For $n > 1$, is the following ratio of gamma functions increasing: $\dfrac{\Gamma(2n - \frac{1.25506n}{\ln n})}{\Gamma(n)^2}$
I suspect that it is at some point where $n > 1$.
I would like figure out if the derivative is increasing or not and if increasing, from what point?
I had hoped that this series ψ would be sufficient with:
$$\frac{d}{dx}(\ln\Gamma(x)) = \frac{\psi(x)}{dx} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \frac{1}{k + x})$$
So, my goal would be to show that the following is increasing for $n \ge 1$: $$\ln\Gamma(2n - \dfrac{1.25506n}{\ln n}) - 2\ln\Gamma(n)$$
This got me to:
$$\frac{d}{dx}\left(\ln\Gamma(2n - \dfrac{1.25506n}{\ln n}) - 2\ln\Gamma(n)\right) = \frac{\psi(2n - \frac{1.25506n}{\ln n})}{2 - \frac{1.25506}{\ln n} + \frac{1.25506}{\ln^2 n}} - 2\psi(n)$$
When I tried to apply the last part, I was at a loss.
How would I complete the argument to determine whether there exists a real $n > 0$ where the function is strictly increasing?
Edit 1:
I had a thought. Does the following logic work?
An easier problem is:
$$\frac{d}{dx}(\ln\Gamma(2n) - 2\ln\Gamma(n)) = \frac{\psi(2n)}{2} - 2\psi(n) = \sum\limits_{k=0}^{\infty}\left(\frac{1}{n} - \frac{1}{2n}\right) > 0$$
If I change this to some real constant $c < 1$:
$$\frac{d}{dx}(\ln\Gamma(n(2-c)) - 2\ln\Gamma(n)) = \frac{\psi(n(2-c))}{2-c} - 2\psi(n) = \sum\limits_{k=0}^{\infty}\left(\frac{1}{n} - \frac{1}{n(2-c)}\right) > 0$$
Would it now be sufficient to complete the argument by showing that for $n \ge 4$:
$$\frac{1.25506}{\ln n} < 1$$
and showing that:
$$\frac{d}{dx}\left(\frac{1.25506}{\ln n}\right) = -\frac{1.25506}{n\ln^2(n)}$$ which is decreasing at $n\ge 4$.
Is this enough to establish the conclusion?
Edit 2:
To be clear, it should be:
$$\frac{\Gamma(2n - \frac{1.25506n}{\ln n})}{[\Gamma(n)]^2}$$
Minimum zone, enlarged:
For integer $n$, minimal value of 0.06628307572263 is achieved for $n=107$.