Infinite Limit Involving a Factorial

Question

$$\lim_{x\to\infty} \left(\frac{x+1}{(x!)^{\frac{1}{x-1}}}\right) $$

I'm pretty sure that this approaches infinity, yet I'm having trouble proving it. Hope someone can help, thanks.

Answer 1

As by Stirling's Approximation for $\lim_{n\to\infty},$

$n!\approx\sqrt{2\pi}\left(\dfrac{n^{n+\dfrac12}}{e^n}\right)$

$$\lim_{x\to\infty}\dfrac{x+1}{(x!)^{\dfrac1{x-1}}}$$

$$=\dfrac{\lim_{x\to\infty}(x+1)\lim_{x\to\infty}e^{\dfrac1{1-\dfrac1x}}}{\lim_{x\to\infty}(\sqrt{2\pi})^{\dfrac1{x-1}} \lim_{x\to\infty}x^{1+\frac1{2x}}}$$

$$=\dfrac{e\cdot\lim_{x\to\infty}\left(1+\dfrac1x\right)}{\lim_{x\to\infty}x^{1/2x}}$$

Use this $\lim_{x\to\infty}x^{\dfrac1x}=1$

Answer 2

$$a=\frac{x+1}{(x!)^{\frac{1}{x-1}}}\implies \log(a)=\log(x+1)-\frac{1}{x-1}\log(x!)$$

Using Stirling approximation anf then Taylor $$\log(a)=1+\frac{\log \left(\frac{1}{x^{3/2}}\right)+2-\frac{1}{2} \log (2 \pi )}{x}+O\left(\frac{1}{x^2}\right)$$ Then $\log(a)\to 1$