Question

Use Stirling’s formula to evaluate $\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{{\left(n!\right)}^{{\displaystyle \frac{\mathbf{1}}{\mathbf{n}}}}}{\mathbf{n}}$.

Short answer

The value of the function is$\frac{1}{e}$.

Step-by-step solution

Given Information

The function is$\underset{\mathrm{n}\to \infty }{\lim }\frac{{(\mathrm{n}!)}^{{\displaystyle \frac{1}{\mathrm{n}}}}}{\mathrm{n}}$.

Definition ofthe sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial. $\mathit{n}\mathbf{!}\mathbf{~}{\mathit{n}}^{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}}\sqrt{\mathbf{2}\mathbf{\pi n}}$.

Find the value of the function.

The function is $\underset{\mathrm{n}\to \infty }{\lim }\frac{{(\mathrm{n}!)}^{{\displaystyle \frac{1}{\mathrm{n}}}}}{\mathrm{n}}$.

The sterling’s formula is $n!~n^{\mathrm{n}}{e}^{-\mathrm{n}}\sqrt{2\mathrm{\pi n}}$.

Substitute above values in the equation mentioned below.

$$\begin{gathered} \underset{\mathrm{n}\to \infty }{\lim }\frac{{(\mathrm{n}!)}^{{\displaystyle \frac{1}{\mathrm{n}}}}}{\mathrm{n}}=\underset{\mathrm{n}\to \infty }{\lim }\frac{{\left({\left(\mathrm{n}\right)}^n{e}^{-n}\sqrt{2\mathrm{\pi n}}\right)}^{{\displaystyle \frac{1}{\mathrm{n}}}}}{\mathrm{n}} \\ \underset{\mathrm{n}\to \infty }{\lim }\frac{{(\mathrm{n}!)}^{{\displaystyle \frac{1}{\mathrm{n}}}}}{\mathrm{n}}=\underset{\mathrm{n}\to \infty }{\lim }{e}^{-1}{\left(2\mathrm{\pi n}\right)}^{\frac{1}{2n}} \\ \underset{\mathrm{n}\to \infty }{\lim }\frac{{(\mathrm{n}!)}^{{\displaystyle \frac{1}{\mathrm{n}}}}}{\mathrm{n}}=\frac{1}{e} \end{gathered}$$

Hence, the value of the function is $\frac{1}{e}$.