Question

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

Short answer

The limit evaluates to \( \frac{1}{\sqrt{ \pi}} \).

Step-by-step solution

Stirling's Approximation

Use Stirling's approximation, which states that for large values of n, \[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \]

Apply Stirling's Approximation to Each Factor

Substitute Stirling's approximation into each factorial term in the limit expression. Start with \[(2n)! \approx \sqrt{2\pi (2n)} \left( \frac{2n}{e} \right)^{2n} \] and \[ n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \]

Substitute the Approximations into the Original Expression

Replace each factorial in the limit expression with its approximation: \[ \frac{(2n)! \sqrt{n}}{2^{2n} (n!)^2} \approx \frac{ \sqrt{2\pi(2n)} \left( \frac{2n}{e} \right)^{2n} \cdot \sqrt{n} }{ 2^{2n} \left( \sqrt{2\pi n} \left( \frac{n}{e} \right)^n \right)^2 } \]

Simplify the Expression

Combine and simplify the terms in the numerator and the denominator. Begin by simplifying the factorial components: \[ \frac{ \sqrt{2\pi (2n)} \left( \frac{2n}{e} \right)^{2n} \sqrt{n} }{ 2^{2n} (2\pi n) \left( \frac{n}{e} \right)^{2n} } = \frac{ \sqrt{2\pi (2n)} \cdot (2n)^{2n} \cdot \sqrt{n} \cdot e^{2n} }{ 2^{2n} \cdot 2\pi n \cdot n^{2n} \cdot e^{2n} } \]

Cancel Like Terms

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

Further Simplify

Further simplify the remaining terms: \[ \frac{ \sqrt{2\pi(2n)} \cdot \sqrt{n} }{ 2 \pi n } = \frac{ \sqrt{2\pi (2n) n} }{ 2 \pi n } = \frac{ \sqrt{2 \pi (2n) \cdot n} }{2 \pi n } = \frac{ \sqrt{2 \pi \cdot 2n^2} }{2 \pi n } \]

Final Simplification

Simplify the square root and divide to get the final limit: \[ \frac{ \sqrt{4 \pi n^2} }{ 2 \pi n } = \frac{ 2 \sqrt{ \pi n^2} }{ 2 \pi n } = \frac{ 2n \sqrt{ \pi}}{ 2 \pi n} = \frac{ \sqrt{ \pi}}{ \pi } = \frac{1}{\sqrt{ \pi}} \]

Key concepts

Factorial Simplification

Understanding factorials is crucial in many mathematical contexts, particularly in combinatorics and asymptotic analysis. The factorial of a number, denoted as \({{\text{{n}}!}}\), is the product of all positive integers up to that number. Simplifying factorials can be cumbersome, especially for large numbers. One useful tool for this is Stirling's Approximation, an estimation formula: \[ n! \approx \sqrt{2\pi n} \(\frac{n}{e}\)^{n} \] where \(e\) is the base of the natural logarithm. This approximation becomes more accurate as \(n\) increases. By replacing large factorials with their Stirling approximations, we can simplify and evaluate complex expressions more easily.

Limit Evaluation

In calculus, limits are used to describe the behavior of a function as it approaches a particular point, often infinity. Evaluating limits involving factorial expressions can be challenging because factorial values grow very quickly. To handle these, we use Stirling's Approximation to transform the factorials into a more manageable form.
For the given problem, the limit we want to evaluate is:
\( \lim_{n \rightarrow \infty} \frac{(2n)! \sqrt{n}}{2^{2n}(n!)^{2}} \).
By substituting Stirling's Approximation into this limit, we convert the original expression into one involving powers and roots, which is much easier to handle analytically. This allows us to simplify and eventually evaluate the limit as \(n\) approaches infinity.

Asymptotic Analysis

Asymptotic analysis deals with the behavior of functions as the input grows very large. Stirling's Approximation is particularly useful in asymptotic analysis of factorials, as it provides a simple formula to approximate the value of a factorial for large \(n\).
This approximation helps us to understand how factorials behave when entering into large-number limits. By converting the complicated factorial-based expression to a simpler one through asymptotic approximations, we can analyze its asymptotic behavior and draw conclusions about its final value as \(n\) tends towards infinity.
For engineers and scientists, grasping the concept of asymptotic behavior is key to solving problems involving large datasets or resource-intensive computations.

Infinite Limits

Infinite limits are an essential part of mathematical analysis involving functions that grow without bound. When dealing with infinite limits, we often use simplifying techniques to make our calculations easier. Stirling's Approximation is one of these techniques.
For the given limit:
\( \lim_{n \rightarrow \infty} \frac{(2n)! \sqrt{n}}{2^{2n}(n!)^{2}} \),
we initially replace the factorial terms with their Stirling approximations. This substitution reduces the complexity of computing the limit as \(n\) approaches infinity.
Through simplification steps (including canceling out terms and reducing expressions), we eventually determine the behavior of the function, which in this case, converges to \( \frac{1}{\sqrt{\pi}} \). Infinite limits let us understand how functions behave as their inputs grow very large, which is crucial in fields like economics, physics, and computer science.