I am trying to learn about digamma function and it's uses. For example series, I found somewhere this solution: $$\begin{align*} \sum_{n=0}^{\infty} \frac{1}{(3n+2)\left ( 3n+3 \right )} &= \sum_{n=0}^{\infty} \left [ \frac{1}{3n+2} - \frac{1}{3n+3} \right ]\\ &=\frac{1}{3} \sum_{n=0}^{\infty} \left [ \frac{1}{n+ \frac{2}{3}} - \frac{1}{n+1} \right ]\\ &=\frac{1}{3} \left [ -\psi^{(0)} \left ( \frac{2}{3} \right ) + \psi^{(0)}(1) \right ]\\ &= \frac{\log 3}{2}- \frac{\pi}{6 \sqrt{3}} \end{align*}$$ I can prove the same result using a different method using integrals, but I am insterested about digamma function here, does this equality hold: $$-\psi(a) =\sum_{n=0}^{\infty} \frac{1} {n+a} ?$$ Now I know that the RHS diverges so this can't be true. My thought is that this hold if the series is not alone(we must have like in the solution from above two parts or more inside the series). Can you help me with a proof for this?
Also here:http://mathworld.wolfram.com/PolygammaFunction.html It's shown a general form for this, but of course this doesnt hold for digamma function right?
Weierstrass Formula: $$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^\infty \frac{e^{\frac{z}{n}}}{1+\frac{z}{n}}$$
$$\log\Gamma(z)=-\gamma z-z+\sum_{n=1}^\infty\frac{z}{n}-\log(1+\frac{z}{n})$$
Differentiating:
$$\frac{\Gamma'(z)}{\Gamma(z)}=\psi(z)=-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{\frac{1}{n}}{1+\frac{z}{n}}$$ $$\psi(z)=-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{z+n}$$
If $z\in\Bbb N^+$, this becomes a simple telescoping sum and $\psi(z)=-\gamma-1+H_z$ where $H_z$ is the $z$th harmonic number.
Furthermore, $\psi(z)\neq\sum_{n=0}^\infty\frac{1}{n+z}$ due to the divergence for any given $z$.
Just because $$\psi(z)-\psi(s)=(-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{z+n})-(-\gamma-1+\sum_{n=1}^\infty\frac{1}{n}-\frac{1}{s+n})=\sum_{n=1}^\infty\frac{1}{s+n}-\frac{1}{z+n}$$
doesn't mean that $\psi(z)=\sum_{n=1}^\infty\frac{1}{z+n}$ and $\psi(s)=\sum_{n=1}^\infty\frac{1}{s+n}$.