Evaluate $\lim_{x\to \infty}\ (x!)^{1/x}$

Solution 1:

Hint:

For any fixed $a>0$, we eventually have $x!>a^x$, which means eventually $(x!)^{1/x}>(a^x)^{1/x}=a$.

Solution 2:

$$A=(n!)^{\frac{1}{n}}\\ \ln(A)=\frac{1}{n}\ln(n!)$$ $$\lim_{n \rightarrow \infty} \ln(A)=\\\lim_{n \rightarrow \infty} \frac{1}{n}(ln(n!))=\\ \lim_{n \rightarrow \infty} \frac{ln(n)+ln(n-1)+ln(n-2)+....}{n}=\\ hop \\ \lim_{n \rightarrow \infty} \frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+...=\\ \lim_{n \rightarrow \infty}\sum_{k=2}^\infty \frac{1}{k}\to \infty \\ \ln(A) \to \infty \\A \to \infty$$

second way

we know $$\color{red} {\lim_{n \rightarrow \infty}\frac{1}{n}(n!)^\frac{1}{n}=\frac{1}{e}} \to \\$$so $$\lim_{n \rightarrow \infty}(n!)^\frac{1}{n}=\lim_{n \rightarrow \infty}\frac{n}{e}=\infty$$

Solution 3:

It is easy to show that $n!>\left(\frac n2\right)^{n/2}$. In fact, $(2n)!>n!\,n^n$.

Therefore, given any number $B>0$, however large, we have

$$\begin{align} \left(n!\right)^{1/n}&>\left(\frac n{2}\right)^{1/2}\\\\ &>B \end{align}$$

whenever $n>2B^2$. And we are done!

Solution 4:

Let $y = (x!)^{\frac{1}{x}}$. Then, $\ln y = \frac{\ln x!}{x}$. Using Sterlings approximation of $\ln x! \sim x\ln x - x$ gives $\ln y \sim \ln x-1$ as $x \to \infty$. This means $y \to \infty$.