Where is the error in this proof : Prove that:
$$\frac {2\Gamma'(2z)}{\Gamma(2z)}-\frac {\Gamma'(z)}{\Gamma(z)}-\frac {\Gamma \prime(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} =2 \log 2$$
But I obtain this equal zero. 
My proof:
From Weierstrass definition of Gamma we have 
$$\frac{1}{\Gamma(z)}=z.e^{\gamma z}.\prod_{n=1}^{\infty}[ (1+\frac{z}{n}).e^{\frac{-z}{n}} ] $$
$$ -log( \Gamma(z))=log (z)+{\gamma z}+\sum_{n=1}^{\infty}[ log((1+\frac{z}{n}))-{\frac{z}{n}} ] $$
$$ log( \Gamma(z))=-log (z)-{\gamma z}-\sum_{n=1}^{\infty}[ log((1+\frac{z}{n}))-{\frac{z}{n}} ] $$
$$\frac {\Gamma'(z)}{\Gamma(z)} =-\frac{1}{z} -{\gamma }-\sum_{n=1}^{\infty}[ ((\frac{1}{z+n}))-{\frac{1}{n}} ]$$
This implies >> 
$$\frac {2\Gamma'(2z)}{\Gamma(2z)} - \frac {\Gamma'(z)}{\Gamma(z)} - \frac {\Gamma'(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})}$$
$$= \sum_{n=0}^\infty \frac{1}{z+\frac{1}{2}+n} + \sum_{n=1}^\infty \frac{1}{z+n}-\sum_{n=1}^\infty \frac{2}{2z+n} $$     
$$= \sum_{n=0}^\infty \frac{1}{z+\frac{1}{2}+n} + \sum_{n=0}^\infty \frac{1}{z+n+1}-\sum_{n=0}^\infty \frac{2}{2z+n+1}$$ $$=  
(\frac{1}{z+\frac{1}{2}})+(\frac{1}{z+1})-(\frac{2}{2z+1})+(\frac{1}{z+\frac{3}{2}})+(\frac{1}{z+2})-(\frac{2}{2z+2})+(\frac{1}{z+\frac{5}{2}})+(\frac{1}{z+3})-(\frac{2}{2z+3})+... $$
$$ = (\frac{1}{z+\frac{1}{2}})+(\frac{1}{z+1})-(\frac{1}{z+\frac{1}{2}})+(\frac{1}{z+\frac{3}{2}})+(\frac{1}{z+2})-(\frac{1}{z+1})+(\frac{1}{z+\frac{5}{2}})+(\frac{1}{z+3})-(\frac{1}{z+\frac{3}{2}})+...=0  $$
Where is the error ?
 A: Let's rewrite your equation :
$$\tag{1}\frac {2\Gamma'(2z)}{\Gamma(2z)}-\frac {\Gamma'(z)}{\Gamma(z)}-\frac {\Gamma \prime(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} =2 \log 2$$
as
$$\left[\log\Gamma(2z)-\log\Gamma(z)-\log \Gamma\left(z+\frac 12\right)\right]'=2 \log 2$$
or 
$$\tag{2}\left[\log\frac{\Gamma(2z)}{\Gamma(z)\,\Gamma\left(z+\frac 12\right)}\right]'=2 \log 2$$
which is easily deduced from the "duplication formula" $\;\displaystyle \Gamma(z)\,\Gamma\left(z+\frac 12\right)=2^{1-2z}\sqrt{\pi}\;\Gamma(2z)$.

Now let's see your derivation. I can follow you until :
$$\tag{3}\psi(z):=\frac {\Gamma'(z)}{\Gamma(z)} =-\frac 1{z} -\gamma-\sum_{n=1}^{\infty}\left[ \frac 1{z+n}-\frac 1n\right]$$
After that (as explained by MPW) you can't remove the $\,\dfrac 1n\,$ parts since you would obtain divergent series. 
Let's continue the computations at this point (without expanding the $\psi(2z)$ part) :
\begin{align}
&f(z):=2\,\psi(2z)-\psi(z)-\psi\left(z+\frac 12\right)\\
&=2\,\psi(2z)+\frac 1{z}+\frac 1{z+1/2}+2\gamma+\sum_{n=1}^{\infty}\left[\frac 1{z+n}+\frac 1{z+1/2+n}-\frac 2{n}\right]\\
&=2\,\left[\psi(2z)+\frac 1{2z}+\gamma+\frac 1{2z+1}+\sum_{n=1}^{\infty}\left[\frac 1{2z+2n}+\frac 1{2z+2n+1}-\frac 1{2n}-\frac 1{2n+1}+\\\frac 1{2n+1}-\frac 1{2n}\right]\right]\\
&=2\,\left[\psi(2z)+\frac 1{2z}+\gamma+\sum_{m=1}^{\infty}\left[\frac 1{2z+m}-\frac 1m\right]+\frac 11+\sum_{n=1}^{\infty}\left[\frac 1{2n+1}-\frac 1{2n}\right]\right]\\
&=2\,\left[\psi(2z)-\psi(2z)-\sum_{m=1}^{\infty}\frac{(-1)^m}m\right]\\
&=2\,\log(1+1)\\
\end{align}
(with $m$ combining the $2n$ and $2n+1$ terms)
