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?
 A: The functional equation tells us:
$$\frac{1}{z+n} -\frac{1}{n}=-\left(
\Psi \left( z+n+1 \right) -\Psi \left( z+n \right)\right) +\left(
\Psi \left( n+1 \right) -\Psi \left( n \right)\right)$$
and so we can form the partial sum:
$$-\sum _{n=1}^{N} \frac{1}{z+n} -\frac{1}{n}=-\sum _{n=1}^{N}\left(
\Psi \left( z+n+1 \right) -\Psi \left( z+n \right)\right) +\sum _{n=1}^{N}\left(
\Psi \left( n+1 \right) -\Psi \left( n \right)\right)$$
and by noting that:
$$\sum _{n=1}^{N}\Psi \left( z+n+1 \right) -\Psi \left( z+n \right) =
\sum _{n=2}^{N+1}\Psi \left( z+n \right) -\sum _{n=1}^{N}\Psi \left( z
+n \right) =\Psi \left( z+N+1 \right) -\Psi \left( z+1 \right)$$
the partial sum becomes:
$$-\sum _{n=1}^{N} \frac{1}{z+n} -\frac{1}{n}=-\Psi \left( z+N+1 \right) +\Psi \left( N+1 \right) +\Psi \left( z+1
 \right) +\Psi(1)$$
If we denote the starting point of the recursion relation $\Psi(1)=\gamma$  and take $N\rightarrow \infty$ we then have:
$$\Psi \left( z+1
 \right)=-\lim_{N\to \infty}\left(-\Psi \left( z+N+1 \right) +\Psi \left( N+1 \right)\right)-\gamma-\lim_{N\to\infty}\sum _{n=1}^{N}\left( \frac{1}{z+n} -\frac{1}{n}\right)$$
So, ultimately we find the functional equation alone would not quite do the job as we also have to prove the limit:
$$\lim_{n\to \infty}\left(-\Psi \left( z+N+1 \right) +\Psi \left( N+1 \right)\right)=0$$
Proving the limit is finite proves the convergence of the sum, proving it vanishes proves the desired result.

To prove the limit we could consider the integral representation of the Digamma function for $\mathfrak{R} (x)>0$:
$$\Psi(x)=\int_{0}^{\infty}\frac{e^{-t}}{t}-\frac{e^{-xt}}{1-e^{-t}}{dt}$$
A: 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}
