Problem 38 - Thomas Calculus

Question

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{3^{n-1}+1}{3^{n}} $$

Step-by-Step Solution

Answer

The series converges as it is a geometric series with ratio \( r = \frac{1}{3} \) (\(|r| < 1\)).

Step 1: Simplify the General Term

First, let's simplify the general term of the series. The term can be written as \( \frac{3^{n-1} + 1}{3^n} = \frac{3^{n-1}}{3^n} + \frac{1}{3^n} \). This further simplifies to \( \frac{1}{3} \frac{1}{3^{n-1}} + \frac{1}{3^n} = \frac{1}{3^n} + \frac{1}{3^n} \). Hence, the term simplifies to \( 2 \frac{1}{3^n} \).

Step 2: Recognize the Type of Series

Now that we have simplified the term to \( 2 \left( \frac{1}{3} \right)^n \), recognize this as a geometric series where \( a = 2 \) and \( r = \frac{1}{3} \).

Step 3: Apply the Geometric Series Convergence Test

A geometric series \( \sum_{n=1}^{\infty} ar^n \) converges if the absolute value of the common ratio, \( r \), is less than 1. Here, \( r = \frac{1}{3} \), which is less than 1. Thus, the series converges.

Key Concepts

Geometric Series

A geometric series is a series of the form \( \sum_{n=1}^{\infty} ar^n \) where \( a \) is the first term and \( r \) is the common ratio between consecutive terms. This type of series is extensively used due to its simple form which makes it easier to apply tests of convergence. When you notice terms in a series that involve a consistent exponent of a variable, it's a good indication you might be dealing with a geometric series.
The geometric series is characterized by each term being the result of multiplying the previous term by the common ratio \( r \). For example, if your series is \( 1 + 3 + 9 + 27 + \ldots \), here \( a = 1 \) and \( r = 3 \). This is different from an arithmetic sequence, where terms have a constant difference rather than a constant ratio.

Convergence Test

To determine if a geometric series converges, we use the convergence test for geometric series. This is a simple yet powerful tool. It states that a geometric series \( \sum_{n=1}^{\infty} ar^n \) converges if the absolute value of the common ratio \( r \) is less than 1, that is \( |r| < 1 \).
If the series converges, the sum can be found using the formula:

• \( S = \frac{a}{1 - r} \), where \( S \) is the sum of the series.

If \( |r| \geq 1 \), the series diverges, meaning it does not settle towards a finite limit. This test is useful because it simplifies evaluating the behavior of terms at the tail end of an infinite series. By focusing just on the common ratio, we can quickly assess a key aspect of the series' behavior.

Geometric Series Ratio

The common ratio \( r \) in a geometric series is crucial as it determines the behavior of the entire series. In the context of convergence, if \( |r| < 1 \), each successive term \( ar^n \) gets smaller and smaller, leading the series to converge.
The ratio is found by dividing a term in the series by its preceding term. For instance, in the sequence \( 2, \frac{2}{3}, \frac{2}{9}, \ldots \), you find the ratio by doing \( \frac{\frac{2}{3}}{2} = \frac{1}{3} \). Each term is smaller than the one before, pointing to the convergence of the series.
Understanding the geometric ratio not only helps in deciding convergence but also provides insights about the speed at which terms decrease, which further affects the calculation of the series sum.