Question

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit. $$ a_{n}=\frac{(n !)^{2}}{(2 n) !} $$

Short answer

The sequence converges to 0.

Step-by-step solution

Understand the Sequence

The sequence given is \(a_n = \frac{(n!)^2}{(2n)!}\). We need to identify if this sequence converges or diverges and, if it converges, determine its limit.

Apply Stirling's Approximation

Stirling's approximation states for large \(n\), \(n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n\). Using this for both \(n!\) and \((2n)!\), we have:- \((n!)^2 \approx (\sqrt{2\pi n} \left(\frac{n}{e}\right)^n)^2 = (2\pi n) \left(\frac{n}{e}\right)^{2n}\)- \((2n)! \approx \sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n} = 2^{2n} n^{2n} \sqrt{4\pi n}\)

Simplify the Sequence Using Approximation

Substituting the approximations into our sequence, we get:\[a_n = \frac{(2\pi n) \left(\frac{n}{e}\right)^{2n}}{2^{2n} n^{2n} \sqrt{4\pi n}} = \frac{2\pi n \cdot n^{2n} \cdot e^{-2n}}{2^{2n} \cdot n^{2n} \cdot \sqrt{4\pi n}}\]This simplifies to:\[a_n = \frac{e^{-2n} \cdot 2 \pi n}{2^{2n} \cdot \sqrt{4\pi n}}\]

Analyze the Limit of Simplified Sequence

Further simplifying gives us:\[a_n = \frac{(2\pi n)}{2^{2n} \sqrt{4\pi n}} \cdot e^{-2n} = \frac{\sqrt{\pi} \cdot n^{3/2}}{2^{2n}}\]As \(n\) becomes very large, the exponential term \(2^{2n}\) in the denominator grows much faster than the polynomial \(n^{3/2}\) in the numerator. Therefore, the sequence tends to 0.

Conclusion of Limit

Since the exponential growth in the denominator overtakes the polynomial growth in the numerator as \(n\) approaches infinity, the sequence converges to 0.

Key concepts

Stirling's approximation

Stirling's approximation is a useful tool in mathematics, especially when dealing with large numbers. It helps to approximate the factorial of a number, denoted as \( n! \). The formula used for Stirling's approximation is:

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

This approximation becomes more accurate as \( n \) increases. It's particularly helpful for simplifying problems that involve large factorials.
In the given exercise, Stirling's approximation allows us to simplify complex expressions like \((n!)^2\) and \((2n)!\). By substituting these approximations, it becomes easier to analyze the sequence and its behavior as \( n \) tends towards infinity.
The benefit of using Stirling's approximation in this context is that it transforms an otherwise very challenging problem into a more manageable form, enabling further simplification and analysis of the sequence.

Factorials in sequences

Factorials in sequences often appear in problems dealing with combinations, permutations, or series. They are a product of all positive integers up to a given number \( n \). The expression \( n! \) grows very quickly as \( n \) increases.
When factorials occur in sequences, they can dominate the behavior of the sequence, often making it difficult to analyze directly. In the original exercise, the sequence is defined by \( a_n = \frac{(n!)^2}{(2n)!} \). This involves both \( n! \) and \( (2n)! \), significantly impacting the nature of the sequence.
To handle sequences involving factorials, simplification techniques like Stirling's approximation become essential. These allow a deeper analysis by converting the factorials into expressions that can be more readily compared or simplified.

Asymptotic analysis

Asymptotic analysis is a powerful technique to understand the behavior of sequences or functions as they progress towards a limit, often as \( n \) approaches infinity. It focuses on approximating the main term or growth rates involved.
In mathematical problems, particularly with sequences, the approach often taken is to simplify the terms in a way that reveals the leading contributions that define the limit behavior.

• For instance, in the original problem, \( a_n = \frac{(n!)^2}{(2n)!} \), asymptotic analysis was used to determine that the growth rate of the denominator, \( 2^{2n} \), outpaces the numerator, resulting in the term approaching zero as \( n \) increases.

By focusing on the dominant terms, asymptotic analysis allows for more straightforward predictions about convergence or divergence without requiring an exact solution for each case. In the context of the exercise, it led us to conclude that the sequence converges to zero, illustrating both convergence and the capacity of asymptotic methods to simplify complex calculations.