Problem 69 - Calculus: Early Transcendentals

Question

For what values of \( r \) is the sequence \( \left\\{ nr^n \right\\} \) convergent?

Step-by-Step Solution

Answer

The sequence is convergent for \(|r| < 1\).

Step 1: Understand the Criteria for Convergence of a Sequence

For a sequence to be convergent, its terms must approach a specific finite number as \( n \) approaches infinity. For the sequence \( \{ nr^n \} \), we need the terms to become smaller and smaller (ultimately reaching 0 or another finite number) as \( n \to \infty \).

Step 2: Examine the Term Structure

The term of the sequence is given by \( nr^n \). Here, \( r^n \) is an exponential term that depends on the base \( r \) and the exponent \( n \). As \( n \to \infty \), \( r^n \) will increase or decrease depending on \( r \). The factor \( n \) is a linear term.

Step 3: Explore Different Cases for \( r \)

1. If \( |r| < 1 \), then \( r^n \to 0 \) as \( n \to \infty \). So \( nr^n \to 0 \cdot n = 0 \), making the sequence convergent.2. If \( |r| = 1 \), then \( r^n \) remains constant, and \( nr^n \to \infty \). 3. If \( |r| > 1 \), then \( r^n \to \infty \), and \( nr^n \to \infty \).

Step 4: Deduce the Value of \( r \)

From the analysis, \( r \) must satisfy \( |r| < 1 \) for the sequence \( \{ nr^n \} \) to be convergent. In any case where \( |r| \geq 1 \), the terms of the sequence will not approach a finite limit.

Key Concepts

Exponential Growth

Exponential growth is a fascinating concept that plays a significant role in understanding sequences, particularly those involving exponential terms. When we talk about exponential growth, we refer to the rapid increase of a quantity over time. This happens when a number is multiplied by a constant factor in each step, which in mathematical terms is represented by a function like \( r^n \).

For our given sequence \( \{ nr^n \} \), the term \( r^n \) illustrates exponential growth or decay depending on the value of \( r \). If \( |r| > 1 \), the exponential term \( r^n \) grows very fast as \( n \) increases.

However, if \( |r| < 1 \), \( r^n \) actually decays towards zero as \( n \) gets larger. This decay is crucial for some sequences to converge because exponential decay can often outweigh other linear terms, leading the sequence to approach a specific limit.

Criteria for Convergence

The criteria for convergence in sequences involve the behavior of its terms as we head towards infinity. A sequence is said to converge if its terms get closer and closer to a particular finite number. In simple terms, we can imagine the sequence stabilizing or flattening as it progresses.

For the sequence \( \{ nr^n \} \), convergence is largely dependent on the value of \( r \). If the terms are to converge, the product of the exponential term \( r^n \) and the linear term \( n \) must shrink to zero or another finite value as \( n \) grows indefinitely.

The primary criteria that ensure convergence in this context are:

• If \( |r| < 1 \): This ensures \( r^n \) approaches zero, rendering \( nr^n \) also to approach zero.
• If \( |r| \geq 1 \): \( r^n \) either remains constant or increases, making \( nr^n \) diverge to infinity.

This simple boundary effectively determines whether the sequence will stabilize into a convergent sequence.

Sequence Analysis

Analyzing sequences involves breaking down their terms and understanding their behavior over time. In this process, we scrutinize how the sequence progresses based on the mathematical operations involved in its formulation. For\( \{ nr^n \} \), the sequence's behavior is primarily shaped by two components: the exponential factor \( r^n \) and the linear factor \( n \).

Sequence analysis can be divided into evaluating each possible scenario for \( r \):

• If \( |r| < 1 \): \( r^n \) tends towards zero rapidly. Therefore, the sequence converges as \( nr^n \to 0 \).
• If \( |r| = 1 \): The sequence does not converge as \( r^n \) does not shrink, leading \( nr^n \) toward infinity.
• If \( |r| > 1 \): Here, \( r^n \) grows exponentially, and \( nr^n \) will certainly diverge since both \( r^n \) and \( nr^n \) become unbounded.

This logical exploration helps us conclude that for a sequence like \( \{ nr^n \} \), convergence is achievable only when \( r \)'s absolute value remains below 1.