trigamma fuction 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 : $\ \displaystyle\psi_1(z)=\frac{d^2}{dz^2}\ln\Gamma(z)$.

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 !
