Question about a function that is a ratio of gamma functions and appears to be strictly increasing for $x\ge 2$

Solution 1:

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.

Solution 2:

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.