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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 $x \ge 785$:

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

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I have not found any mistakes.


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 $x\ge 785$.

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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

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

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