Problem 56 - Thomas Calculus

Question

Convergence or Divergence Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{(-1)^{n}(3 n) !}{n !(n+1) !(n+2) !}$$

Step-by-Step Solution

Answer

The series converges by the Alternating Series Test.

Step 1: Identify the Series Type

The series given is \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(3n)!}{n!(n+1)!(n+2)!} \), suggesting an alternating series due to the presence of \((-1)^{n}\). Determine if the series is alternating, then check for conditional or absolute convergence.

Step 2: Alternating Series Test

Apply the Alternating Series Test. The series \( \sum (-1)^{n} a_n \) converges if the absolute term \( a_n \) satisfies: 1. \( a_{n+1} < a_n \) for all \( n \). 2. \( \lim_{n \to \infty} a_n = 0 \).First, set \( a_n = \frac{(3n)!}{n!(n+1)!(n+2)!}\) and check these conditions.

Step 3: Check Decreasing Condition

Derive \( a_n \) to determine whether \( a_{n+1} < a_n \). Use the property that factorial terms are significant in growth comparisons. As \(n\) increases, the denominator grows faster than the numerator, indicating that \( a_n \) is decreasing.

Step 4: Take the Limit

Find \( \lim_{n \to \infty} \frac{(3n)!}{n!(n+1)!(n+2)!} \). Simplifying this limit will show that the term approaches zero as \(n\) grows very large.

Step 5: Conclusion Based on Alternating Series Test

Since both conditions of the Alternating Series Test are met (terms \(a_n\) are decreasing and \( \lim_{n \to \infty} a_n = 0 \)), the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(3n)!}{n!(n+1)!(n+2)!} \) converges.

Key Concepts

Alternating Series Test

When dealing with series like \( \sum_{n=1}^{\infty} (-1)^{n} a_n \), the Alternating Series Test is a powerful tool. This test helps us determine the convergence of series where the terms alternate in sign. To apply this test successfully, two conditions need to be satisfied:

• The absolute value of the series terms \( a_n \) must be decreasing, meaning \( a_{n+1} < a_n \) for all \( n \), which indicates that every term is smaller than the preceding one.
• The terms must approach zero as \( n \) goes to infinity, or \( \lim_{n \to \infty} a_n = 0 \).

In the given series, these conditions were verified: the factorial terms in \( a_n \) ensure decreasing terms by their growth pattern, and the limit evaluation confirmed that the terms approach zero. Therefore, the series converges according to this test.

Factorial Growth

Factorials can grow incredibly fast, which plays a key role in evaluating series convergence. The factorial of a number \( n! \) is the product of all positive integers less than or equal to \( n \). This rapid growth often means that when factorials are in both the numerator and the denominator of a fraction, the denominator often grows faster and contributes to the series converging.

In our series \( \sum_{n=1}^{\infty} \frac{(3n)!}{n!(n+1)!(n+2)!} \), observe that the denominator consists of multiple factorial terms compared to a single factorial in the numerator.
This leads to a denominator that can easily surpass the numerator in growth, causing the terms \( a_n \) to decrease as \( n \) increases, contributing to the convergence of the series.

Limit of a Sequence

A crucial aspect of determining convergence is examining whether the terms of a sequence or series approach a certain value as \( n \) increases. This value is known as the limit of the sequence.

For the series in question, we calculate the limit \( \lim_{n \to \infty} \frac{(3n)!}{n!(n+1)!(n+2)!} \). Upon simplifying, the sequence reveals that as \( n \) becomes very large, the terms approach zero. When the sequence terms tend toward zero, it's a strong indicator that the series may converge.
This is one of the conditions verified by the Alternating Series Test, reinforcing the test's conclusion that the series converges.

Absolute Convergence

Absolute convergence refers to a series that converges even when all its terms are made positive. If a series \( \sum a_n \) absolutely converges, then \( \sum |a_n| \) also converges. This is stronger than conditional convergence, which only involves the original series converging.

To test for absolute convergence in the given series, it involves evaluating \( \sum \left| \frac{(-1)^{n}(3n)!}{n!(n+1)!(n+2)!} \right| = \sum \frac{(3n)!}{n!(n+1)!(n+2)!} \).
Given the rapid factorial growth in the denominator, analyzing just the positive terms reveals that they converge slower than the original alternating series. In our situation, we primarily focus on the original series, which indeed converges, but it does not exhibit absolute convergence, as this stronger condition is not met here.

Conditional Convergence

Conditional convergence occurs when a series converges, but its absolute counterpart does not. This means the series naturally depends on the alternating nature of its terms to sum to a finite value.

In this context, the series \( \sum_{n=1}^{\infty} \frac{(-1)^{n}(3n)!}{n!(n+1)!(n+2)!} \) is a perfect candidate for conditional convergence. The Alternating Series Test validated that the original series converges due to its alternating terms decreasing to zero. However, its absolute counterparts (positive terms) alone do not converge.
This is why the series is not absolutely convergent, but it conditionally converges by using the oscillating feature of positive and negative terms to converge to a particular value.