Can someone please help me with the calculation of this limit?
$$\lim_{x \to \infty} \frac{\Gamma(x+1)}{\Gamma(x+1+1/x^2)}$$
I tried wolframalpha and seems to be 1, yet there are no "detailed steps" as to how it reaches the conclusion (I understand there wouldn't be even if I pay, as in other cases it shows me the first ones).
Thanks in advance, Sergio
On
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\lim_{x \to \infty}\,\,{\Gamma\pars{x + 1} \over \Gamma\pars{x + 1 + 1/x^{2}}}} = \lim_{x \to \infty}\,\,{x! \over \pars{x + 1/x^{2}}!} \\[5mm] = &\ \lim_{x \to \infty}\,\,{\root{2\pi}x^{x + 1/2}\,\,\expo{-x} \over \root{2\pi}\pars{x + 1/x^{2}}^{x\ +\ 1/x^{2}\ + 1/2} \,\,\,\,\expo{-x - 1/x^{2}}\,\,} \\[5mm] = &\ \lim_{x \to \infty}\,\,{x^{x + 1/2} \over x^{x\ +\ 1/x^{2}\ + 1/2}\,\,\, \bracks{\pars{1 + 1/x^{3}}^{x^{3}}}^{1/x^{2}\ +\ 1/x^{5}\ + 1/\pars{2x^{3}}} \,\,\,\,}\,\expo{1/x^{2}} \\[5mm] = & \lim_{x \to \infty}x^{-1/x^{2}} = \exp\pars{-\lim_{x \to \infty}{\ln\pars{x} \over x^{2}}} = \exp\pars{-\lim_{x \to \infty}{1/x \over 2x}} \\[5mm] = & \bbx{1} \\ & \end{align}
$$\Gamma(z+\epsilon) =\Gamma(z) + \epsilon \Gamma'(z) + O(\epsilon^2) $$
$$\Gamma'(z) = \Gamma(z) \psi(z) $$
so the limit is
$$\lim_{x \to \infty} \frac1{1+ \psi(x+1)/x^2} = \lim_{x \to \infty} \frac1{1+ (\log{x}-\gamma)/x^2} = 1$$