# Is the following ratio of gamma functions increasing: $\frac{\Gamma(2n - \frac{1.25506n}{\ln n})}{\Gamma(n)^2}$?

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:

$$\frac{\Gamma(2n - \frac{1.25506n}{\ln n})}{[\Gamma(n)]^2}$$

• Where does this mysterious $1.25506$ come from? May 13, 2020 at 2:25
• It is the upper bound for $\pi(n)$, see here May 13, 2020 at 2:26
• Is it $\Gamma(n^2)$ or $\Gamma(n)^2$? Your title and the initial line say the former but later when taking the logarithm you say its the latter. I suspect a typo because the former is easy to show it goes quickly to zero. May 13, 2020 at 4:56
• It should be $\Gamma(n)^2$. I will fix it where I say $\Gamma(n^2)$. Thanks for calling this out. May 13, 2020 at 4:59
• Try taking the limit as $n$ goes to infinity and using Stirlings approximation. Its still difficult, but it at least transforms to an ugly limit that wolfram alpha should be able to solve. May 13, 2020 at 5:31

f[n_] := Gamma[2 n - 1.25506 n/Log[n]]/Gamma[n]^2  For integer $$n$$, minimal value of 0.06628307572263 is achieved for $$n=107$$.