Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series. $$ \sum_{k=1}^{\infty} \frac{4^{k+1}}{7^{k-1}} $$

Step-by-Step Solution

Answer

The series converges and its sum is \( \frac{112}{3} \).

Step 1: Simplify the Terms

The given series is \( \sum_{k=1}^{\infty} \frac{4^{k+1}}{7^{k-1}} \). Begin by simplifying the general term. Rewrite it as follows: \( \frac{4^{k+1}}{7^{k-1}} = \frac{4 \cdot 4^k}{7^{k-1}} = 4 \cdot \frac{4^k}{7^{k-1}} = 4 \cdot \frac{4^k}{7^k} \cdot 7 = 4 \cdot 7 \cdot \left( \frac{4}{7} \right)^k \). This results in the simplified expression: \( 28 \left( \frac{4}{7} \right)^k \).

Step 2: Identify the Type of Series

The series \( \sum_{k=1}^{\infty} 28 \left( \frac{4}{7} \right)^k \) is a geometric series with the first term \( a = 28 \cdot \frac{4}{7} \) and common ratio \( r = \frac{4}{7} \).

Step 3: Determine Convergence

A geometric series converges if the absolute value of the common ratio \( |r| < 1 \). Here, \( r = \frac{4}{7} \) and indeed \( |\frac{4}{7}| < 1 \). Therefore, the series converges.

Step 4: Calculate the Sum of the Series

For a convergent geometric series \( \sum_{k=1}^{\infty} ar^k \), the sum is given by \( \frac{a}{1-r} \). Here, \( a = 28 \cdot \frac{4}{7} = 16 \) and \( r = \frac{4}{7} \). The sum is \( \frac{16}{1-\frac{4}{7}} = \frac{16}{\frac{3}{7}} = 16 \cdot \frac{7}{3} = \frac{112}{3} \). The sum of the series is \( \frac{112}{3} \).

Key Concepts

Series Convergence

When we talk about series convergence, we're trying to see if the series adds up to a finite value as more terms are included. In this exercise, we are given a geometric series, which is a type of series that is easier to analyze for convergence. For a geometric series, we can determine if it converges by examining the common ratio, denoted as \( r \). A geometric series converges if the absolute value of the common ratio \( |r| \) is less than 1. If this condition is met, it means that the series will approach a specific sum rather than increasing indefinitely. In our exercise, the common ratio \( r \) is \( \frac{4}{7} \), and since \( |\frac{4}{7}| < 1 \), the series converges.

Sum of Series

Once we've established that a geometric series converges, we can determine its sum. The formula to find the sum \( S \) of an infinite convergent geometric series is \( \frac{a}{1-r} \), where \( a \) is the first term of the series and \( r \) is the common ratio. This formula allows us to compute the precise value that the series converges to. In this particular example, the series simplifies such that the first term \( a \) is \( 16 \) and the common ratio \( r \) remains \( \frac{4}{7} \). Plugging these values into the formula gives us: \[ S = \frac{16}{1-\frac{4}{7}} = \frac{16}{\frac{3}{7}} = 16 \times \frac{7}{3} = \frac{112}{3} \]. So, the sum of the series is \( \frac{112}{3} \).

Common Ratio

The common ratio is an essential part of a geometric series as it determines the behavior of the series. The series is generated by multiplying each term by this common value \( r \). If \( r \) is between \(-1\) and \(1\), the terms will get progressively smaller, leading to series convergence. In our problem, the common ratio was identified as \( \frac{4}{7} \). A common ratio less than one in absolute value means each subsequent term is smaller than the previous, driving the total sum to remain finite. It's crucial to simplify the series first to correctly identify the common ratio, as done in the exercise where the given expression was rewritten to reveal \( r = \frac{4}{7} \). Understanding the common ratio helps predict whether a series converges and allows us to calculate its sum.