1. Gamma Function logarithmic derivative

Show that $-\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})$

Hint: use Weierstrass product formula $\frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}$

2. Originally Posted by EinStone
Show that $-\frac{\Gamma'}{\Gamma}(z)=\gamma + \frac{1}{z} + \sum_{j=1}^\infty(\frac{1}{z+j}-\frac{1}{j})$

Hint: use Weierstrass product formula $\frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}$
Double post... tisk tisk tisk

3. I first thought posting didnt work in the other thread so I decided to change forum, but now its double and I didnt see how to delete, sry.

4. Originally Posted by EinStone
I first thought posting didnt work in the other thread so I decided to change forum, but now its double and I didnt see how to delete, sry.
Haha, it's ok