Log of Gamma Function is Convex on Positive Reals/Proof 3

Theorem
Let $\Gamma: \R_{>0} \to \R$ be the Gamma function, restricted to the strictly positive real numbers.

Let $\ln$ denote the natural logarithm function.

Then the composite mapping $\ln \circ \operatorname \Gamma$ is a convex function.

Proof
The strategy is to use the Euler Form of the Gamma function and directly calculate the second derivative of $\ln\Gamma\left(z\right)$.


 * $\displaystyle\Gamma \left({z}\right) = \lim_{m \to \infty} \frac {m^z m!} {z \left({z+1}\right) \left({z+2}\right) \dotsb \left({z+m}\right)}$


 * $\displaystyle \dfrac{\mathrm d^2 \ln \Gamma\left(z\right)}{\mathrm d z^2}

= \dfrac{\mathrm d^2}{\mathrm d z^2} \ln\left[ \lim_{m \to \infty} \frac {m^z m!} {z \left({z+1}\right) \left({z+2}\right) \dotsb \left({z+m}\right)}\right] = \lim_{m \to \infty} \dfrac{\mathrm d^2}{\mathrm d z^2} \ln\left[ \frac {m^z m!} {z \left({z+1}\right) \left({z+2}\right) \dotsb \left({z+m}\right)}\right]$

The limit interchange is justified because Gamma Function is Smooth on Positive Reals.


 * $\displaystyle \lim_{m \to \infty} \dfrac{\mathrm d^2}{\mathrm d z^2} \left[z\ln\left(m\right) + \ln\left(m!\right) - \sum_{n \mathop = 0}^{m} \ln\left(z+n\right)\right]

= \lim_{m \to \infty}\sum_{n \mathop = 0}^{m} \dfrac{1}{\left(z+n\right)^2}$


 * $\displaystyle \dfrac{\mathrm d^2 \ln \Gamma\left(z\right)}{\mathrm d z^2} = \sum_{n \mathop = 0}^{\infty} \dfrac{1}{\left(z+n\right)^2} > 0$

Logarithmic convexity then follows from Second Derivative of Strictly Convex Real Function is Strictly Positive.