# 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? – Saad May 13 '20 at 2:25
• It is the upper bound for $\pi(n)$, see here – Larry Freeman May 13 '20 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. – QC_QAOA May 13 '20 at 4:56
• It should be $\Gamma(n)^2$. I will fix it where I say $\Gamma(n^2)$. Thanks for calling this out. – Larry Freeman May 13 '20 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. – QC_QAOA May 13 '20 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 $$n=107$$.