Problem 44 - Thomas Calculus

Question

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{(n-1) !}{(n+2) !}\end{equation}

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{(n-1)!}{(n+2)!} \) converges.

Step 1: Understand the Series

The given series is \( \sum_{n=1}^{\infty} \frac{(n-1)!}{(n+2)!} \). A factorial \( n! \) is the product of all positive integers from 1 to \( n \).

Step 2: Simplify the General Term

We start by simplifying \( \frac{(n-1)!}{(n+2)!} \). Note that \((n+2)! = (n+2)(n+1)n(n-1)!\). Thus, \( \frac{(n-1)!}{(n+2)!} = \frac{1}{(n+2)(n+1)n} \).

Step 3: Apply the Comparison Test

We compare the series with \( \sum_{n=1}^{\infty} \frac{1}{n^3} \), a known convergent p-series with \( p = 3 > 1 \). Since \( \frac{1}{(n+2)(n+1)n} < \frac{1}{n^3} \) for all \( n \geq 1 \), and \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) converges, by the comparison test, the given series also converges.

Key Concepts

Comparison Test

The Comparison Test is a valuable method for determining the convergence or divergence of a series. This test involves comparing the given series to another series whose convergence behavior is already known. If we can establish a relationship between the two, it helps us infer the behavior of the initial series.
For the Comparison Test, consider two series with positive terms:

• Series A: \( \sum a_n \)
• Series B (Comparison Series): \( \sum b_n \)

The test states:

• If \( a_n \leq b_n \) for all \( n \) beyond some point, and if \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• Conversely, if \( a_n \geq b_n \) for all \( n \) beyond some point, and if \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

For our exercise, we compared the series \( \sum_{n=1}^{\infty} \frac{1}{(n+2)(n+1)n} \) with the known p-series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \), which converges because the exponent 3 is greater than 1. Since each term of our series is smaller than the corresponding term of the convergent p-series, the original series is also convergent.

Factorial

A factorial, denoted as \( n! \), is the product of all positive integers from 1 to \( n \). Factorials grow very rapidly as \( n \) increases. Here’s how factorial works:

• \( 0! = 1 \) (special case)
• \( 1! = 1 \)
• \( 2! = 2 \times 1 = 2 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• And so forth.

In the problem, we are given \( \frac{(n-1)!}{(n+2)!} \). We simplify this expression by realizing that \((n+2)! = (n+2)(n+1)n(n-1)! \). Therefore, the factorial expression can be simplified to \( \frac{1}{(n+2)(n+1)n} \). This simplification is crucial as it transforms a complex expression into something much easier to analyze for convergence.

P-Series

P-Series is a specific type of series that can be written in the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). The convergence of such a series depends on the value of \( p \).

• If \( p > 1 \), the series converges.
• If \( p \leq 1 \), the series diverges.

In our example, we used the p-series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) as a comparison. Since the exponent 3 is greater than 1, this p-series converges. Recognizing when you have a p-series allows you to quickly determine convergence or divergence without complex calculations.
By using a p-series in our comparison test, we leveraged known results about convergence, thus simplifying the process of determining the behavior of the original series given in our exercise.