Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(r\) be a fixed number with \(|r|<1 .\) Then it can be shown that \(\sum_{k=1}^{\infty} k r^{k}\) converges, say with sum \(S .\) Use the properties of \(\Sigma\) to show that $$ (1-r) S=\sum_{k=1}^{\infty} r^{k} $$ and then obtain a formula for \(S\), thus generalizing Problem 47 a.

Step-by-Step Solution

Answer

The sum \( S \) of the series is \( \frac{r}{(1-r)^2} \).

Step 1: Understanding the Series

The series given is \( \sum_{k=1}^{\infty} k r^{k} \), where \( |r| < 1 \). The goal is to express this series in terms of \( \sum_{k=1}^{\infty} r^{k} \). The sum \( \sum_{k=1}^{\infty} r^{k} \) is a geometric series with common ratio \( r \) and its sum is \( \frac{r}{1-r} \).

Step 2: Expressing the Series with Sum S

Let's express the series \( \sum_{k=1}^{\infty} k r^{k} \) using a variable, say \( S \), for its sum. We need to manipulate this series to relate it to \( \sum_{k=1}^{\infty} r^{k} \). We are given that \( (1-r) S = \sum_{k=1}^{\infty} r^{k} \).

Step 3: Analyze Both Sides of the Equation

We know the right-hand side \( \sum_{k=1}^{\infty} r^{k} = \frac{r}{1-r} \) is a geometric series. Thus, the equation becomes \( (1-r) S = \frac{r}{1-r} \).

Step 4: Solving for S

To find an expression for \( S \), solve the equation \( (1-r) S = \frac{r}{1-r} \). Divide both sides by \( 1-r \) to isolate \( S \):\[S = \frac{r}{(1-r)^2}\]

Step 5: Generalizing the Solution

The formula for the sum \( S \) of the series \( \sum_{k=1}^{\infty} k r^{k} \) is generalized as \( \frac{r}{(1-r)^2} \), given \( |r| < 1 \). This shows how the series of terms multiplied by their index results in a squared denominator in the sum's formula.

Key Concepts

Understanding a Geometric Series

A geometric series is a sequence where each term is derived by multiplying the previous term by the same constant, known as the common ratio. In mathematical terms, if the first term is denoted by \(a\) and the common ratio is \(r\), then the series can be expressed as \(a, ar, ar^2, ar^3, \ldots \).
For a geometric series, the sum to infinity is given by the formula \(\frac{a}{1-r}\), but only if the absolute value of the common ratio is less than one, i.e., \(|r| < 1\).
In the original exercise, we have a modified form where the series is \(\sum_{k=1}^{\infty} r^k\), a geometric series with common ratio \(r\) and a simple starting term. This series converges to \(\frac{r}{1-r}\).
Knowing how a geometric series behaves is essential in solving the given problem, as it becomes a building block for exploring more complex series.

The Principle of Convergence

The concept of convergence is central to the study of infinite series. In simple terms, convergence means that as you progress through the series, the sum gets closer and closer to a particular value, called the limit.
For geometric series or any series for that matter, convergence is determined by the behavior of the common ratio and the terms of the series.
In the exercise, \(|r| < 1\) ensures convergence, allowing us to meaningfully discuss the sum and manipulate the series to find useful expressions.
Without convergence, the series would diverge, meaning it grows without bound or fails to settle at a fixed number, making it impossible to define a precise sum.

Calculating the Sum of Series

To calculate the sum of an infinite series, especially one involving terms like \(k r^k\), understanding partial sums and other mathematical techniques is important.
In the problem, the series \(\sum_{k=1}^{\infty} k r^{k}\) is manipulated by employing the properties of known geometric series \(\sum_{k=1}^{\infty} r^k\).
By stating that \((1-r) S = \sum_{k=1}^{\infty} r^{k} = \frac{r}{1-r}\), we solve for \(S\).
This has led us to find \(S = \frac{r}{(1-r)^2}\), skillfully employing algebraic manipulation rooted in understanding geometric progressions.

Application of Calculus Solutions

In solving series problems like this one, we often employ calculus techniques to gain deeper insights and solve effectively.
Properties like differentiation and integration of series elements help relate complex forms of series back to simpler, known types such as geometric series.
By differentiating each term of the simple geometric series, we move from \(r^k\) to \(k r^{k-1}\), adjusting the series to match \(\sum_{k=1}^{\infty} k r^k\).
This approach illustrates the versatility and power of calculus, making it possible to generalize solutions across different series by understanding their foundations.