How one can show that the gamma function is a strictly increasing function on the interval $(1.4616,+∞)$.

Question

How one can show that the gamma function is a strictly increasing function on the interval $(1.4616,+∞)$.

Answer 1

$$f(x)=\Gamma (x) \implies f'(x)=\Gamma (x)\, \psi ^{(0)}(x) \implies f''(x)=\Gamma (x) \left(\psi ^{(0)}(x)^2+\psi ^{(1)}(x)\right)$$ If $x>0$, the only root of the digamma function is at $x_*=1.461632144968$ (see here).

You could easily obtain this solution expanding the digamma function as a Taylor series around $x=1$ and use series reversion. Otherwise, use Newton method.

For this value , $f''(x_*) \sim 0.856974$ showing that that it is a minimum.