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 ?
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 $\frac1{N}$.