Problem 56 - Calculus: Early Transcendentals

Question

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 to 0.

Step 1: Understand the Sequence

The given sequence is \( a_n = \frac{(-3)^n}{n!} \). This sequence is defined for each natural number \( n \). Here, the numerator is \((-3)^n\), which means it alternates between positive and negative values, growing quickly. The denominator is \(n!\), which is the factorial of \(n\) and it grows very rapidly as \(n\) increases.

Step 2: Analyze the Behavior of the Sequence

To determine convergence, we analyze the behavior of the sequence as \( n \rightarrow \infty \). Notice that the denominator, \(n!\), grows much faster than the numerator \((-3)^n\). This growth means that the fraction \( \frac{(-3)^n}{n!} \) gets smaller as \(n\) increases.

Step 3: Use the Ratio Test for Series

The Ratio Test is useful for determining convergence of series and can also be applied here to confirm the sequence's behavior. Calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). For this sequence, \( a_n = \frac{(-3)^n}{n!} \), so \( a_{n+1} = \frac{(-3)^{n+1}}{(n+1)!} \). The ratio becomes:\[\frac{a_{n+1}}{a_n} = \frac{(-3)^{n+1}}{(n+1)!} \cdot \frac{n!}{(-3)^n} = \frac{-3}{n+1}.\] 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 is 0, which is less than 1, the sequence converges.

Step 4: Conclude the Limit of the Sequence

Since the Ratio Test confirmed that the sequence converges, and since the exponential growth of \(n!\) in the denominator outpaces any contribution from \((-3)^n\) in the numerator, the sequence effectively approaches 0. Therefore, the limit of the sequence as \( n \to \infty \) is 0.

Key Concepts

Understanding Factorial Growth

Factorial growth is a fascinating and crucial concept in mathematics, especially when analyzing sequences and series. The factorial of a number, denoted as \(n!\), represents the product of all positive integers from 1 to \(n\). This means \(n! = n \times (n - 1) \times (n - 2) \times \ldots \times 2 \times 1\).
\(n!\) grows incredibly quickly, much faster than exponential growth. As \(n\) increases, the factorial value increases at a rapid speed. For example, \(3! = 6\), but \(5! = 120\), and by the time you reach \(10!\), you get 3,628,800.
This rapid increase makes factorial a powerful denominator in sequences, often causing the sequence to shrink or converge as \(n\) becomes very large, due to the overwhelming size of \(n!\) compared to other factors in the sequence.
In the sequence \(a_n = \frac{(-3)^n}{n!}\), the denominator \(n!\) grows much faster than the numerator \((-3)^n\), leading the sequence to converge to zero.

Exploring the Ratio Test

The Ratio Test is a method used to determine whether a series converges or diverges by examining the ratio of successive terms. Although it's primarily used for series, it is also handy for sequences in some contexts. The test works by comparing the absolute value of the ratio between consecutive terms. For a series or sequence \(a_n\), you calculate \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\).
If this limit is less than 1, the series or sequence is said to converge. If it is greater than 1, it diverges, and if it equals 1, the test is inconclusive.
In our sequence \(a_n = \frac{(-3)^n}{n!}\), applying the Ratio Test involves computing:

• \(a_{n+1} = \frac{(-3)^{n+1}}{(n+1)!}\)
• \(\frac{a_{n+1}}{a_n} = \frac{(-3)^{n+1}/(n+1)!}{(-3)^n/n!} = \frac{-3}{n+1}\)
• The limit: \(\lim_{n \to \infty} \left| \frac{-3}{n+1} \right| = \lim_{n \to \infty} \frac{3}{n+1} = 0\)

Since the limit is 0, which is less than 1, the sequence converges.

Sequence Analysis for Convergence

Sequence analysis helps in understanding whether a sequence converges to a limit or diverges. It's a process of examining the components of a sequence to predict its long-term behavior as \(n\) approaches infinity.
In sequence \(a_n = \frac{(-3)^n}{n!}\), the analysis starts by considering the behavior and relative size of the terms in the sequence:

• The numerator \((-3)^n\): This grows in magnitude as \(n\) increases, but alternates between positive and negative values.
• The denominator \(n!\): As previously explored, grows extremely fast, much faster than any polynomial or exponential function like \((-3)^n\).

As \(n\) becomes larger, the factorial growth in the denominator dominates the behavior of the sequence, overpowering the influence of the numerator. This imbalance means that each term of the sequence gets smaller, pulling the whole sequence towards zero. We can thus conclude that the sequence converges, and the limit is 0. This keen insight is confirmed by both behavioral analysis and the Ratio Test, ensuring a solid understanding of the sequence's convergence.