Question on series $\frac {\Gamma'(z)}{\Gamma(z)}$ 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:
$$\frac {2\Gamma'(2z)}{\Gamma(2z)} - \frac {\Gamma'(z)}{\Gamma(z)} - \frac {\Gamma'(z+\frac{1}{2})}{\Gamma(z+\frac{1}{2})} = 0$$
What's the correct answer? $0$    or   $2\log 2$?
Can anyone help?

Since
$$\Gamma(z)\cdot \Gamma(z+1/2)=2\sqrt{\pi}\cdot 4^z\cdot \Gamma(2z) $$
by considering the logarithmic derivative of both sides we get:
$$\frac{\Gamma'}{\Gamma}(z)+\frac{\Gamma'}{\Gamma}(z+1/2)=2\frac{\Gamma'}{\Gamma}(2z)+2\log 2.$$
