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I was surprised to discover that the following function appears to be strictly increasing for $x \ge 2$:

$$f(x) = \frac{\Gamma(x+1)}{\Gamma(\frac{x}{2}+1)\Gamma(\frac{x}{3}+1)\Gamma(\frac{x}{5}+1)}$$

when I tested out different values of $x$ using Excel.

I had assumed that it would be decreasing since $x < \frac{x}{2} + \frac{x}{3} + \frac{x}{5}$.

I wanted to verify this is true by checking the derivative.

It seemed to me that the right way to do this is to use this series of the digamma function so that:

$$\frac{d}{dx}\left(\ln \Gamma(x+1) - \ln \Gamma(\frac{x}{2}+1) - \ln\Gamma(\frac{x}{3}+1) - \ln\Gamma(\frac{x}{5}+1)\right) =$$

$$ \psi(x+1) - \frac{\psi(\frac{x}{2}+1)}{2} - \frac{\psi(\frac{x}{3}+1)}{3}-\frac{\psi(\frac{x}{5}+1}{5}=$$

$$\frac{\gamma}{6}+ \sum_{k=0}^{\infty}\left(\frac{-1}{30k+30}-\frac{1}{k+x+1} + \frac{1}{2k+x+2} + \frac{1}{3k+x+3} + \frac{1}{5k+x+5}\right)$$

It is not obvious to me that this derivative is greater than $0$ for $x \ge 2$

How would I complete the derivative to reach my conclusion about whether this function is strictly increasing or not for $x \ge 2$?

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  • $\begingroup$ Just in case, here's a simple R algorithm for digamma function, which uses its asymptotic approximation and functional equation: zm <- 20; psi <- function(z){if(z < zm){k <- floor(zm-z)}else{k <- 0}; return(log(z+k+1)-1/2/(z+k+1)-1/12/(z+k+1)^2-sum(1/(z+0:k)))}; you can increase the accuracy by increasing zm. The R environment is free and open source. $\endgroup$ – Yuriy S Nov 13 '18 at 5:52
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Stirling's approximation says that $\log \Gamma(z) \sim z \log z + O(z)$. So $$\log \Gamma(\alpha x + 1)\sim (\alpha x + 1)\log(\alpha x + 1)+O(x)\sim \alpha x \log x + O(x),$$ and so certainly $\log f(x)\sim -\frac{1}{30}x\log x + O(x)$: your function must, eventually, start decreasing asymptotically to zero. But eventually is important. Let's try to estimate where this happens.

Consider the ratio $f(30y+30) / f(30y)$, where $y$ is an integer. You have $$ \frac{f(30y+30)}{f(30y)}=\frac{\Gamma(30y+31)\Gamma(15y+1)\Gamma(10y+1)\Gamma(6y+1)}{\Gamma(30y+1)\Gamma(15y+16)\Gamma(10y+11)\Gamma(6y+7)}=\frac{(30y+30)!(15y)!(10y)!(6y)!}{(30y)!(15y+15)!(10y+10)!(6y+6)!}=\frac{(30y+30)(30y+29)\cdots(30y+1)}{(15y+15)(15y+14)\cdots(15y+1)\cdot(10y+10)\cdots(10y+1)\cdot(6y+6)\cdots(6y+1)}\approx\frac{30^{30}}{15^{15}10^{10}6^{6}}\frac{1}{y}=\frac{1.008\times10^{12}}{y}, $$ where the $\approx$ is arrived at by pulling out the prefactors, counting powers of $y$, and ignoring the additive terms (which is justified when $y$ is large). This is going to be less than $1$ eventually... when $y$ exceeds a trillion. Your function will start decreasing when $x$ is greater than about $3\times 10^{13}.$ By that point, $f(x)$ will have many trillions of decimal digits, which I dare say is a bit more than Excel can handle.

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  • $\begingroup$ Thanks very much! I appreciate your analysis. $\endgroup$ – Larry Freeman Nov 14 '18 at 22:28
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To take it from your derivative expression:

If you actually perform the sum of the fractions (which is quite tedious, but Mathematica or equivalent could help), you will find that you have:

$\frac{\gamma}{6} + \sum_{k=0}^{\infty} x\frac{-x^3 + 49(k+1)x^2 + 359(k+1)^2 x + 539(k+1)^3}{denominator}$

(Please verify the precise coefficients.) The denominator is a 4th-order polynomial in $x$ (and 5th order in $k$) with all coefficients positive. So it is positive for x>0 and increasing fast with $k$.

The numerator is a 3rd order polynomial with high positive coefficients except for the highest-order term which is -x^3. For small enough x, the numerator will be positive. For large x, it can eventually become negative. For very high $k$ it will not, but then the denominator will be so large that the terms will be suppressed. Then we are in agreement with the conclusion of @mjqxxxx which they reached by a different approach.

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