Problem 57 - Calculus: Early Transcendentals

Question

Find the values of \( x \) for which the series converges. Find the sum of the series for those values of \( x. \) \( \displaystyle \sum_{n = 1}^{\infty} (-5)^n x^n \)

Step-by-Step Solution

Answer

The series converges for \(-\frac{1}{5} < x < \frac{1}{5}\); the sum is \(\frac{(-5)x}{1+5x}\).

Step 1: Recognize the Series

The given series is \( \sum_{n = 1}^{\infty} (-5)^n x^n \). This is a geometric series with the first term \( a = (-5)x \) and common ratio \( r = (-5)x \). Geometric series converge when the absolute value of the common ratio is less than 1.

Step 2: Find the Convergence Interval

To find for which values of \( x \) the series converges, we ensure the absolute value of the common ratio is less than 1: \[ |(-5)x| < 1 \]Solve the inequality:\[ 5|x| < 1 \]\[ |x| < \frac{1}{5} \].Thus, the series converges when \( -\frac{1}{5} < x < \frac{1}{5} \).

Step 3: Calculate the Sum of the Series

For a convergent geometric series, the sum is given by \( \frac{a}{1 - r} \), where \( a = (-5)x \) and \( r = (-5)x \). So, the sum of the series is:\[ \frac{(-5)x}{1 - (-5)x} = \frac{(-5)x}{1 + 5x} \].This formula is valid for \( x \) in the interval \( -\frac{1}{5} < x < \frac{1}{5} \).

Key Concepts

Series Convergence

When we talk about series convergence, we're discussing whether the sum of an infinite series actually settles on a specific number. For geometric series, convergence depends on the common ratio. If the absolute value of the common ratio is less than 1, the series will converge. If not, it diverges, meaning it won't sum up to a particular value. So for the series \( \sum_{n=1}^{\infty} (-5)^n x^n \) to converge, the condition \(|r| < 1\) must be satisfied. This ensures we're adding smaller and smaller pieces, allowing the infinite series to have a finite sum.

Common Ratio

A geometric series has terms that are multiplied by a constant term to get from one term to the next; this constant term is known as the common ratio. In the series \( \sum_{n = 1}^{\infty} (-5)^n x^n \), the common ratio \( r \) is calculated as \((-5)x\).

• If \(|r| < 1\), the series will converge.
• If \(|r| \geq 1\), the series will diverge.

For our given problem, we observe that in order for the series to converge, \(|(-5)x| < 1\) should hold true, leading to the condition \(5|x| < 1\), or equivalently \(|x| < \frac{1}{5}\). This is key in determining when the series will converge.

Interval of Convergence

The interval of convergence tells us the range of values where the series will converge. For a geometric series, this interval is tied directly to the common ratio. After deriving \(|x| < \frac{1}{5}\) from our common ratio \((-5)x\), we find the interval of convergence to be \(-\frac{1}{5} < x < \frac{1}{5}\). Within this interval, all \(x\) values will ensure that the series converges. It’s like finding the sweet spot where the series behaves well.

• The endpoints \(-\frac{1}{5}\) and \(\frac{1}{5}\) are not included, because at these points the series does not converge.
• This interval is a straight stretch on the x-axis where you can guarantee convergence.

Sum of a Series

The sum of a convergent geometric series can be found using the formula \( \frac{a}{1 - r} \), where \(a\) is the first term and \(r\) is the common ratio. For our series \( \sum_{n=1}^{\infty} (-5)^n x^n \), both \(a\) and \(r\) are equal to \((-5)x\). So, the sum is \(\frac{(-5)x}{1 + 5x} \), only when \(x\) is within the interval \(-\frac{1}{5} < x < \frac{1}{5}\).This formula helps us sum up all those infinite terms into one single number, as long as \(x\) is within the right interval. It’s remarkable because it simplifies an infinite process into a simple calculation, provided the series converges.