I am working on understanding the following function:
$$g(x) = \ln\Gamma\left(\frac{x}{4}\right) - \ln\Gamma\left(\frac{x}{5}+\frac{1}{2}\right) - \ln\Gamma\left(\frac{x}{20}+\frac{1}{2}\right) - 2(1.03883)\left(\sqrt{\frac{x}{4}}\right) - (1.03883)\left(\frac{x}{8}\right) + \ln\Gamma \left(\frac{x}{10}\right) - \ln\Gamma\left(\frac{x}{12}+\frac{1}{2}\right) - \ln\Gamma \left(\frac{x}{60}+\frac{1}{2}\right)$$
Using this series for the digamma, I am using:
$$\frac{d}{dx}(\ln\Gamma(x)) = \psi(x) = -\gamma + \sum_{k=0}^{\infty}\left(\frac{1}{k+1} - \frac{1}{k+x}\right) $$
So that:
$$g'(x) = 2\gamma + \sum_{k=0}^{\infty}(\frac{-2}{k+1} + \frac{1}{k+\frac{x}{5}+\frac{1}{2}} + \frac{1}{k + \frac{x}{20} + \frac{1}{2}} - \frac{1}{k+\frac{x}{4}} + \frac{1}{k+\frac{x}{12} + \frac{1}{2}} + \frac{1}{k + \frac{x}{60} + \frac{1}{2}} - \frac{1}{k+\frac{x}{10}}) - \frac{0.519415}{\sqrt{x}} -0.12985375$$
On the surface, I can't see how I would prove this is increasing for $x \ge 214$? Did I make a mistake in the derivative? Is there some trick to show that this function increases for $x \ge 214$?
Yet, when I look at Wolfram Alpha at $214 \le x \le 1000$, the graph, the function is increasing.
Does it continue to increase for $x > 1000$?
One thought that occurs to me is that I can analyze the following function:
$$h(k) = \frac{-2}{k+1} + \frac{1}{k+\frac{x}{5}+\frac{1}{2}} + \frac{1}{k + \frac{x}{20} + \frac{1}{2}} - \frac{1}{k+\frac{x}{4}} + \frac{1}{k+\frac{x}{12} + \frac{1}{2}} + \frac{1}{k + \frac{x}{60} + \frac{1}{2}} - \frac{1}{k+\frac{x}{10}}$$
I will see if I can prove that at a certain point, the sum is increasing. If I make progress with this approach, I will post my thinking as an answer.
Thanks very much,
-Larry
Edit: I added a link to the graph based on a comment. The function starts increasing before $x$ gets to $214$.
Let's evaluate the asymptotic expansion of the derivative.
We may use the well known asymptotic expansion for $\psi$ (with the error bounded by the first neglected term) : $$\psi(x)\sim \ln(x)-\frac 1{2\,x}-\frac 1{12\,x^2}$$ to get : $$\frac {\psi\left(\frac x4\right)}4 - \frac {\psi\left(\frac x5 + \frac 12\right)}5 - \frac {\psi\left(\frac x{20} + \frac 12\right)}{20}\sim -\frac {\ln(4)}4 + \frac{\ln(5)}5 +\frac {\ln(20)}{20}-\frac 1{2\,x}-\frac {11}{8\,x^2}$$ and $$\frac {\psi\left(\frac x{10}\right)}{10} - \frac {\psi\left(\frac x{12} + \frac 12\right)}{12} - \frac {\psi\left(\frac x{60} + \frac 12\right)}{60}\sim -\frac {\ln(10)}{10} +\frac {\ln(12)}{12} + \frac {\ln(60)}{60}-\frac 1{2\,x}-\frac {23}{6\,x^2}$$ while (your $\dfrac{0.519415}{x^2}$ in the derivative should be $\dfrac{0.519415}{\sqrt{x}}$) : $$\left(- 2\cdot 1.03883\,\sqrt{\frac{x}{4}} - 1.03883\frac{x}{8}\right)'=-{1.03883}\left(\frac 1{2\sqrt{x}}+\frac 18\right)$$ Adding these three terms we get the asymptotic expansion of $g'(x)$ : $$g'(x)\sim C-\frac{1.03883}{2\sqrt{x}}-\frac 1x-\frac {125}{24\,x^2}$$ with \begin{align} C&=-\frac {\ln(4)}4 + \frac{\ln(5)}5 +\frac {\ln(20)}{20}-\frac {\ln(10)}{10} +\frac {\ln(12)}{12} + \frac {\ln(60)}{60}-\frac{1.03883}8\\ &=-\frac {3\ln(2)}{10}+\frac {\ln(3)}{10}+\frac{\ln(5)}6-\frac{1.03883}8\\ &\approx 0.040303 \end{align}
Now, setting $\,\displaystyle z:=\frac 1{\sqrt{x}}\,$, you may solve $\displaystyle C-\frac{1.03883}2z-z^2=0$ and get $\,x\approx 212.8\ $
Or better observe that for $\,x>214\,$ we get $\ \displaystyle C-\frac{1.03883}{2\sqrt{x}}-\frac 1x>\frac{125}{24\,x^2}\ $ and remember that the error in the asymptotic expansion of $\psi$ is bounded by the first neglected term so that $\ g'(x)>0$ for $x>214\ $ as wished !