Let:
- $\pi(x)$ be the prime counting function
- $\psi(x)$ be the digamma function
Given that $\pi(x) < \dfrac{1.25506n}{\ln n}$ (see here), let:
$$f(n) = \frac{\Gamma\left(2n + 1 – \frac{1.25506n}{\ln n} \right)}{[\Gamma(n+1)]^2}$$
I am trying to determine if $f(n)$ is increasing or decreasing for $n \ge 631$
While it has been suggested that Stirling's Approximation is the way to go, I decided to see what happens by using the Digamma Function.
Here are my assumptions. Please let me know if any of these assumptions are wrong or any of steps are wrong.
(1) $f(n)$ is increasing if and only if $\ln f(n)$ is increasing where:
$$\ln f(n) = \ln\Gamma\left(2n + 1 – \frac{1.25506n}{\ln n}\right) – 2\ln\Gamma(n+1)$$
(2) $\psi(x) = \frac{d}{dx}\left(\ln \Gamma(x)\right)$ so that after applying the Chain Rule:
$$\frac{d}{dn}(\ln f(n)) = \psi\left(2n + 1 – \frac{1.25506n}{\ln n} \right)\left(2 – \frac{1.25506}{\ln(x)} + \frac{1.25006}{(\ln x)^2}\right) – 2\psi(n+1)$$
(3) The following Series equation (from Inequalities here), applies:
$$\psi(z+1) = -\gamma + \sum_{n=1}^{\infty}\left(\frac{z}{n(n+z)}\right),\,\,\,\,\,\,\,\,\,\,\, z \ne -1, -2, -3, \dots$$
Here's my attempt to resolve the question. Unfortunately, I am ending up with a counterintuitive conclusion that $\frac{d}{dn}(\ln f(n)) < 0$
(1) $\frac{d}{dn}(\ln f(n))$ can be restated to:
$$\frac{d}{dn}(\ln f(n)) = \psi\left(2n + 1 – \frac{1.25506n}{\ln n} \right)\left(2 – \frac{1.25506}{\ln(x)} + \frac{1.25006}{(\ln x)^2}\right) – 2\psi(n+1) = \left(2 – \frac{1.25506}{\ln(x)} + \frac{1.25006}{(\ln x)^2}\right)\left[-\gamma + \sum_{k=1}^{\infty}\left(\frac{2n – \frac{1.25506n}{\ln n}}{k\left(k+2n – \frac{1.25506n}{\ln n}\right)}\right)\right] – 2\left[-\gamma + \sum_{k=1}^{\infty}\left(\frac{n}{k(k+n)}\right)\right]$$
(2) For $x \ge 631$:
$$2 – \frac{1.25506}{\ln(x)} + \frac{1.25006}{(\ln x)^2} > 1.8$$
So that:
$$\frac{d}{dn}(\ln f(n)) < -1.9\gamma + \sum_{k=1}^{\infty}\left(\frac{3.8n – \frac{2.384614n}{\ln n}}{k\left(k+2n – \frac{1.25506n}{\ln n}\right)}\right) +2\gamma – \sum_{k=1}^{\infty}\left(\frac{2n}{k(k+n)}\right)$$
(3) The conclusion is shown if for $k \ge 1$,
$$\left(\frac{3.8n – \frac{2.384614n}{\ln n}}{k\left(k+2n – \frac{1.25506n}{\ln n}\right)}\right) < \frac{2n}{k(k+n)}$$
(4) Here is the argument:
$$ 0.9k + \frac{0.062753n}{\ln n} < 0.1n + \left(\frac{1.192307k}{\ln n}\right)$$
$$ 1.9k + 1.9n + \frac{0.062753n}{\ln n} < k + 2n + \left(\frac{1.192307k}{\ln n}\right)$$
$$\left(1.9k – \frac{1.192307k}{\ln n}\right)+ \left(1.9n – \frac{1.192307n}{\ln n}\right) < \left(k+2n – \frac{1.25506n}{\ln n}\right)$$
$$\left(1.9 – \frac{1.192307}{\ln n}\right)(k+n) < \left(k+2n – \frac{1.25506n}{\ln n}\right)$$
$$\left(3.8 – \frac{2.384614}{\ln n}\right)(k+n) < (2)\left(k+2n – \frac{1.25506n}{\ln n}\right)$$
$$\left(3.8n – \frac{2.384614n}{\ln n}\right)[k(k+n)] < (2n)\left[k\left(k+2n – \frac{1.25506n}{\ln n}\right)\right]$$
Edit:
I change the point I am checking to $631$ which reflects the point where Excel is suggesting that the result is greater than $1$.
