# Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$

Does it follow for $$x \ge 785$$, that Gautschi's Inequality implies that $$\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$$

Here's my reasoning. Please let me know if I made any mistakes or made any jumps in my logic.

(1) From Gautschi's Inequality, from any real $$z$$ and any real $$s$$ where $$0 < s < 1$$, it follows that:

$$z^s > \frac{\Gamma(z+s)}{\Gamma(z)} > (z)(z+1)^{s-1}$$

(2) Setting $$z = 2x+2$$ gives us:

$$(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+2)} > (2x+2)(2x+3)^{s-1}$$

(3) Multiplying $$2x+1$$ to both sides:

$$(2x+1)(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{s-1}$$

(4) Since $$\dfrac{1.25506}{\ln x} < 1$$ for $$x \ge 4$$, setting $$s = 1 - \dfrac{1.25506}{\ln x}$$ gives us:

$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{-\frac{1.25506}{\ln x}} = \frac{4x^2+6x+2}{(2x+3)^{\frac{1.25506}{\ln x}}}$$

(5) Since for $$x \ge 785$$ (see here for details), $$(2x+3)^{\frac{1.25506}{\ln x}} < 4$$, it follows that for $$x \ge 785$$:

$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > x^2$$

In the following, let us prove, with the help of WolframAlpha, that if $$x \ge 785$$, then $$(2x+3)^{\frac{1.25506}{\ln x}} < 4$$ since it seems that you forgot to show the details.

The inequality is equivalent to $$f(x)\gt 0$$ where $$f(x)=(\ln 4)(\ln x)-1.25506\ln(2x+3)$$ with $$f'(x)=\frac{(\ln(16)-2.51012)x+3\ln 4}{x(2x+3)}\gt 0$$ since $$\ln(16)-2.51012\gt 0\$$ (see here).

Since $$f(x)$$ is increasing with $$f(785)=(\ln 4)(\ln 785)-1.25506\ln(1573)\gt 0$$ (see here), it follows that $$f(x)\gt 0$$ for $$x\ge 785$$.

It seems to me that the inequality holds even for smaller values of $$x$$.

Considering that we look for the zero of function $$f(x)=\log \left(\frac{\Gamma \left(2x+3-\frac{a}{\log (x)}\right)}{\Gamma (2 x+1)}\right)-2\log(x)$$ and using Stirling approximation plus Taylor series $$f(x)=-\left(\frac{a \log (2)}{\log (x)}+a-2\log (2)\right)+\frac{(3 \log (x)-a) (2 \log (x)-a )}{4 x \log ^2(x)}+O\left(\frac{1}{x^2}\right)$$ an overestimate of the solution is given by $$2^{-\frac{a}{a-2 \log (2)}}$$ which, for $$a=1.25506$$ gives $$756.66$$.

Newton iterates are

$$\left( \begin{array}{cc} n & x_n \\ 0 & 756.000 \\ 1 & 455.485 \\ 2 & 512.524 \\ 3 & 517.233 \\ 4 & 517.260 \end{array} \right)$$