Find the limit $\lim _{n \to \infty}(n!)^{{\frac{1}{n^2}}}$ 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.
 A: Let$$y=\lim _{n \to \infty}(n!)^{\large {\frac{1}{n^2}}}$$
Therefore
$$\ln y=\lim _{n \to \infty}\frac{1}{n^2}\sum_{r=0}^{n-1}\ln(n-r)$$
$$\ln y=\lim _{n \to \infty}\frac{\ln n}{n^2} - \frac1{n}\sum_{r=0}^{n-1}\frac1n\ln(1-\frac{r}{n})$$
$$\ln y=0- \lim _{n \to \infty}\frac1{n}\int_{0}^{1}\ln(1-x)dx$$
$$\ln y=0- \lim _{n \to \infty}\frac1{n}(-1)$$
$$\ln y=0- 0$$
$$y=e^0=\boxed{1}$$
A: 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
$$ \lim_{n\rightarrow\infty} (n!)^{\frac{1}{n^2}} = \lim_{n\rightarrow\infty} e^{ \frac{\ln (n!)}{n^2}} = e^0 = 1$$
A: $$(n!)^{\frac{1}{n^2}} \le (n^n)^{\frac{1}{n^2}}=n^{\frac{1}{n}}=e^{\frac{\ln n}{n}}$$
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 $\lim \limits_{n \to \infty} (n!)^{\frac{1}{n^2}}=1$
A: You can just squeeze it:
$$
1\leq(n!)^{1/n^2}\leq (n^n)^{1/n^2}=n^{1/n}\to 1.
$$
So $\lim_{n\to\infty} (n!)^{1/n^2}=1$.
A: 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$\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}} = 0$.
$\lim\limits_{n \to \infty}{L} = e^{\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}}}=e^0=1$
