Question

Use Stirling's formula to evaluate \(\lim _{n \rightarrow \infty} \frac{(2 n) ! \sqrt{n}}{2^{2 n}(n !)^{2}}.\)

Short answer

\(\frac{1}{\sqrt{\pi}}\)

Step-by-step solution

Understand Stirling's Formula

Stirling's formula approximates factorials and is given by: \[ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \]

Apply Stirling's Formula to Factorials

Apply Stirling's formula to each factorial in the expression: \[ (2n)! \approx \sqrt{2 \pi (2n)} \left( \frac{2n}{e} \right)^{2n} \] \[ n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \]

Substitute into the Given Expression

Substitute the approximations into the given limit: \[ \frac{(2n) ! \sqrt{n}}{2^{2n}(n !)^{2}} \approx \frac{\sqrt{2 \pi (2n)} \left( \frac{2n}{e} \right)^{2n} \sqrt{n}}{2^{2n} \left( \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n \right)^2} \]

Simplify the Expression

Simplify the expression: \[ \frac{\sqrt{2 \pi (2n)} \left( \frac{2n}{e} \right)^{2n} \sqrt{n}}{2^{2n}(2 \pi n) \left( \frac{n}{e} \right)^{2n}} \approx \frac{\sqrt{2 \pi (2n)} \cdot 2^{2n} n^{2n} e^{-2n} \cdot \sqrt{n}}{2^{2n} \cdot 2 \pi n \cdot n^{2n} e^{-2n}}\]

Cancel Out Common Terms

Cancel out common terms in the numerator and the denominator: \[ \approx \frac{\sqrt{2 \pi (2n)} \cdot \sqrt{n}}{2 \pi n} \]

Further Simplify the Expression

Combine and simplify the remaining terms: \[ \approx \frac{\sqrt{2 \pi } \sqrt{2n} \sqrt{n}}{2 \pi n} = \frac{\sqrt{2 \pi } \sqrt{2} \sqrt{n^2}}{2 \pi n} = \frac{\sqrt{2 \pi } \sqrt{2} \cdot n}{2 \pi n} = \frac{\sqrt{4 \pi }}{2 \pi} = \frac{2 \sqrt{\pi}}{2 \pi} = \frac{1}{\sqrt{\pi}} \]

Evaluate the Limit

Evaluate the simplified limit as \(n \to \infty\): \[ \lim _{n \rightarrow \infty} \frac{1}{\sqrt{\pi}} = \frac{1}{\sqrt{\pi}} \]

Key concepts

Factorial Approximation

In mathematics, factorials are a special product represented by the symbol '!'. Factorials grow very quickly as numbers increase, which makes them important but challenging to work with directly. Stirling's formula is a key tool for approximating factorials, especially when dealing with large numbers. It allows us to simplify factorial expressions using an approximation: \( n! \approx\sqrt{ 2 \pi n } \left( \frac{ n }{ e } \right)^ n \) This approximation becomes extremely useful in various mathematical contexts, such as limit evaluations and asymptotic analysis. By substituting this formula into complex expressions involving factorials, we can make otherwise intractible problems much more manageable.

Limit Evaluation

When dealing with limits, especially involving factorials, it is often useful to use approximations like Stirling's formula. For the given exercise: \(\lim_{ n \to \infty } \frac{ (2 n)!\sqrt{ n } }{ 2^{2 n} (n! )^2 }\), we first substitute Stirling's approximations into the factorial expressions:

• (2n)! \approx\sqrt{ 2 \pi (2n) } \left( \frac{ 2n }{ e } \right)^ {2n}
• n! \approx\sqrt{ 2 \pi n } \left( \frac{ n }{ e } \right)^ n

Then, substituting these into the limit, allows for simplification: \frac{ \sqrt{ 2 \pi (2n) } \left( \frac{ 2n }{ e } \right)^ {2n} \sqrt{ n } }{ 2^{2n} \left( \sqrt{ 2 \pi n } \left( \frac{ n }{ e } \right)^ n \right)^ 2 }. This substitution and subsequent cancellation and simplification lead us to evaluate the limit successfully.

Asymptotic Analysis

Asymptotic analysis involves studying the behavior of functions as variables approach certain limits, often infinity. When n becomes very large, Stirling's formula provides a way to understand the factorial growth rates in a simpler manner. In our problem, evaluating the following limit: \(\lim_{ n \to \infty } \frac{ (2 n)!\sqrt{ n } }{ 2^{2 n} (n! )^2 }\) by substituting Stirling's formula helps to perform this asymptotic analysis. By expressing the factorial terms using Stirling's approximations, we can simplify the expression and focus on the leading terms. This approach not only simplifies the limit evaluation but also highlights how terms behave relatively to each other as n grows.

Simplifying Algebraic Expressions

To solve complex mathematical problems, it is crucial to simplify algebraic expressions effectively. In our exercise, once we have substituted Stirling's approximations for the factorials, the expression initially looks daunting. However, recognizing common elements in the numerator and the denominator allows us to cancel or combine terms step by step. After substituting, we carefully combine square roots and cancel matching exponential terms to simplify the expression. Ultimately, we reduce it to: \( \frac{ \sqrt{ 2 \pi (2n) } \left( \frac{ 2n }{ e } \right)^ {2n} \sqrt{ n } }{ 2^{2n}(2 \pi n ) \left( \frac{ n }{ e } \right)^ {2n }} \approx \frac{ \sqrt{ 2 \pi (2n) } \sqrt{ n } }{ 2 \pi n } \). Breaking down this simplification into more digestible steps makes the process clearer and demonstrates the power of algebraic manipulation in solving complex problems.