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Find the limit $$\lim _{n \to \infty}(n!)^{{\frac{1}{n^2}}}$$

I have tried it using Stirling's approximation and then using L'Hospital's Rule and get answer $1$

Is there any other easy method to find this limit ?

Any hint. Thanks in advance.

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marked as duplicate by YuiTo Cheng, Misha Lavrov, Lord Shark the Unknown, Shogun, Lee David Chung Lin Jun 11 at 1:46

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  • $\begingroup$ The denominator is $2n$ or $n^2$? $\endgroup$ – Adam Latosiński Jun 8 at 8:58
  • $\begingroup$ try integration .. method using summation .. $\endgroup$ – learningstudent Jun 8 at 8:59
  • $\begingroup$ @AdamLatosiński its $n^2$ $\endgroup$ – Eklavya Jun 8 at 9:05
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You can just squeeze it: $$ 1\leq(n!)^{1/n^2}\leq (n^n)^{1/n^2}=n^{1/n}\to 1. $$ So $\lim_{n\to\infty} (n!)^{1/n^2}=1$.

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Let $L=({n!})^{\frac{1}{n^2}}$

$L=e^\frac{\log_{e}{n!}}{n^2}=e^\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}$

Now $0\leq\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}\leq \frac{n\log(n)}{n^2}=\frac{\log(n)}{n}$ which tends to $0$ as $n$ tends to $\infty$. Hence$\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}} = 0$.

$\lim\limits_{n \to \infty}{L} = e^{\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}}}=e^0=1$

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Let$$y=\lim _{n \to \infty}\Large(n!)^{\large {\frac{1}{n^2}}}$$ Therefore $$\ln y=\lim _{n \to \infty}\frac{1}{n^2}\sum_{r=0}^{n-1}\ln(n-r)$$ $$\ln y=\lim _{n \to \infty}\frac{\ln n}{n^2} - \frac1{n}\sum_{r=0}^{n-1}\frac1n\ln(1-\frac{r}{n})$$ $$\ln y=0- \lim _{n \to \infty}\frac1{n}\int_{0}^{1}\ln(1-x)dx$$ $$\ln y=0- \lim _{n \to \infty}\frac1{n}(-1)$$ $$\ln y=0- 0$$ $$y=e^0=\boxed{1}$$

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Using Stolz–Cesàro theorem:

\begin{align} \lim_{n\rightarrow\infty} \frac{\ln (n!)}{n^2} &= \lim_{n\rightarrow\infty} \frac{\ln ((n+1)!) - \ln(n!)}{(n+1)^2 - n^2} = \\ &= \lim_{n\rightarrow\infty} \frac{\ln (n+1)}{2n+1} = \\ &= \lim_{n\rightarrow\infty} \frac{\ln(n+2) - \ln (n+1)}{(2n+3) - (2n+1)} = \\ &= \lim_{n\rightarrow\infty} \frac{\ln(1+\frac{1}{n+1})}{2} = 0\end{align} So $$ \lim_{n\rightarrow\infty} (n!)^{\frac{1}{n^2}} = \lim_{n\rightarrow\infty} e^{ \frac{\ln (n!)}{n^2}} = e^0 = 1$$

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$$(n!)^{\frac{1}{n^2}} \le (n^n)^{\frac{1}{n^2}}=n^{\frac{1}{n}}=e^{\frac{\ln n}{n}}$$ Since $$1 \le (n!)^{\frac{1}{n^2}} \le e^{\frac{\ln n}{n}}$$ and $\lim \limits_{n \to \infty} e^{\frac{\ln n}{n}}=1$ then by squeeze theorem there is a limit $\lim \limits_{n \to \infty} (n!)^{\frac{1}{n^2}}=1$

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