trigamma fuction

Question

Explain me please why the equality $\ \displaystyle\psi_1(z)=\sum_{n=0}^\infty \frac 1{(z+n)^2}\ $ is right if we define trigamma function like this : $\ \displaystyle\psi_1(z)=\frac{d^2}{dz^2}\ln\Gamma(z)$.

Answer 1

There is an easy way to come at the first equation if you note that : $$\Gamma(z+1)=z\;\Gamma(z)$$ so that $$\ln\Gamma(z+1)=\ln(z)+\ln\Gamma(z)$$ Differentiation gives $$\psi(z+1)=\frac 1z+\psi(z)$$ Differentiating again : $$\psi_1(z+1)=-\frac 1{z^2}+\psi_1(z)$$ so that $$\psi_1(z)=\frac 1{z^2}+\psi_1(z+1)$$ $$\psi_1(z)=\frac 1{z^2}+\frac 1{(z+1)^2}+\psi_1(z+2)$$ $$\psi_1(z)=\frac 1{z^2}+\frac 1{(z+1)^2}+\frac 1{(z+2)^2}+\cdots$$

I hope that this will be a help too for your intuition !