Deriving the series representation of the digamma function from the functional equation By repeatedly using the functional equation $ \displaystyle\psi(z+1) = \frac{1}{z} + \psi(z)$, I get that $$ \psi(z) = \psi(z+n)  - \frac{1}{z+n-1} - \ldots - \frac{1}{z+1} - \frac{1}{z}$$  
or $$\psi(z+1) = \psi(z+n+1) - \frac{1}{z+n} - \ldots - \frac{1}{z+2} - \frac{1}{z+1} . $$
Is it possible to derive the series representation $ \displaystyle \psi(z+1) = - \gamma - \sum_{n=1}^{\infty} \Big( \frac{1}{z+n} - \frac{1}{n} \Big)$ from that?

This is very late, but I would like to share a slightly simpler method of deriving that series. 
\begin{align}
\Psi(z)
&=-\gamma+H_{z-1}\\
&=-\gamma+\int^1_0\frac{1-t^{z-1}}{1-t}{\rm d}t\\
&=-\gamma+\sum_{n \ge 0}\int^1_0(t^n-t^{n+z-1}){\rm d}t\\
&=-\gamma+\sum_{n \ge 0}\left(\frac{1}{n+1}-\frac{1}{n+z}\right)
\end{align}
