Problem 10 - Calculus: Early Transcendentals

Question

Use the Ratio Test to determine whether the series is convergent or divergent. \( \displaystyle \sum_{n = 0}^{\infty} \frac {( - 3)^n}{(2n + 1)!} \)

Step-by-Step Solution

Answer

The series converges by the Ratio Test.

Step 1: Write the general term of the series

The general term of the series is given by \( a_n = \frac{(-3)^n}{(2n+1)!} \). This is the function we will analyze using the Ratio Test.

Step 2: Apply the Ratio Test

To use the Ratio Test, compute the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). We begin by finding \( a_{n+1} = \frac{(-3)^{n+1}}{(2(n+1)+1)!} = \frac{(-3)^{n+1}}{(2n+3)!} \).

Step 3: Compute the Ratio

Formulate the ratio \( \frac{a_{n+1}}{a_n} = \frac{(-3)^{n+1}}{(2n+3)!} \cdot \frac{(2n+1)!}{(-3)^n} = \frac{-3 \cdot (-3)^n}{(2n+3)(2n+2)(2n+1)!} \cdot \frac{(2n+1)!}{(-3)^n} \). Simplify to get \( \frac{-3}{(2n+3)(2n+2)} \).

Step 4: Take the limit

Now compute the limit: \( L = \lim_{n \to \infty} \left| \frac{-3}{(2n+3)(2n+2)} \right| = \lim_{n \to \infty} \frac{3}{(2n+3)(2n+2)} \). As \( n \to \infty \), the denominator grows without bound, causing \( L \to 0 \).

Step 5: Determine convergence from the ratio

According to the Ratio Test, if \( L < 1 \), then the series converges. Here, \( L = 0 < 1 \), indicating that the series converges.

Key Concepts

Series Convergence

When studying infinite series, it is crucial to determine if a series converges or diverges. Convergence means that as we add more and more terms of the series, the sum approaches a specific number. If a series diverges, its sum does not settle at any value but instead grows indefinitely or oscillates.

Here, we utilize the **Ratio Test**, a popular method for assessing convergence. The Ratio Test is effective when the series involves factorials or exponential sequences. It examines the limit of the ratio of successive terms.

To apply it, consider a series with terms labeled as \(a_n\). If - the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges.- If \( L > 1 \), the series diverges.- If \( L = 1 \), the test is inconclusive.

In the given exercise, the series converges because the limit \( L = 0 \), which is definitely less than 1. Understanding convergence helps us determine the behavior and potential sums of infinite series.

Factorial in Series

Factorials often appear in series, especially in mathematical and engineering contexts. They are represented as \((n!)\), which means the product of all positive integers up to \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow rapidly as \(n\) increases.

This characteristic makes factorials particularly influential in series, affecting the convergence of a series.

In the exercise at hand, the term of the series \(a_n = \frac{(-3)^n}{(2n+1)!}\) includes a factorial in the denominator. This factorial controls the growth of the terms as \(n\) increases, eventually making them very small. This rapid decrease in term size due to the factorial is essential in determining the series convergence.

• Factorials can make terms shrink faster than exponential growth in the numerator.
• Often results in series converging, as confirmed in our example.

Limit of a Sequence

A fundamental concept in calculus, the **limit of a sequence** involves finding what value, if any, the terms of a sequence approach as the sequence progresses indefinitely. A sequence is a list of numbers in a specific order, and the limit reflects the behavior of these numbers as you go further into the list.

In sequences where the terms are defined using functions with \(n\), as \(n\) approaches infinity, we check whether these terms settle towards a particular value.

Taking limits helps in determining convergence for series, especially when using the Ratio Test. In our exercise, after computing the ratio of terms \(\frac{a_{n+1}}{a_n}\), the next step was to find the limit \( L = \lim_{n \to \infty} \left| \frac{-3}{(2n+3)(2n+2)} \right| \).

With the denominator increasing infinitely, the fraction becomes smaller and smaller, dragging the limit closer to zero.

• This limiting behavior confirms convergence when \( L < 1 \).
• Helps in analyzing how sequences and series behave as they extend infinitely.

Understanding these limits is valuable in calculus, providing insights into infinite processes and their outcomes.