Problem 63 - Calculus: Early Transcendentals

Question

Find the values of \( x \) for which the series converges. Find the sum of the series for those values of \( x. \) \( \displaystyle \sum_{n = 0}^{\infty} e ^{nx} \)

Step-by-Step Solution

Answer

The series converges for \( x < 0 \) with sum \( \frac{1}{1 - e^x} \).

Step 1: Identify the Series Type

The given series is \( \sum_{n=0}^{\infty} e^{nx} \). This is a geometric series where the first term \( a = 1 \) and the common ratio \( r = e^x \).

Step 2: Establish Convergence Criteria for Geometric Series

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of the common ratio is less than 1, i.e., \( |r| < 1 \). In this series, the common ratio is \( r = e^x \), so we need \( |e^x| < 1 \) for convergence.

Step 3: Solve Inequality for Convergence

Since \( e^x > 0 \) for all \( x \), the condition \( |e^x| < 1 \) simplifies to \( e^x < 1 \). The exponential function \( e^x \) is less than 1 when \( x < 0 \).

Step 4: Calculate the Sum of the Convergent Series

If \( |r| < 1 \), the sum of a geometric series is given by \( S = \frac{a}{1 - r} \). In our series, \( a = 1 \) and \( r = e^x \). Thus, the sum is \( S = \frac{1}{1 - e^x} \) for \( x < 0 \).

Key Concepts

Geometric Series

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.
For example, in a series like 2 + 4 + 8 + 16 + ..., the common ratio is 2. In the context of the exercise, the series given is \( \sum_{n=0}^{\infty} e^{nx} \), where \( e^{x} \) acts as the common ratio:

• The first term \( a \) is understood to be 1.
• The common ratio \( r \) is \( e^x \).

Understanding this series involves recognizing how the terms escalate depending on the value of \( x \). The nature of the geometric series allows it to either converge or diverge, depending on whether the ratio meets certain criteria.

Exponential Function

The exponential function, denoted as \( e^x \), is a mathematical function with important properties in calculus and complex analysis. It is defined as:

• Natural exponential function \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \), where \( e \approx 2.71828 \) is the base of natural logarithms.

This function grows rapidly as \( x \) increases and is always positive, which affects its behavior within a series.
In the series we are considering, \( e^{nx} \) acts as the common ratio. Since \( e^{nx} \) is typically greater than 1 when \( x > 0 \), it's crucial to determine conditions where \( e^{nx} \) would result in convergence. Consequently, finding values where \( e^x \) remains below 1 is significant to ensure the geometric series converges.

Convergence Criteria

To determine whether a geometric series converges, we need to evaluate the convergence criteria.
A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of the common ratio \( r \) is less than 1.
In our exercise:

• The common ratio \( r \) is \( e^x \).
• Convergence occurs if \( | e^x | < 1 \).

Since exponential functions are always positive, \( | e^x | = e^x \). For the series to converge, \( e^x < 1 \) is required, which occurs precisely when \( x < 0 \).
Understanding these criteria helps solve when given series converge, ensuring calculations focus only on meaningful series.

Series Sum Formula

Once we verify the series' convergence, we can utilize the formula to find the sum of the geometric series.
The formula to find the sum \( S \) of an infinite geometric series is:

• \( S = \frac{a}{1 - r} \), where \( |r| < 1 \).

For our series, \( a = 1 \) and \( r = e^x \).
The sum becomes \( S = \frac{1}{1 - e^x} \). However, this formula only holds when \( x < 0 \) as per our convergence criteria, ensuring the series converges in this domain.
This formula is a powerful tool, allowing us to sum series that may seem infinitely large into a manageable number, provided the criteria are met.