Question

Use Stirling’s formula to evaluate$\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{(}\mathbf{n}\mathbf{+}{\displaystyle \frac{3}{2}}\mathbf{)}\mathbf{!}}{\sqrt{\mathbf{n}}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{!}}$.

Short answer

The value of the function is $e^{\frac{-3}{2}}$.

Step-by-step solution

Given Information

The function is$\underset{\mathrm{n}\to \infty }{\lim }\frac{(\mathrm{n}+{\displaystyle \frac{3}{2}})!}{\sqrt{\mathrm{n}}(\mathrm{n}+1)!}$.

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 $\underset{\mathrm{n}\to \infty }{\lim }\frac{(\mathrm{n}+{\displaystyle \frac{3}{2}})!}{\sqrt{\mathrm{n}}(\mathrm{n}+1)!}$.

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

$\begin{array}{rcl}\left(n+\frac{3}{2}\right)!& \approx & {\left(n+\frac{3}{2}\right)}^{n+\frac{\mathrm{\pi }}{2}}{e}^{-\left(n+\frac{3}{2}\right)}\sqrt{2\mathrm{\pi }\left(\mathrm{n}+\frac{3}{2}\right)}\\ \left(n!\right)& \approx & \left(n^n{e}^{-n}\sqrt{2\mathrm{\pi n}}\right)\end{array}$

Substitute above valuesin the equation mentioned below.

$$\begin{gathered} \underset{\mathrm{n}\to \infty }{\lim }\frac{(\mathrm{n}+{\displaystyle \frac{3}{2}})!}{\sqrt{\mathrm{n}}(\mathrm{n}+1)!}=\frac{{\left(n+{\displaystyle \frac{3}{2}}\right)}^{n+{\displaystyle \frac{\mathrm{\pi }}{2}}}{e}^{-\left(n+{\displaystyle \frac{3}{2}}\right)}\sqrt{2\mathrm{\pi }\left(\mathrm{n}+{\displaystyle \frac{3}{2}}\right)}}{n^n{e}^{-n}\sqrt{2}} \\ \underset{\mathrm{n}\to \infty }{\lim }\frac{(\mathrm{n}+{\displaystyle \frac{3}{2}})!}{\sqrt{\mathrm{n}}(\mathrm{n}+1)!}=e^{\frac{-3}{2}} \end{gathered}$$

Hence, the value of the function is $e^{\frac{-3}{2}}$.