Question

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{(2 n) !}{n^{2 n}} $$

Short answer

The series diverges by the ratio test.

Step-by-step solution

Understand the Ratio Test

The ratio test states that for a series \( \sum_{n=1}^{\infty} a_n \), where \( a_n > 0 \), compute the limit \( L = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \). If \( L < 1 \), the series converges absolutely. If \( L > 1 \) or \( L \) is infinite, the series diverges. If \( L = 1 \), the test is inconclusive.

Define the Terms

For the given series \( \sum_{n=1}^{\infty} \frac{(2n)!}{n^{2n}} \), identify \( a_n = \frac{(2n)!}{n^{2n}} \) and \( a_{n+1} = \frac{(2(n+1))!}{(n+1)^{2(n+1)}} \).

Set Up the Ratio

Set up the ratio for the test: \[ \frac{a_{n+1}}{a_n} = \frac{\frac{(2(n+1))!}{(n+1)^{2(n+1)}}}{\frac{(2n)!}{n^{2n}}} = \frac{(2(n+1))! \cdot n^{2n}}{(2n)! \cdot (n+1)^{2n+2}} \].

Simplify the Ratio

Simplify the expression: \[ \frac{(2(n+1))(2n+1)! \cdot n^{2n}}{(2n)! \cdot (n+1)^{2n+2}} = \frac{(2n+2)(2n+1)n^{2n}}{(2n)!(n+1)^{2n+2}} \].

Take the Limit as n Approaches Infinity

Find the limit \( L = \lim_{n \to \infty} \frac{(2n+2)(2n+1)n^{2n}}{(2n)!(n+1)^{2n+2}} \). Use Stirling's approximation and algebraic simplification as necessary to find that the limit approaches infinity.

Conclude Using the Ratio Test

Since \( L \) is greater than 1 (in fact, it is infinite), the original series \( \sum_{n=1}^{\infty} \frac{(2n)!}{n^{2n}} \) diverges.

Key concepts

Convergence of Series

When we discuss the convergence of a series like \( \sum_{n=1}^{\infty} a_n \), we're essentially asking whether the sequence of partial sums \( S_n = a_1 + a_2 + \ldots + a_n \) approaches a specific value as \( n \) becomes very large. A convergent series results in such a sequence that approaches a finite limit.
A divergent series, on the other hand, has partial sums that either grow without bound or do not approach any specific number.
The Ratio Test is a common method used to determine convergence. By exploring the limit \( L = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} \):

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L \) is infinite, the series diverges.
• If \( L = 1 \), the test is inconclusive.

This tool proves especially handy for series involving complex algebraic or factorial structures.

Factorials in Series

Factorials, denoted by \( n! \), are products of all positive integers up to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). They often appear in series due to their presence in combinatorial and algebraic expressions.
In the given problem, we encounter \( (2n)! \), which significantly affects the growth rate of the terms in the series. Factorials grow very quickly, much faster than exponential sequences.
This rapid growth impacts the convergence or divergence of a series, as factorials can heavily dominate other components like powers of \( n \) or \( e^n \). Managing these terms effectively, sometimes with the help of approximations, is crucial in analyzing series behavior.

Infinity in Limits

Limits provide a way to explore the behavior of functions as their inputs become very large or very small. In this exercise, we used a limit in the Ratio Test:
\( L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} \).
When analyzing limits leading to infinity, we think about how the components of the expression behave for very large \( n \).

• Are they increasing or decreasing?
• Are certain terms growing more rapidly than others?

The outcomes help us decide if the series converges or diverges. If the limit yields infinity, one part of the series overtakes others, causing the terms to grow unbounded, leading to divergence. Understanding the component interactions in limits, always highlights the dominant factors.

Stirling's Approximation

Stirling's Approximation is a handy tool used to estimate factorials for large numbers, expressed as:
\( n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \).
This approximation simplifies the complex expressions involving factorials, making calculations in series analysis more manageable.
When factorials skyrocket in size quickly, a direct computation is unwieldy. Stirling's formula allows us to approximate and break down these massive terms in relation to powers and other functions, making limits easier to handle.
Applying Stirling's approximation helped us to efficiently predict how the terms of our series change as \( n \) grows very large, leading to more accurate conclusions about the series' convergence or divergence. It's like having a magnifying glass to inspect the intricate behaviors of factorial-term series!