How find the limit of $\lim\limits_{n\to \infty }\left(n-1\right)!$

Question

I have the next limit: $$\lim\limits_{n\to \infty }\left(\sqrt[n]{\left(\frac{n!-1}{n!+1}\right)^{\left(n+1\right)!}}\right)$$

I had done some steps and simplified it to: $$\lim\limits_{n\to \infty }\left(1-\frac{2}{n!+1}\right)^{(n+1)(n-1)!}=\\ \lim\limits_{n\to \infty }\left(1-\frac{1}{(n(n-1)!+1)\cdot 0.5}\right)^{(n+1)(n-1)!}$$

And my final result is:

$$\lim\limits_{n\to \infty }\left(\frac{1}{e}\right)^{\frac{3n+2+\frac{1}{(n-1)!}}{2}}$$

My question is what happens to $\frac{1}{(\infty -1)!}$? Is it 0?

Answer 1

$$\lim _{n\to \infty }\left(\sqrt[n]{\left(\frac{n!-1}{n!+1}\right)^{\left(n+1\right)!}}\right)\\=\lim _{n\to \infty }\left({\left(1-\frac{2}{n!+1}\right)^{n!+1}}\right)^{\frac{\left(n+1\right)!}{n!+1}\frac1{n}}\\=\lim _{n\to \infty }\left({\left(1-\frac{2}{n!+1}\right)^{n!+1}}\right)^{\frac{1}{1+1/n!}\frac{n+1}{n}}=e^{-2}$$

Answer 2

We have that

$${\left(\frac{n!-1}{n!+1}\right)^{\left(n+1\right)!}}={\left(1-\frac{2}{n!+1}\right)^{\left(n+1\right)!}}=e^{{\left(n+1\right)!}\log\left(1-\frac{2}{n!+1}\right)}=e^{-\frac{2(n+1)!}{n!+1}+O\left(\frac{(n+1)!}{(n!+1)^2}\right)}\sim e^{-2n}$$

therefore

$$\left(\sqrt[n]{\left(\frac{n!-1}{n!+1}\right)^{\left(n+1\right)!}}\right) \sim\sqrt[n]{e^{-2n}}\to \frac1{e^2}$$

Answer 3

Rewrite:

$[\dfrac{(1-1/n!)^{n!}}{(1+1/n!)^{n!}}]^{(1+1/n)}.$

$a_n:=\dfrac{(1-1/n!)^{n!}}{(1+1/n!)^{n!}}.$

Then $(a_n)(a_n)^{1/n}$.

Note :

$\lim_{n \rightarrow \infty} a_n= $

$\dfrac{\lim_{ n \rightarrow \infty}(1-1/n!)^{n!}}{\lim_{ n \rightarrow \infty}(1+1/n!)^{n!}}=$

$\dfrac{e^{-1}}{e^{1}} =e^{-2}$, i.e. bounded.

For $n$ large enough: $L < a_n < U$, where $L,U >0$, real, are bounds.

Then

$L^{1/n}a_n \lt a_n (a_n)^{1/n} < a_n U^{1/n}$.

Take the limit.

Used: $\lim_{x \rightarrow \infty}(1+a/x)^x= e^a$, $a$ real,

and $\lim_{n \rightarrow \infty}A^{(1/n)} =1$ , $A >0$, real.