Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1+\frac{x+1}{2}+\frac{(x+1)^{2}}{2^{2}}+\frac{(x+1)^{3}}{2^{3}}+\cdots $$

Step-by-Step Solution

Answer

The power series converges for \(-3 < x < 1\).

Step 1: Identify the General Term of the Series

The given power series is a geometric series. It can be expressed in the standard form of a series sum: \[ a_n = \frac{{(x+1)^n}}{2^n} \]where the first term is 1 (which can be noted as \((x+1)^0/2^0 = 1\)) and the nth term is \( a_n = \frac{{(x+1)^n}}{2^n} \).

Step 2: Apply the Ratio Test

The Absolute Ratio Test involves taking the limit of \( |a_{n+1}/a_n| \) as \( n \to \infty \). For this series:- \( a_n = \frac{{(x+1)^n}}{2^n} \)- \( a_{n+1} = \frac{{(x+1)^{n+1}}}{2^{n+1}} \)Calculate the ratio:\[ \left| \frac{{a_{n+1}}}{a_n} \right| = \left| \frac{{(x+1)^{n+1}/2^{n+1}}}{(x+1)^n/2^n} \right| = \left| \frac{{x+1}}{2} \right| \] according to the Ratio Test, for convergence: \[ \left| \frac{{x+1}}{2} \right| < 1 \]

Step 3: Solve the Inequality for Convergence

To find the convergence set, solve the inequality:\[ \left| \frac{{x+1}}{2} \right| < 1 \]Simplifying this gives:\[ |x+1| < 2 \]This implies:\[ -2 < x+1 < 2 \]Subtracting 1 from each side, we get:\[ -3 < x < 1 \]

Step 4: Check the Endpoints for Convergence

Now check the convergence at the endpoints \( x = -3 \) and \( x = 1 \):1. For \( x = -3 \): the series becomes: \( \sum \frac{(-2)^n}{2^n} = \sum (-1)^n \) which diverges.2. For \( x = 1 \): the series becomes: \( \sum \frac{2^n}{2^n} = \sum 1 \), which diverges.Both endpoints \( x = -3 \) and \( x = 1 \) do not lead to convergence, so they are not included in the interval.

Key Concepts

Power Series

A power series is a sum of terms in the form \( a_n (x - c)^n \), where \( a_n \) are coefficients and \( c \) is a constant. The power series's purpose is to represent functions as infinite sums that are easy to compute. In this exercise, the series given is

• \( 1 + \frac{x+1}{2} + \frac{(x+1)^{2}}{2^{2}} + \cdots \)

This expression can be rewritten using the general formula for a geometric series term:

• \( a_n = \frac{(x+1)^n}{2^n} \)

In this formula, the base \( x+1 \) allows the series to effectively 'shift' the center, making this series center around \( x = -1 \). Understanding the structure of a power series helps in forming a strategy to evaluate its convergence.

Ratio Test

The Ratio Test is a method used to determine whether a given series converges or diverges by comparing successive terms. For a series \( \sum a_n \), you calculate the limit

• \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \)

Consider the series in question:

• \( a_n = \frac{(x+1)^n}{2^n} \)
• \( a_{n+1} = \frac{(x+1)^{n+1}}{2^{n+1}} \)

The ratio \( \left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{x+1}{2} \right| \) measures the relationship between the terms as \( n \) grows. The Ratio Test states that the series converges if \( L < 1 \). Thus, finding \( L \) assists in determining the range of \( x \) for which the series converges.

Inequality Solving

Solving the inequality derived from the Ratio Test tells us where the series converges.The inequality for this series was:

• \( \left| \frac{x+1}{2} \right| < 1 \)

This translates to \( |x+1| < 2 \). To solve this inequality, break it down:

• \( -2 < x+1 < 2 \)

Subtracting \( 1 \) from each component yields:

• \( -3 < x < 1 \)

The solution \(-3 < x < 1\) is the interval of convergence without considering endpoints, where we will further analyze if the series converges or diverges at \( x = -3 \) and \( x = 1 \).

Endpoint Analysis

Endpoint analysis involves testing the values of \( x \) obtained from the inequality whose endpoints might change the convergence behavior of the series. In this problem, the range found was \(-3 < x < 1\). Therefore, we check:

• \( x = -3 \): If substituted in, the series becomes \( \sum (-1)^n \), an alternating series whose terms do not approach zero, thus it diverges.
• \( x = 1 \): With substitution, the series becomes \( \sum 1 \), a well-known divergent series, because every term is a constant 1 and the sum grows infinitely.

Since neither endpoint results in a convergent series, we conclude that the interval of convergence for the power series is \(-3 < x < 1\) and does not include the endpoints. Endpoint analysis is crucial because convergence intervals without checked endpoints might give misleading results.