Chapter 11: Q11.4P (page 554)

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

Short Answer

Expert verified

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

Step by step solution

Given Information

The function is $\lim_{n \to \infty} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}}$ .

Definition of the sterling’s formula.

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

$n! □ n^{n} e^{- n} \sqrt{2 πn}$ .

Find the value of the function.

The function is $\lim_{n \to \infty} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}}$ .

The sterling’s formula is $n! □ n^{n} e^{- n} \sqrt{2 πn}$ .

$\begin{array}{rcl} (2 n)! & \approx & {(2 n)}^{n} e^{- 2 n} \sqrt{4 πn} \\ {(n!)}^{2} & \approx & (n) \end{array}$ ⁿe^-n2πn2

Substitute above valuesin the equation mentioned below.

$\begin{array}{rcl} \underset{n \to \infty}{l i m} \frac{(2 n)! \sqrt{n}}{2^{2 n} {(n!)}^{2}} & = & \frac{{(2 n)}^{n} e^{- 2 n} \sqrt{4 πn}}{{(n^{n} e^{- n} \sqrt{2 πn})}^{2}} \\ = & \frac{1}{\sqrt{π}} \end{array}$

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

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

Short Answer

Step by step solution

Given Information

Definition of the sterling’s formula.

Find the value of the function.

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Use Stirling’s formula to evaluatelimn→∞(2n)!n22n(n!)2.

Short Answer

Step by step solution

Given Information

Definition of the sterling’s formula.

Find the value of the function.

One App. One Place for Learning.

Most popular questions from this chapter

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