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Chapter 9: Problem 25

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

Short Answer

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

Step by step solution

01

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 \( a_n = \frac{{(x+1)^n}}{2^n} \).
02

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: \[ \left| \frac{{x+1}}{2} \right| < 1 \]
03

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:\[ -3 < x < 1 \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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