Question about $\frac {\Gamma'(z+1)}{\Gamma(z+1)}$ If $\psi (z)= \log\Gamma(z+1)$ 
Prove that : $$\psi(n)+\gamma=1+\frac{1}{2}+\cdots+\frac{1}{n}$$  
My Proof :
$$\psi (z)= \frac {\Gamma'(z+1)}{\Gamma(z+1)}=-\frac{1}{z+1}-\gamma + \sum_{n=1}^\infty \frac{1}{n} - \frac{1}{n+z+1}  $$
$$= \sum_{n=0}^{\infty}\frac{1}{n+1}\frac{1}{n+z+1} -\gamma  \ $$
at $z=n$
$$ \psi(n)+\gamma = \sum_{n=0}^{\infty} \frac{1}{n+1}-\frac{1}{2n+1} $$
$$ \psi(n)+\gamma =(1-1)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\cdots$$
$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}+\cdots $$
$$\psi(n)+\gamma=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}+\cdots $$
where is the error in my proof ? 
 A: One error:
From 
$\psi(z) = \sum_{n=0}^{\infty}\left(\frac{1}{n+1}-\frac{1}{n+z+1}\right) -\gamma  $
you set $z = n$
to get
$\psi(n)+\gamma = \sum_{n=0}^{\infty} \left(\frac{1}{n+1}-\frac{1}{2n+1}\right)$,
You use $n$ both as an argument of $\psi$ and as an index of summation. 
You can't do that.
(added later)
From this,
if you set
$z = m$
where $m$ is a positive integer,
$\psi(m)  +\gamma 
= \sum_{n=0}^{\infty}(\frac{1}{n+1}-\frac{1}{n+m+1} )
= \sum_{n=0}^{m-1}\frac{1}{n+1}
= \sum_{n=1}^{m}\frac{1}{n}
$,
since all terms from
$n=m+1$
get cancelled out.
(added even later)
Here is why the cancellation happens:
Suppose $N$ is a large integer.
Then
$\begin{array}\\
\sum_{n=0}^{N}(\frac{1}{n+1}-\frac{1}{n+m+1} )
&=\sum_{n=0}^{N}\frac{1}{n+1}-\sum_{n=0}^{N} \frac{1}{n+m+1}\\
&=\sum_{n=1}^{N+1}\frac{1}{n}-\sum_{n=m+1}^{N+m+1} \frac{1}{n}\\
&=\left(\sum_{n=1}^{m}\frac{1}{n}+\sum_{n=m+1}^{N+1}\frac{1}{n}\right)
-\left(\sum_{n=m+1}^{N+1} \frac{1}{n}+\sum_{n=N+2}^{N+m+1} \frac{1}{n}\right)\\
&=\sum_{n=1}^{m}\frac{1}{n}+\sum_{n=N+2}^{N+m+1} \frac{1}{n}\\
\end{array}
$
and the right hand sum
is less than
$\frac{m}{N}$
since there are $m$ terms
each of which is
less than
$\frac1{N}$.
