Question on series $\frac {\Gamma'(z)}{\Gamma(z)}$

Question

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?

Answer 1

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: $$\frac{\Gamma'}{\Gamma}(z)+\frac{\Gamma'}{\Gamma}(z+1/2)=2\frac{\Gamma'}{\Gamma}(2z)+2\log 2.$$