Problem 79 - Essential Calculus

Question

Find all values of \(x\) for which the series converges. For these values of \(x,\) write the sum of the series as a function of \(x\). $$ \sum_{n=0}^{\infty}(-1)^{n} x^{n} $$

Step-by-Step Solution

Answer

The series converges for all \(x\) in the interval \(-1 < x < 1\), and the sum of the series for these \(x\) values is \(S = \frac{1}{1 + x}\).

Step 1: Convergence Condition

The given series is geometric with a common ratio of \(-x\). The absolute value of the common ratio must be less than 1 for the series to converge. So in this case, we must have \(|-x| < 1\). This translates to \(-1 < x < 1\). Hence, the infinite series converges for all \(x\) in the interval \(-1 < x < 1\).

Step 2: Sum of Geometric Series

The formula to find the sum \(S\) of an infinite geometric series where \(|r| < 1\) is: \(S = \frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the common ratio. For this series, the first term \(a = 1\) and the common ratio \(r = -x\). Substituting these we get, \(S = \frac{1}{1 - (-x)} = \frac{1}{1 + x}\).

Step 3: Final Answer

So, for the values of \(x\) in the interval \(-1 < x < 1\), the sum of the series is \(S = \frac{1}{1 + x}\).

Key Concepts

Geometric Series

A geometric series is a series of numbers in which each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. This type of series takes the form:

• First term: \(a\)
• Common ratio: \(r\)

To visualize it better, consider a simple example with \(a = 1\) and \(r = \frac{1}{2}\). The series would look like this: \[1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots\]For a geometric series to converge, the absolute value of \(r\) must be less than one (\(|r| < 1\)). This condition ensures that each successive term contributes less to the sum, which allows the series to approach a finite limit. In our original exercise, we identified \(r = -x\), hence the convergence condition applies as \(|-x| < 1\). This simplifies to the interval \(-1 < x < 1\). Understanding this fundamental characteristic is essential when working with any geometric series.

Convergence Interval

The convergence interval of an infinite series is the range of values for which the series converges to a finite sum. In the context of a geometric series, the series converges when the common ratio \(r\) satisfies \(|r| < 1\). For the geometric series given in the exercise, where the common ratio is \(-x\), we derived the inequality \(|-x| < 1\). Simplifying this gives us the interval: \(-1 < x < 1\)Within this range of \(x\), the terms of the series decrease in magnitude as they progress, ensuring that the series adds up to a definite number. Once \(x\) leaves this interval, the series no longer approaches any fixed value and is said to diverge. This behavior is critical to understanding series convergence, ensuring that solutions are accurate and applicable to real-life problems. Identifying this interval allows us to express solutions as functions within a defined range.

Infinite Series Sum

When dealing with infinite series, particularly geometric ones, determining their sum when they converge can be very insightful. The sum of an infinite geometric series is calculated using the formula: \[S = \frac{a}{1 - r}\]where \(S\) is the sum, \(a\) is the first term, and \(r\) is the common ratio.
In the exercise, we were given a series with \(a = 1\) and \(r = -x\). Substituting these values into the formula produces:\[S = \frac{1}{1 + x}\]This expression gives us the sum of the series for any \(x\) within the convergence interval \(-1 < x < 1\). The ability to express the series as a simple function of \(x\) is powerful.
It allows us to compute the sum quickly for any specific \(x\) within the interval. This concept highlights the elegance of geometric series, enabling precise calculations while also providing a deeper understanding of how series behave as \(x\) changes within its convergence range.