Problem 26 - Calculus: Early Transcendentals

Question

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum. \( \displaystyle \sum_{n = 1}^{\infty} \frac {6 \cdot 2^{2n - 1}}{3^n} \)

Step-by-Step Solution

Answer

The series is divergent because the common ratio \( \frac{4}{3} \) is greater than 1.

Step 1: Identify the general term

The given series is the infinite sum \( \sum_{n=1}^{\infty} \frac{6 \cdot 2^{2n-1}}{3^n} \). We start by simplifying the general term: \( a_n = \frac{6 \cdot 2^{2n-1}}{3^n} \). Simplifying \(2^{2n-1}\) gives \(2 \cdot 4^{n-1}\). Thus, \( a_n = \frac{6 \cdot 2 \cdot 4^{n-1}}{3^n} = \frac{12 \cdot 4^{n-1}}{3^n} \).

Step 2: Express the series in standard geometric form

Notice the powers of 4 and 3 suggest the common ratio involves these. Write \( a_n = \frac{12}{3} \cdot \left( \frac{4}{3} \right)^{n-1} \). Simplifying gives \( a_n = 4 \cdot (\frac{4}{3})^{n-1} \), which is in the standard form \( a \cdot r^{n-1} \) for a geometric series, with \( a = 4 \) and \( r = \frac{4}{3} \).

Step 3: Determine convergence using the common ratio

A geometric series \( a \cdot r^{n-1} \) converges if the absolute value of the common ratio \(|r| < 1\). Here, \( r = \frac{4}{3} \), so \(|r| = \frac{4}{3} > 1\). Thus, the series does not converge.

Key Concepts

Series Convergence

The convergence of a series is a core concept in understanding whether an infinite series approaches a specific limit. For a geometric series specifically, convergence depends heavily on the common ratio, denoted by \( r \).

Generally, a geometric series takes the form \( a + ar + ar^2 + \, \ldots \) and it converges only if the absolute value of the common ratio \( |r| < 1 \). If \( |r| \) is 1 or greater, the terms do not decrease in magnitude, which means the series is divergent and does not have a sum.

In practical terms, if a series converges, it means that as we keep adding more terms, the total tends towards a particular finite number. For instance, if you've seen \( \sum_{n=1}^\infty \left( \frac{1}{2} \right)^n \), this series converges because its common ratio \( r = \frac{1}{2} \) is less than 1, indicating its sum is finite.

Common Ratio

The common ratio \( r \) is a vital component of a geometric series. Understanding the common ratio helps us to determine both the pattern of the series and its convergence properties.

The common ratio \( r \) is found by dividing any term in the series by the previous term, such as \( r = \frac{a_{n+1}}{a_n} \). It reflects how each term in the series is related to the one before it.

When examining series convergence, the magnitude of this ratio is critical:

• If \(|r| < 1\), the series converges to a finite value.
• If \(|r| \geq 1\), the series diverges, growing indefinitely as more terms are added.

In our original exercise, we determined \( r = \frac{4}{3} \). This value, being greater than 1, indicates that the series diverges.

Geometric Progression

A geometric progression, or geometric sequence, 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 \( r \).

The general form of a geometric progression looks like this: \( a, ar, ar^2, ar^3, \ldots \) Here, \( a \) is the initial term, and each successive term is obtained by multiplying the previous one by \( r \).

Geometric progressions are everywhere around us:

• They are used to model exponential growth and decay, such as populations or radioactive decay.
• They help in financial calculations involving repeated interest payments.

Understanding the basic pattern of a geometric progression is essential for identifying geometric series in mathematical problems, like in the original exercise, where we expressed the terms in the standard geometric form using the simplified expression.