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 : $\ \displaystyle\psi_1(z)=\frac{d^2}{dz^2}\ln\Gamma(z)$.
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This depends on your definition of the gamma function... do you know about the Weierstrass product definition? – J. M. ain't a mathematician May 30 '13 at 19:38
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No. Can you give me link on some literature where i can read about my question? – Alex May 30 '13 at 19:45
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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 !
Raymond Manzoni
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(Out of votes today, so an IOU for now.) Repeatedly applying the functional equation is very much expedient. :D – J. M. ain't a mathematician May 30 '13 at 20:05
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