Problem 75 - Thomas Calculus

Question

In Exercises \(73-78,\) find the values of \(x\) for which the given geometric series converges. Also, find the sum of the series (as a function of \(x )\) for those values of \(x .\) $$ \sum_{n=0}^{\infty}(-1)^{n}(x+1)^{n} $$

Step-by-Step Solution

Answer

The series converges for \(-2 < x < 0\). The sum is \(\frac{1}{x+2}\).

Step 1: Identify the Series Form

The given series \( \sum_{n=0}^{\infty}(-1)^{n}(x+1)^{n} \) is a geometric series. A geometric series has the general form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term, and \( r \) is the common ratio. Here, \( a = 1 \) and \( r = -(x+1) \).

Step 2: Determine Convergence Criterion

A geometric series converges if the common ratio satisfies \( |r| < 1 \). Therefore, we need \( |-(x+1)| < 1 \). This simplifies to \( |x+1| < 1 \).

Step 3: Solve Inequality for Convergence

The inequality \( |x+1| < 1 \) can be further broken down to two inequalities: \( -1 < x+1 < 1 \). Solving these, we subtract 1 from all parts: \( -2 < x < 0 \).

Step 4: Find the Sum of the Series

The sum \( S \) of a convergent geometric series is given by \( S = \frac{a}{1-r} \). Substituting \( a = 1 \) and \( r = -(x+1) \), the sum becomes \( S = \frac{1}{1 - (-(x+1))} = \frac{1}{1 + x + 1} = \frac{1}{x + 2} \).

Step 5: Conclusion

The series converges for \( -2 < x < 0 \), and for these values of \( x \), the sum of the series is \( \frac{1}{x + 2} \).

Key Concepts

Series Convergence

When dealing with series, understanding convergence is crucial. In mathematics, a series is simply a sum of terms from a sequence. The big question is whether this sum approaches a finite number as we add more and more terms. If it does, we say that the series converges. If the sum keeps growing without bound or oscillates indefinitely, the series diverges.

To determine if a series converges, we often use tests that depend on the specific type of series you're examining. For a geometric series, there's a very intuitive condition: the series converges if the absolute value of the common ratio is less than 1.

• Convergence: Sum approaches a fixed number as more terms are added.
• Divergence: Sum grows infinitely large or fluctuates indefinitely.

This core concept of convergence helps us to unlock whether a series is useful in practical applications or just a mathematical curiosity.

Geometric Series Convergence

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This special type of series has a straightforward test for convergence.

For a geometric series to converge, the absolute value of the common ratio must be less than 1. Mathematically, if the series is expressed as \( \sum_{n=0}^{\infty} ar^n \), then it converges if \( |r| < 1 \).

Let's look at an example:

• Given: \( \sum_{n=0}^{\infty}(-1)^{n}(x+1)^{n} \)
• First Term \((a)\): 1
• Common Ratio \((r)\): \(-(x+1)\)
• Convergence Condition: \(|-(x+1)| < 1\)
• Solve: \(|x+1| < 1\), leading to \(-2 < x < 0\)

This condition ensures that the terms of the series are getting smaller in a controlled way, allowing the overall sum to settle at a finite value.

Sum of Series

Once we've determined that a geometric series converges, we can calculate its sum. This is an exciting part because it allows you to find a single value that represents the entire series. The sum of an infinite geometric series \( \sum_{n=0}^{\infty} ar^n \) is given by the elegant formula \( S = \frac{a}{1-r} \), provided that the series converges.

Let's apply this to our series:

• First term, \( a = 1 \)
• Common ratio, \( r = -(x+1) \)
• Substitute in formula: \( S = \frac{1}{1 - (-(x+1))} = \frac{1}{x+2} \)

Thus, for values \(-2 < x < 0\), the sum of the series is \( \frac{1}{x+2} \). This concise formula not only demonstrates the power of series convergences but also gives insight into how a seemingly infinite process can boil down to a simple expression.