Problem 21 - Thomas Calculus

Question

Which of the series in Exercises \(11-40\) converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{2^{n}}{3^{n}} $$

Step-by-Step Solution

Answer

The series converges since it's a geometric series with \( r = \frac{2}{3} < 1 \).

Step 1: Recognize the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{2^n}{3^n} \). This is a geometric series, where \( a_n = \left( \frac{2}{3} \right)^n \). A geometric series can be expressed in the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.

Step 2: Identify the Common Ratio

In the series \( \sum_{n=1}^{\infty} \frac{2^n}{3^n} \), the common ratio \( r \) is \( \frac{2}{3} \). This can be deduced from the rewritten series form \( \sum_{n=0}^{\infty} \left( \frac{2}{3} \right)^n \).

Step 3: Apply the Geometric Series Convergence Test

A geometric series \( \sum_{n=0}^{\infty} ar^n \) converges if the absolute value of the common ratio \( |r| < 1 \), and diverges if \( |r| \geq 1 \). Here, we have \( |r| = \left| \frac{2}{3} \right| \), and since \( \frac{2}{3} < 1 \), the series converges.

Step 4: Determine the Sum (Optional)

Since the series converges and can be expressed in the form \( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n \), we can calculate the sum using the formula \( S = \frac{a}{1 - r} \) for a geometric series. Here \( a = \frac{2}{3} \) and \( r = \frac{2}{3} \). Thus, the sum is \( \frac{\frac{2}{3}}{1 - \frac{2}{3}} = 2 \).

Key Concepts

Series Convergence

Understanding whether a series converges or diverges is crucial in grasping series calculations. A series converges when its sum approaches a finite number as the number of terms goes to infinity. Conversely, if the sum grows indefinitely, it diverges.

In the case of geometric series, there's a simple criterion for determining convergence. The series will converge if the absolute value of its common ratio, noted as \( |r| \), is less than 1. This means the terms get progressively smaller, approaching zero, and the sum approaches a finite number.

In our exercise, we identified \( r = \frac{2}{3} \) for the series \( \sum_{n=1}^{\infty} \left( \frac{2}{3} \right)^n \), meaning the series converges because \( |\frac{2}{3}| = \frac{2}{3} < 1 \). This tells us that, as we add more terms, the partial sums of the series will get closer and closer to a certain value.

Geometric Series Sum

When a geometric series converges, not only can we determine it, but we can also calculate its sum. The formula for the sum of a converging geometric series \( S \) is \( S = \frac{a}{1 - r} \), where \( a \) is the first term, and \( r \) is the common ratio.

The initial term in our series, derived from \( \sum_{n=1}^{\infty} \left(\frac{2}{3}\right)^n \), is \( a = \frac{2}{3} \). Substituting into the sum formula gives us:

• \( S = \frac{\frac{2}{3}}{1 - \frac{2}{3}} \)


• Simplifying further, we get \( S = \frac{\frac{2}{3}}{\frac{1}{3}} = 2 \).

This sum tells us that even as we add infinitely many terms of the sequence, their total does not exceed 2. It is a neat demonstration of how infinite processes can sometimes yield finite results.

Common Ratio

The common ratio in a geometric series is an important parameter that dictates the behavior of the series. It is defined as the factor by which we multiply each term to get the next one in the series.

To find the common ratio \( r \), we can compare consecutive terms in the series. In our exercise, we have each term of the series given by \( \left( \frac{2}{3} \right)^n \). From this, the common ratio \( r \) directly takes the value \( \frac{2}{3} \).

This ratio reveals crucial information about the series:

• If \( |r| < 1 \), the series converges as the terms increasingly tend towards zero.
• If \( |r| \geq 1 \), the series diverges, meaning it doesn't settle to a fixed value.

In our series, since \( |r| = \frac{2}{3} < 1 \), the subsequent terms are diminishing in size, ensuring the series converges.