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 $\frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}$
$\frac{1}{\Gamma (s)}=se^{\gamma s} \prod_{n=1}^{\infty}(1 + \frac{s}{n})e^{-s/n}$

Now take $\log$'s of both sides.

$-\log\Gamma(s) = \log s +\gamma s+\sum_{n=1}^\infty \left(\log\left(1+\frac{s}{n}\right)-\frac{s}{n}\right)$

Now differentiate both sides.

$-\frac{\Gamma'(s)}{\Gamma(s)}=\frac{1}{s} + \gamma + \sum_{n=1}^\infty(\frac{1}{s+n}-\frac{1}{n})$

This process is called logarithmic differentiation.

3. Oh nice it was easy, somehow I havent heard that term before. Here yet another problem:

Show that, for real z > 0, we have $\frac{\Gamma'}{\Gamma}(z) = O(log|z|)$ as $z \to \infty$

4. Originally Posted by EinStone
Oh nice it was easy, somehow I havent heard that term before. Here yet another problem:

Show that, for real z > 0, we have $\frac{\Gamma'}{\Gamma}(z) = O(log|z|)$ as $z \to \infty$
$\frac{\Gamma'(z)}{\Gamma(z)} = O(1) + \sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{z+n}\right) = O\left(\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{\lfloor z\rfloor+n}\right)\right)$

$= O\left(\sum_{n=1}^{\lfloor z \rfloor} \frac{1}{n}\right)$ since the previous sum is telescoping.

Now it's a well known fact that for $x\in\mathbb{R}$ and $x>0$, $\log(x) = \sum_{n=1}^{\lfloor x \rfloor} \frac{1}{n}+O(1)$. (Ask if you'd like to see why.)

Thus $\frac{\Gamma'(z)}{\Gamma(z)} = O(\log(z))$ as $z\to \infty$. (I omitted the absolute value since $z > 0$.)

5. Ok thanks, could you say something about the log x thing? A reference is enough or just give the general idea, I dont need details.

6. Originally Posted by EinStone
Ok thanks, could you say something about the log x thing? A reference is enough or just give the general idea, I dont need details.
Perform LRAM and RRAM with step size $1$ to approximate $\int_{1}^x \frac{1}{t}dt$, where $x\in\mathbb{N}$.
We know that $\int_{1}^x \frac{1}{t}dt = \log(x)$, but since $\frac{1}{t}$ is monotonically decreasing, RRAM $<\log(x)<$ LRAM.

Now RRAM $= \sum_{n=2}^x \frac{1}{n}$
LRAM $= \sum_{n=1}^{x-1} \frac{1}{n}$

Another way to prove this is by applying the Euler Maclaurin formula to $\sum_{n\leq x} \frac{1}{n}$, if you are familiar.