Inequality involving the Gamma function : $e<\frac{\Gamma(x)e^x}{(x-1)^{(x-1)}\sqrt{x}}<e\sqrt{2\pi}$

Question

it's an inequality found with WA .

Claim

Let $x>1$ then we have : $$e<\frac{\Gamma(x)e^x}{(x-1)^{(x-1)}\sqrt{x}}<e\sqrt{2\pi}$$

My attempt

The RHS is a limit as $x\to \infty$ so I have tried the Hospital's rule we have :

$f(x)=\Gamma(x)e^x$ then :

$$f'(x)= \Gamma(x)e^x(\psi(x)+1)$$

On the other hand :

$g(x)=(x-1)^{(x-1)}\sqrt{x}$ then :

$$g'(x)=\frac{((x-1)^{(x-1)}(2x+2x\ln(x-1)+1))}{2\sqrt{x}}$$

But it gets nothing good .

I have tried also to differentiate the function :

$$h(x)=\frac{\Gamma(x)e^x}{(x-1)^{(x-1)}\sqrt{x}}$$

And prove that it's positive we have :

$$h'(x)=-\frac{(e^x(x-1)^{(1-x)}\Gamma(x)(2x\ln(x-1)-2x\psi(x)+1))}{(2x^{\frac{3}{2}})}$$

Now I think we can use the relation :

$$\psi(x+1)-\psi(x)=\frac{1}{x}$$ and use this link http://www.kurims.kyoto-u.ac.jp/EMIS/journals/JIPAM/images/139_05_JIPAM/139_05.pdf .

Question :

Can someone achieve this or have an alternative proof ?

Thanks in advance !

Regards Max .

Answer 1

You can use the fact that $$ \psi (x) > \log x - \frac{1}{{2x}} - \frac{1}{{12x^2 }} $$ for all $x>0$ (follows from http://dlmf.nist.gov/5.11.E2 and http://dlmf.nist.gov/5.11.ii). Hence, for $x>1$, $$ 2x\log (x - 1) - 2x\psi (x) + 1 < 2x\log \left( {1 - \frac{1}{x}} \right) + \frac{1}{{6x}} + 2 \\ < 2x\left( { - \frac{1}{x} - \frac{1}{{2x^2 }}} \right) + \frac{1}{{6x}} + 2 = - \frac{5}{{6x}} < 0. $$ This shows that $h'(x)>0$ for $x>1$, i.e., $h(x)$ is increasing. Now just note that $$ \mathop {\lim }\limits_{x \to 1 + } h(x) = e,\quad \mathop {\lim }\limits_{x \to + \infty } h(x) = e\sqrt {2\pi } $$ to finish the proof.