Problem 32 - Essential Calculus Early Transcendentals

Question

\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_{n}=\frac{(-3)^{n}}{n !}$$

Step-by-Step Solution

Answer

The sequence converges and its limit is 0.

Step 1: Understand the Type of Sequence

The given sequence is \( a_n = \frac{(-3)^n}{n!} \). This sequence involves powers and factorials. The main task is to determine whether it converges or diverges as \( n \to \infty \), and if it converges, find its limit.

Step 2: Apply Ratio Test

One effective method to determine convergence of sequences involving factorials and powers is the Ratio Test. This test states that for \( a_n \), if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \) and \( L < 1 \), the series converges.

Step 3: Calculate Ratio of Successive Terms

Compute \( a_{n+1} \) as \( \frac{(-3)^{n+1}}{(n+1)!} \). Then find \( \frac{a_{n+1}}{a_n} \):\[\frac{a_{n+1}}{a_n} = \frac{(-3)^{n+1}}{(n+1)!} \div \frac{(-3)^n}{n!} = \frac{(-3)^{n+1} \cdot n!}{(-3)^n \cdot (n+1)!} = \frac{-3}{n+1}.\]

Step 4: Perform Limit Calculation

Calculate the limit:\[\lim_{n \to \infty} \left| \frac{-3}{n+1} \right| = \lim_{n \to \infty} \frac{3}{n+1} = 0.\]Since the limit \( 0 < 1 \), the sequence converges by the Ratio Test.

Step 5: Determine the Limit of the Sequence

Since the sequence converges to 0 (as shown in the previous step), the limit of the sequence \( a_n = \frac{(-3)^n}{n!} \) is determined as:\[\lim_{n \to \infty} a_n = 0.\]

Key Concepts

Understanding the Ratio Test

The ratio test is a powerful tool in determining whether a sequence or a series converges or diverges. It's particularly useful when dealing with factorials and powers, as these can significantly affect the behavior of a sequence. Here's how it works:

• You start with a sequence, say \( a_n \).
• The test involves finding the ratio of successive terms in the sequence: \( \left| \frac{a_{n+1}}{a_n} \right| \).
• You then compute the limit of this ratio as \( n \) approaches infinity: \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• If this limit \( L \) is less than 1, the sequence converges. If \( L \) is greater than 1, the sequence diverges. If \( L = 1 \), then the test is inconclusive.

The elegance of the ratio test lies in its simplicity, especially with sequences that grow or shrink rapidly due to exponentials and factorials.

Factorials and Powers in Sequences

Understanding factorials and powers is crucial for analyzing sequences. A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).Factorials grow very quickly as \( n \) increases, often outpacing powers. This is a fundamental reason why many sequences involving factorials tend to shrink over time.Powers, on the other hand, are of the form \( x^n \), where \( x \) is raised to the power of \( n \). This can lead to rapid growth, but not as rapidly as factorials.

• In the sequence \( a_n = \frac{(-3)^n}{n!} \), the factorial in the denominator helps in making the sequence converge because it increases much faster than the exponential term in the numerator.
• This property makes factorials powerful in sequence convergence as they dominate the behavior by causing the terms to become smaller.

Performing the Limit Calculation

Calculating limits is a fundamental part of determining convergence in sequences. When applying limits, we often want to find out how terms behave as \( n \) approaches infinity.In the sequence \( a_n = \frac{(-3)^n}{n!} \), after applying the ratio test, you are left with the expression \( \frac{-3}{n+1} \). To determine the convergence:

• Consider the absolute value: \( \left| \frac{-3}{n+1} \right| = \frac{3}{n+1} \).
• Given that \( n \) is going to infinity, \( n+1 \) grows very large, making \( \frac{3}{n+1} \) approach 0.
• Mathematically, the limit is calculated as \( \lim_{n \to \infty} \frac{3}{n+1} = 0 \), which shows that the sequence converges to 0.

This process helps us conclude that the terms of the sequence diminish as \( n \) increases, thus confirming convergence.