Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 7–12, show that each series converges absolutely. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{n !} $$

Step-by-Step Solution

Answer

The series converges absolutely.

Step 1: Identify the Series Components

We are given the series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{n!} \). The series has terms of the form \((-1)^{n+1} \frac{2^{n}}{n!}\). Notice the presence of \((-1)^{n+1}\), indicating it is an alternating series.

Step 2: Form the Absolute Series

To check absolute convergence, we need to consider the series formed by the absolute values of the terms: \( \sum_{n=1}^{\infty} \left| (-1)^{n+1} \frac{2^{n}}{n!} \right| = \sum_{n=1}^{\infty} \frac{2^{n}}{n!} \).

Step 3: Apply the Ratio Test for Absolute Convergence

The ratio test states that a series \( \sum_{n=1}^{\infty} a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \). For our series, \( a_n = \frac{2^{n}}{n!} \). Compute the ratio:\[\frac{a_{n+1}}{a_n} = \frac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}} = \frac{2^{n+1} \cdot n!}{2^n \cdot (n+1)!} = \frac{2 \cdot n!}{(n+1) \cdot n!} = \frac{2}{n+1}.\]

Step 4: Analyze the Ratio Test Limit

Find the limit \( \lim_{n \to \infty} \frac{2}{n+1} \). As \( n \to \infty \), \( \frac{2}{n+1} \to 0 \), which is less than 1. Therefore, the absolute series \( \sum_{n=1}^{\infty} \frac{2^{n}}{n!} \) converges.

Step 5: Conclude Absolute Convergence

Since the series of absolute values \( \sum_{n=1}^{\infty} \frac{2^{n}}{n!} \) converges, the original series \( \sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n}}{n!} \) converges absolutely.

Key Concepts

Alternating Series

An alternating series is a special kind of mathematical series where the signs of the terms alternate between positive and negative. This is often indicated by a factor like \((-1)^{n+1}\) in the series term. Let's break it down further:

• In the given series, each term has a factor of \((-1)^{n+1}\), causing the sign of the terms to switch between positive and negative as \('n'\) changes.
• Such series can converge even if the absolute series does not. This convergence is due to the canceling effect of alternating positive and negative terms.

To determine the convergence of an alternating series, we often examine the sequence of absolute values of the terms. If these terms steadily decrease to zero and the series satisfies other conditions, it converges.

Ratio Test

The ratio test is a powerful method used to determine the convergence of a series. It is especially useful for series containing factorials or exponential terms, such as the one of interest here. Applying this test:

• For a series \(\sum_{n=1}^{\infty} a_n\), compute the ratio: \(|a_{n+1}/a_n|\)
• Find the limit of this ratio as \('n'\) approaches infinity.
• If the limit is less than 1, the series converges absolutely.

In our example, computing the ratio for \(a_n = \frac{2^n}{n!}\) gives us \(\frac{2}{n+1}\). Taking the limit as \('n'\) goes to infinity results in 0, which clearly lies below 1, confirming absolute convergence.

Series Convergence

Convergence of a series is when the sum of all its terms approaches a specific value. There are generally three main types:

• Absolute Convergence: A series \(\sum a_n\) is absolutely convergent if the series of absolute values \(\sum |a_n|\) converges.
• Conditional Convergence: Occurs if \(\sum a_n\) converges while \(\sum |a_n|\) does not.
• Divergence: If the sum does not approach a finite value, the series is said to diverge.

The series given, after applying the ratio test, converges absolutely. Absolute convergence is a stronger condition than standard convergence and implies very stable behavior in terms of overall sum control.

Exponential Series

An exponential series involves terms that contain exponential expressions, often of the form \(b^n\) where \(b\) is a constant. These series are fundamentally important in mathematics:

• In this exercise, the series \(\sum \frac{2^n}{n!}\) features an exponential component \(2^n\).
• Such series can often be related to the exponential function \(e^x\), especially when factorial components are involved.
• The relation comes from the expansion of \(e^x\) as \(\sum \frac{x^n}{n!}\), showing strong ties between mathematics of growth phenomena and these series.

Recognizing a series of exponential nature helps in predicting behavior, as these are often part of solutions in differential equations and understanding limits.