induction.

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