Problem 78 - 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 \(x\). $$ \sum_{n=0}^{\infty} 4\left(\frac{x-3}{4}\right)^{n} $$

Step-by-Step Solution

Answer

The series converges for \(x\) in the open interval (1, 5), and the sum of the series as a function of \(x\) is \(S=16/(5-x)\).

Step 1: Identify the common ratio

The power series presented is basically the geometric series where the common ratio, \(r=(x-3)/4\). The geometric series converges if and only if \(|r| < 1\).

Step 2: Find the interval of convergence

To find the interval of convergence, we solve the following inequality \(|(x-3)/4| < 1\). By multiplying out the modulus and rearranging, this inequality simplifies to -1 < (x-3)/4 < 1 which further simplifies to 1 < x < 5. Therefore, the series converges for \(x\) in the open interval (1, 5).

Step 3: Find the function representing the sum of series

The formula for the sum of a geometric series, when it converges, is \(S = a / (1 - r)\) where \(a\) is the first term and \(r\) is the common ratio. Here, \(a=4\) and \(r=(x-3)/4\), so the sum may be written as \(S = 4/(1 - (x-3)/4)\). This simplifies to \(S=16/(5-x)\).

Key Concepts

Geometric Series

A geometric series is a sum of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the given problem, the series is written as \( \sum_{n=0}^{\infty} 4\left(\frac{x-3}{4}\right)^{n} \). This illustrates a geometric series with the first term \( a = 4 \) and the common ratio \( r = \left(\frac{x-3}{4}\right) \).
This simple structure allows us to predict how the series behaves based on the value of \( r \). When the absolute value of the common ratio, \( |r| \), is less than 1, the series converges to a particular sum. In this scenario, the convergence depends critically on the ratio, which we will explore further in the next sections.

Interval of Convergence

The interval of convergence is a set of values for which the series will converge. For a geometric series, the critical aspect for convergence is finding out when \( |r| < 1 \). From the solution, we know that our common ratio is \( r = (x-3)/4 \).
To determine the interval where our series converges, we solve the inequality \( \left|(x-3)/4\right| < 1 \). Solving this step by step, we arrive at the inequality \(-1 < (x-3)/4 < 1\). Solving further, it simplifies to \(1 < x < 5\). Thus, the series converges for values of \( x \) within this open interval. It ensures the terms of the series are progressively getting smaller, ultimately resulting in convergence.

Sum of series

When a geometric series converges, it not only does so to a single value but it can be expressed as a function of its variables. The sum formula for an infinite geometric series is \( S = \frac{a}{1-r} \) where \( a \) is the first term, and \( r \) is the common ratio. In our exercise, we identify \( a = 4 \) and \( r = \frac{x-3}{4} \).
Plugging these in, we find the sum of the series as \( S = \frac{4}{1 - \frac{x-3}{4}} \). Simplifying this further results in \( S = \frac{16}{5-x} \) for all \( x \) in the interval \( (1, 5) \). This formula shows how the sum of the entire series depends on \( x \), giving a neat expression to understand how the series behaves within its domain of convergence.