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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?

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  • $\begingroup$ Stirling's approximation? $\endgroup$ – tonychow0929 Dec 15 '18 at 9:52
  • $\begingroup$ I have not learned that yet, is there any other solution? $\endgroup$ – violettagold Dec 15 '18 at 9:53
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    $\begingroup$ If you only need $\lim_{n \to \infty} \frac{1}{(n-1)!}$, the limit is indeed $0$ because $0 < \frac{1}{(n-1)!} < \frac{1}{n-1} \, \forall n>2$. Now apply Squeeze Theorem. $\endgroup$ – tonychow0929 Dec 15 '18 at 9:58
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$$\lim _{n\to \infty }\left(\sqrt[n]{\left(\frac{n!-1}{n!+1}\right)^{\left(n+1\right)!}}\right)\\=\lim _{n\to \infty }\left({\left(1-\frac{2}{n!+1}\right)^{n!+1}}\right)^{\frac{\left(n+1\right)!}{n!+1}\frac1{n}}\\=\lim _{n\to \infty }\left({\left(1-\frac{2}{n!+1}\right)^{n!+1}}\right)^{\frac{1}{1+1/n!}\frac{n+1}{n}}=e^{-2}$$

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  • $\begingroup$ Thank you very much for the help, but I have just one question: Why $\frac{1}{1+\frac{1}{n!}}\cdot \frac{n+1}{n}$ is equal to 1 when $n→∞$? $\endgroup$ – violettagold Dec 15 '18 at 10:15
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    $\begingroup$ @violettagold Oh, I missed your question, you can write $1/n!=e^{-\ln{n!}}=e^{-\sum{\ln{n}}}$ to find your answer. $\endgroup$ – Nanayajitzuki Dec 15 '18 at 10:23
  • $\begingroup$ Thanks a lot :) $\endgroup$ – violettagold Dec 15 '18 at 10:24
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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

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

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    $\begingroup$ You are, once more, too fast for the old man ! Cheers and $+1$. $\endgroup$ – Claude Leibovici Dec 15 '18 at 10:06
  • $\begingroup$ @ClaudeLeibovici I wasn't so fast that time...and you are not old...we all are young on MSE...you are just very expert! :) $\endgroup$ – gimusi Dec 15 '18 at 10:12
  • $\begingroup$ Thank you very much :) $\endgroup$ – violettagold Dec 15 '18 at 10:19
  • $\begingroup$ @violettagold You are welcome! Bye $\endgroup$ – gimusi Dec 15 '18 at 10:24
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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.

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