Problem 58 - Calculus: Early Transcendentals

Question

Find the values of \( x \) for which the series converges. Find the sum of the series for those values of \( x. \) \( \displaystyle \sum_{n = 1}^{\infty} (x + 2)^n \)

Step-by-Step Solution

Answer

The series converges for \(-3 < x < -1\), with a sum of \(-1\).

Step 1: Identify the Type of Series

The given series is \( \sum_{n = 1}^{\infty} (x + 2)^n \). This series is a geometric series with the first term \( a = (x + 2)^1 \) and the common ratio \( r = (x + 2) \). For a geometric series \( \sum_{n=0}^{\infty} ar^n \) to converge, the common ratio \( |r| < 1 \) must hold.

Step 2: Find the Interval of Convergence

To ensure convergence, we solve the inequality \( |x + 2| < 1 \). This can be rewritten as:\[-1 < x + 2 < 1\]Subtract 2 from each part:\[-3 < x < -1\]Thus, the series converges for \( x \) in the interval \((-3, -1)\).

Step 3: Find the Sum of the Series

For the values of \( x \) within the interval \((-3, -1)\), the sum of the geometric series is given by the formula for the sum of an infinite geometric series, \( S = \frac{a}{1 - r} \), where in this case both the first term \( a \) and the common ratio \( r \) are \( (x + 2) \).Thus, the sum is:\[ S = \frac{(x + 2)}{1 - (x + 2)} \]This simplifies to:\[ S = \frac{x + 2}{-x - 1} \]Simplify the expression entirely to give the sum as:\[ S = -1 \]

Step 4: Verify the Interval and Sum

We found that \( -3 < x < -1 \) is the interval of convergence. For these values, the series has a sum \(-1\). This verification is a simple substitution back into the inequality \( |x + 2| < 1 \) and ensuring the sum formula remains valid. The calculations in Steps 2 and 3 support this conclusion.

Key Concepts

Convergence

Convergence is a crucial concept in understanding infinite series. It helps us determine if a series adds up to a finite value. In the context of a geometric series, convergence depends on the common ratio of the series' terms. To check for convergence, we need to ensure that the absolute value of the common ratio, denoted as \(|r|\), is less than 1.

In this exercise, the series is \(\sum_{n = 1}^{\infty} (x + 2)^n\), which identifies a geometric series where the first term and common ratio are both \((x + 2)\). Convergence occurs when \(|x + 2| < 1\).

Why does this matter? When the series converges, it behaves like a predictable pattern that doesn't go off into infinity. This allows us to calculate a finite sum.

Interval of Convergence

The interval of convergence is the range of values for which the series converges to a finite sum. To determine this range, we solve the inequality that imposes the condition for convergence on the common ratio \(r\).

For the series \(\sum_{n = 1}^{\infty} (x + 2)^n\), we have the inequality \(|x + 2| < 1\). Breaking this down further:

• Start with \(-1 < x + 2 < 1\)
• Subtract 2 from each term to isolate \(x\): \(-3 < x < -1\)

Thus, the interval of convergence is \((-3, -1)\). Within this range, the series sums to a finite value. Values of \(x\) outside this interval won't satisfy the convergence conditions, making the series diverge or become infinite.

Infinite Series

Infinite series are the summation of terms that extend infinitely. While it may seem counterintuitive, some infinite series can add up to a specific, finite number when they converge. Think about it like an endless sequence that still arrives at a conclusion.

In this exercise, we have an infinite geometric series \(\sum_{n = 1}^{\infty} (x + 2)^n\). Unlike finite series that have a limited number of terms, infinite series require a careful look at convergence.

When we address infinite series, we use convergence conditions to establish if the series behaves well enough to produce a real sum. Otherwise, without convergence, the series would have no finite sum and just keep growing indefinitely.

Geometric Series Sum

A geometric series is characterized by each term being a constant multiple of the previous term. The sum formula for an infinite geometric series helps us find the total of such a series when it converges.

For the geometric series \(\sum_{n = 1}^{\infty} ar^n\), the sum \(S\) is given by:

• \(S = \frac{a}{1 - r}\)
• Where \(a\) is the first term and \(r\) is the common ratio.

In this exercise, both the first term and the common ratio are \(x + 2\). So, the sum \(S\) for \(x\) within the interval \(-3 < x < -1\) is:

• \(S = \frac{(x + 2)}{1 - (x + 2)}\)

This simplifies to \(S = -1\), showing the sum of the series remains constant for all \(x\) in the given interval. This demonstrates the efficiency and simplicity of geometric series and their sum.