Question

Use Stirling’s formula to evaluate$\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{\left(}\mathbf{2}\mathbf{n}\mathbf{\right)}\mathbf{!}\sqrt{\mathbf{n}}}{{\mathbf{2}}^{\mathbf{2}\mathbf{n}}{\mathbf{(}\mathbf{n}\mathbf{!}\mathbf{)}}^{\mathbf{2}}}$.

Short answer

The value of the function is$\frac{1}{\sqrt{\mathrm{\pi }}}$.

Step-by-step solution

Given Information

The function is$\underset{\mathrm{n}\to \infty }{\lim }\frac{\left(2\mathrm{n}\right)!\sqrt{\mathrm{n}}}{2^{2\mathrm{n}}{(\mathrm{n}!)}^2}$.

Definition of the sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial.

$\mathit{n}\mathbf{!}\mathbf{\square }{\mathit{n}}^{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}}\sqrt{\mathbf{2}\mathbf{\pi n}}$.

Find the value of the function.

The function is $\underset{\mathrm{n}\to \infty }{\lim }\frac{\left(2\mathrm{n}\right)!\sqrt{\mathrm{n}}}{2^{2\mathrm{n}}{(\mathrm{n}!)}^2}$.

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

$\begin{array}{rcl}\left(2n\right)!& \approx & {\left(2n\right)}^n{e}^{-2n}\sqrt{4\mathrm{\pi n}}\\ {(n!)}^2& \approx & {\left(nne-n\sqrt{2\mathrm{\pi n}}\right)}^2\end{array}$

Substitute above valuesin the equation mentioned below.

$\begin{array}{rcl}\underset{n\to \infty }{lim}\frac{\left(2n\right)!\sqrt{n}}{2^{2n}{\left(n!\right)}^2}& =& \frac{{\left(2n\right)}^n{e}^{-2n}\sqrt{4\mathrm{\pi n}}}{{\left(n^n{e}^{-n}\sqrt{2\mathrm{\pi n}}\right)}^2}\\ & =& \frac{1}{\sqrt{\mathrm{\pi }}}\end{array}$

Hence, the value of the function is $\begin{array}{rcl}& & \frac{1}{\sqrt{\mathrm{\pi }}}\end{array}$.