Proof trigamma recurrence relation

Question

How can I prove the digamma and trigamma recurrence relations, $$\psi(x+1)=\psi(x)+\frac{1}{x} \,?$$ I tried some advice for common recurrence relations, but it didn't open for me.

Answer 1

Starting with the Gamma function property $$\Gamma(x + 1) = x \, \Gamma(x)$$ then by differentiation with respect to $x$ of the logarithm of the relation it is seen that \begin{align} \ln(\Gamma(x + 1)) &= \ln(\Gamma(x)) + \ln(x) \\ \frac{\Gamma^{'}(x+1)}{\Gamma(x + 1)} &= \frac{\Gamma'(x)}{\Gamma(x)} + \frac{1}{x}. \end{align} Using the defined relation for the digamma function, namely, $\Gamma(x) \, \psi(x) = \Gamma'(x)$ then the digamma relation is $$\psi(x+1) = \psi(x) + \frac{1}{x}.$$ Applying differentiation again leads to the following trigamma relation: \begin{align} \psi'(x+1) = \psi'(x) - \frac{1}{x^{2}}. \end{align}