Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 9-28, 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 convergence set for the series is \([-3, 1)\).

Step 1: Identify the General Term

We need to identify the general term \( a_n \) of the power series. The given series \( 1 + \frac{x+1}{2} + \frac{(x+1)^2}{2^2} + \cdots \) can be written as \( \sum_{n=0}^{\infty} \frac{(x+1)^n}{2^n} \). Therefore, the general term is \( a_n = \frac{(x+1)^n}{2^n} \).

Step 2: Apply the Absolute Ratio Test

Apply the Ratio Test to find the radius of convergence \( R \). Compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \):\[ \frac{a_{n+1}}{a_n} = \frac{(x+1)^{n+1}}{2^{n+1}} \cdot \frac{2^n}{(x+1)^n} = \frac{x+1}{2}. \]Thus, the limit is \( \left| \frac{x+1}{2} \right| \).

Step 3: Evaluate the Radius of Convergence

For the series to converge, the limit from the ratio test must be less than 1:\[ \left| \frac{x+1}{2} \right| < 1. \]This inequality simplifies to \(-2 < x+1 < 2\), or \(-3 < x < 1\).

Step 4: Determine the Interval of Convergence

The interval of convergence is \(-3 < x < 1\). We should also check convergence at the endpoints \( x = -3 \) and \( x = 1 \) separately.

Step 5: Test Endpoint \( x = -3 \)

Substitute \( x = -3 \) into the series: \(1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots\). This is a geometric series with first term \( a = 1 \) and common ratio \( r = -\frac{1}{2} \). Since \(|r| < 1\), this series converges.

Step 6: Test Endpoint \( x = 1 \)

Substitute \( x = 1 \) into the series: \(1 + \frac{2}{2} + \frac{4}{4} + \frac{8}{8} + \cdots\). This is a geometric series with first term \( a = 1 \) and common ratio \( r = 1 \). Since \(|r| = 1\), this series diverges.

Step 7: Write the Convergence Set

Combining the results, the convergence set is \([-3, 1)\), as the series converges at \( x = -3 \) but not at \( x = 1 \).

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when studying power series, as it determines the interval on which the series converges. To find the radius of convergence, we generally start by identifying the general term of the series. Here, the series is given as \( \sum_{n=0}^{\infty} \frac{(x+1)^n}{2^n} \). The general term \( a_n = \frac{(x+1)^n}{2^n} \) is extracted from this expression. The next step is to apply the Ratio Test, which helps us find the radius of convergence \( R \) by calculating the limit of the absolute value of the ratio of consecutive terms.The Ratio Test states:

• Calculate \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
• Set the result less than 1 to find the domain where the series converges.

In this example, this leads to the inequality \( \left| \frac{x+1}{2} \right| < 1 \) which simplifies to \(-2 < x+1 < 2\). Therefore, the radius of convergence \( R \) is 2, indicating that the series converges when \( x \) is within a distance of 2 from \(-1\). This forms the basis for determining the interval of convergence.

Interval of Convergence

The interval of convergence provides the actual set of \( x \)-values for which a given power series converges. After determining the radius of convergence, the series converges in the interval \( (-3, 1) \) after simplifying the radius inequality. This indicates that the series converges for all \( x \) values between -3 and 1, excluding the endpoints initially.However, to completely define the interval of convergence, we must test the behavior of the series at both endpoints \( x = -3 \) and \( x = 1 \). Substituting \( x = -3 \) into the series gives a convergent geometric series: \( 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots \), which converges since its common ratio is \(-\frac{1}{2} \), and \(|r| < 1\).For \( x = 1 \), substituting into the series results in a geometric series: \( 1 + 1 + 1 + \cdots \), which does not converge as \( r = 1 \). Thus, based on these evaluations, the series converges at \( x = -3 \) but not at \( x = 1 \), giving the final interval of convergence \([-3, 1)\).

Ratio Test

The Ratio Test is a fundamental tool in evaluating the convergence of a power series. It is especially helpful when dealing with expressions involving exponentials or factorials. The test involves taking the limit of the absolute value of the ratio of consecutive terms:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]For convergence, this limit must be less than 1.In our example, the series term is \( a_n = \frac{(x+1)^n}{2^n} \). By comparing terms \( a_{n+1} \) and \( a_n \), the ratio simplifies to:\[\frac{a_{n+1}}{a_n} = \frac{x+1}{2} \]Taking the limit as \( n \) approaches infinity doesn't change this expression since it is independent of \( n \), leading us to: \[\left| \frac{x+1}{2} \right| < 1\]This condition helps narrow down when the power series will converge, forming the basis for calculating the radius and interval of convergence. The Ratio Test is a practical way to quickly identify where a power series will behave properly, either converging or diverging, given specific values of \( x \).