Problem 39 - Thomas Calculus

Question

Which of the series Converge absolutely, which converge, and which diverge? Give reasons for your answers. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{(2 n) !}{2^{n} n ! n} $$

Step-by-Step Solution

Answer

The series converges conditionally by the alternating series test.

Step 1: Understanding the Problem

We need to determine whether the given series \( \sum_{n=1}^{\infty}(-1)^{n} \frac{(2 n) !}{2^{n} n ! n} \) converges absolutely, converges conditionally, or diverges. To do this, we can examine the terms of the series and apply appropriate convergence tests.

Step 2: Apply the Alternating Series Test

The series has the form \((-1)^n a_n\) with \(a_n = \frac{(2n)!}{2^n n! n}\). According to the alternating series test, this series will converge if \(a_n\) is a decreasing sequence and \(\lim_{n \to \infty} a_n = 0\).

Step 3: Check \(a_n\) Decreasing

Compute \(a_{n+1} / a_n\) to check if \(a_n\) is decreasing:\[\frac{a_{n+1}}{a_n} = \frac{(2(n+1))!}{2^{n+1} (n+1)! (n+1)} \cdot \frac{2^n n! n}{(2n)!}\]. Simplifying, we find that the ratio becomes less than 1 for large \(n\), confirming \(a_n\) is decreasing.

Step 4: Check the Limit of \(a_n\)

Calculate \(\lim_{n \to \infty} a_n\):Using Stirling's approximation, for large \(n\), \( (2n)! \approx \sqrt{4\pi n} \left( \frac{2n}{e} \right)^{2n} \) and \(n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^{n} \). This shows that \( a_n \to 0 \) as \(n \to \infty\).

Step 5: Determine Type of Convergence

Since \(a_n\) is decreasing and approaches 0, the alternating series converges. Next, check for absolute convergence: determine if \(\sum_{n=1}^{\infty} |a_n|\) converges. Simplify \(|a_n| = \frac{(2n)!}{2^n n! n}\) and use the ratio test.Compute\[\frac{|a_{n+1}|}{|a_n|} = \frac{(2n+2)(2n+1)}{4(n+1)} < 1 \text{ for large } n,\]indicating the series does not converge absolutely, as the terms grow. Thus, it converges conditionally.

Key Concepts

Alternating series test

The Alternating Series Test is a handy tool to check if a series converges. When you have a series that alternates signs, like \[\sum (-1)^n a_n\]you can use this test. This works especially well if you suspect the series might converge, but aren't sure if it converges absolutely or just conditionally. To apply the test, you need two main things. First, the sequence \(a_n\) should be decreasing. This means each term in \(a_n\) gets smaller as \(n\) increases. Secondly, the sequence \(a_n\) should approach 0 as \(n\) increases indefinitely.Here's the procedure:

• Determine if \(a_n\) decreases. You can do this by checking \(\frac{a_{n+1}}{a_n} < 1\) for large \(n\).
• Check whether \(\lim_{n \to \infty} a_n = 0\).

If both conditions are met, your alternating series converges! Remember, this only tells you about conditional convergence, not absolute convergence. That means the series converges, but the sum of absolute values might still diverge. For the series in our problem, both conditions are satisfied, confirming that the series converges by the Alternating Series Test.

Absolute convergence

Absolute convergence is a step up from regular convergence. A series \[\sum_{n=1}^{\infty} a_n\]converges absolutely if the series of absolute values\[\sum_{n=1}^{\infty} |a_n|\]converges. If this is the case, it means the series doesn't depend on the arrangement of terms to converge, which is a stronger condition than just converging conditionally.To check absolute convergence for the provided series, we removed the alternating sign and applied the Ratio Test to \[|a_n| = \frac{(2n)!}{2^n n! n}\]By computing the ratio\[\frac{|a_{n+1}|}{|a_n|}\]and simplifying, it was found that for large \(n\), the result was greater than 1. This means the terms of the series grow rather than shrink. Therefore, the series fails the test for absolute convergence. So, while it converges conditionally via the Alternating Series Test, it does not converge absolutely.

Stirling's approximation

Stirling's approximation is a great mathematical tool that helps simplify factorials, especially for large numbers. It's expressed as:\[n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n\]This is particularly useful in handling complex expressions involving large factorials. When you're dealing with complex series, Stirling's approximation can be a lifesaver to evaluate asymptotic behavior.For our problem, we used Stirling's approximation to simplify expressions like \[(2n)!\]Knowing:\[(2n)! \approx \sqrt{4\pi n} \left( \frac{2n}{e} \right)^{2n}\]And comparing it to:\[n! \approx \sqrt{2\pi n} \left( \frac{n}{e} \right)^n\]This approximation shows how quickly \(a_n\) approaches zero. It acted as a confirmation step when checking the convergence of our series. By applying Stirling’s approximation, we better understood why the terms \(a_n\) diminish to zero as \(n\) becomes very large, helping us conclude the nature of the series's convergence efficiently.