See my Bounty explanation for details.
Edit 2:
Made fix suggested by Integrand to apply the Chain Rule
Best Answer
As I mentioned in a comment, the derivative in (2) is $$ \frac{d}{dx} \ln(f(x)) = \psi\left(2x+1-\frac{1.25506 x}{\log(x)}\right)\color{red}{\cdot \left(2+\frac{1.25506}{\log ^2(x)}-\frac{1.25506}{\log (x)}\right)}-2 \psi(x+1); $$this follows from the Chain Rule. Let $k$ be an arbitrary constant and let $c=1.25506$ be the constant you mentioned. For brevity, I'll denote $g_k(x)=2x+1-\frac{k x}{\ln x}$ and $f_k(x)=\Gamma(g_k(x))/\Gamma(x+1)^2$.
A partial resolution: $f_c(x)$ is eventually increasing.
Start with $\ln(f_k(x))$: $$ \ln(f_k(x)) = \ln\left(\Gamma\big(g_k(x)\big)\right)-2\ln(\Gamma(x+1)) $$Since everything is positive, we can use Binet's log-gamma formula: $$ \ln(\Gamma(z))=-z+\left(z-\frac{1}{2}\right) \log (z)+\frac{1}{2} \log (2 \pi )+2 \int_0^{\infty } \frac{\arctan\left(\frac{t}{z}\right)}{\exp (2 \pi t)-1} \, dt $$Of note: the integral is decreasing as $z$ increases, has a very small value, and approaches $0$ in the limit as $z\to\infty$. So we have $$ \ln(f_k(x)) = 1-\frac{1}{2} \log (2 \pi )+2 \int_{0}^{\infty} \frac{\arctan(t/g_k(x))-\arctan(t/(x+1))}{e^{2\pi t}-1}\,dt $$ $$\underbrace{-2 x \log (x+1)+\frac{k x(1- \log \left(g_k(x))\right)}{\log (x)}+2 x \log \left(g_k(x)\right)-\log (x+1)+\frac{1}{2} \log \left(g_k(x)\right)}_{F_k(x)} $$Here's the good part: if we disregard the integral (to be justified momentarily), then a messy but routine calculation gives: $$ \lim_{x\to\infty}\left(\frac{d}{dx}F_k(x)\right) = -k +\log(4) $$In particular, when $k=c$, the value is approximately $0.131234>0$. So, assuming we can explain away that pesky integral, then $f_c(x)$ is eventually increasing. Note that for $x>100$, say, we have $1<x+1<g_1(x)$. Further, $g_1(x)$ and $x+1$ are continuous, positive, and increasing without bound. Then for some $\xi=\xi(x)>x$, we have $$ 2 \int_{0}^{\infty} \frac{\arctan(t/g_1(x))-\arctan(t/(x+1))}{e^{2\pi t}-1}\,dt $$ $$ =-2 \int_{0}^{\infty} \frac{\arctan(t/\xi)}{e^{2\pi t}-1}\,dt $$This integral is increasing in $x$ and approaches $0$ as $x\to\infty$. At this point, we have shown that $f_c(x)$ is eventually increasing.
Some things I 'know' are obvious but cannot prove right now: