Jtdylktuy

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:

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

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

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

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 $x ge 2$?

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 $log f(x)sim -frac{1}{30}xlog x + O(x)$: 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
$$
frac{f(30y+30)}{f(30y)}=frac{Gamma(30y+31)Gamma(15y+1)Gamma(10y+1)Gamma(6y+1)}{Gamma(30y+1)Gamma(15y+16)Gamma(10y+11)Gamma(6y+7)}=frac{(30y+30)!(15y)!(10y)!(6y)!}{(30y)!(15y+15)!(10y+10)!(6y+6)!}=frac{(30y+30)(30y+29)cdots(30y+1)}{(15y+15)(15y+14)cdots(15y+1)cdot(10y+10)cdots(10y+1)cdot(6y+6)cdots(6y+1)}approxfrac{30^{30}}{15^{15}10^{10}6^{6}}frac{1}{y}=frac{1.008times10^{12}}{y},
$$
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 $3times 10^{13}.$ By that point, $f(x)$ will have many trillions of decimal digits, which I dare say is a bit more than Excel can handle.