Problem 44 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{(n-1) !}{(n+2) !} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{(n-1)!}{(n+2)!} \) converges by the comparison test with a convergent p-series.

Step 1: Identify the General Term

The general term of the series is \( a_n = \frac{(n-1)!}{(n+2)!} \). Simplifying it gives \( a_n = \frac{1}{(n+2)(n+1)n} \).

Step 2: Apply the Ratio Test

The ratio test involves finding the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Substitute \( a_{n+1} = \frac{n!}{(n+3)!} = \frac{1}{(n+3)(n+2)(n+1)} \). Then compute the ratio \[ \frac{a_{n+1}}{a_n} = \frac{1}{(n+3)(n+2)(n+1)} \times \frac{(n+2)(n+1)n}{1} = \frac{n}{(n+3)} \].

Step 3: Take the Limit of the Ratio

Calculate \( L = \lim_{n \to \infty} \frac{n}{n+3} \). As \( n \to \infty \), \( \frac{n}{n+3} \to 1 \). Thus, \( L = 1 \).

Step 4: Conclude with the Ratio Test

Since \( L = 1 \), the ratio test is inconclusive. We cannot determine convergence or divergence from the ratio test directly.

Step 5: Apply the Alternating Series Test

We shift to the alternating series test because of inconclusive ratio test results. However, the given series is not alternating nor easily applicable to alternating series test directly.

Step 6: Use Comparison Test

Let's compare \( a_n \) with \( \frac{1}{n^3} \), as we have \[ \frac{1}{(n+2)(n+1)n} \approx \frac{1}{n^3} \text{ for large } n \]. The series \( \sum \frac{1}{n^3} \) is a p-series with \( p = 3 > 1 \), which converges.

Step 7: Decide Using Comparison Test

Since \( \frac{1}{(n+2)(n+1)n} < \frac{1}{n^3} \) for all \( n \geq 1 \) and \( \sum \frac{1}{n^3} \) converges, by the comparison test, the original series \( \sum \frac{(n-1)!}{(n+2)!} \) also converges.

Key Concepts

Ratio Test

The ratio test is a handy tool to determine whether a series converges or diverges. When applying the ratio test, we compute a limit, denoted as \( L \), which is the limit of the absolute value of the ratio of successive terms. To break it down:

• Compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \).
• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L = \infty \), the series diverges.
• If \( L = 1 \), the test is inconclusive.

In this instance, after finding \( L = 1 \), the test couldn't help us determine convergence or divergence. This outcome requires us to explore other methods, as the ratio test alone cannot always provide a definitive answer.

Comparison Test

The comparison test is another valuable technique when analyzing the convergence behavior of a series. It's practical when comparing a complex series to another "test" series that is easier to understand. The steps include:

• Identify a series \( b_n \) that is similar to \( a_n \) and whose convergence is known.
• If \( 0 \leq a_n \leq b_n \) for all large \( n \), and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( a_n \geq b_n \geq 0 \) for all large \( n \), and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

In our original series, we compared it to the simpler \( \frac{1}{n^3} \) series, which is known to converge as it is a p-series with \( p = 3 > 1 \). The comparison with this convergent series confirmed that our challenging series must also converge.

P-series

A p-series is a specific and common type of series characterized by the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \). Understanding the behavior of a p-series is essential because it often serves as a benchmark for comparison in series tests. The rules are simple:

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

The understanding of p-series heavily informed the decision-making in our solution. By identifying a suitable p-value from a simpler comparable series, we established a clear convergence benchmark. In this instance, the choice of comparing with \( \frac{1}{n^3} \) (a p-series with \( p = 3 \)) provided a straightforward path to the conclusion due to its convergence properties.