Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Determine convergence or divergence for each of the series. Indicate the test you use. \(\sum_{n=1}^{\infty} \frac{5}{3^{n}+1}\)

Step-by-Step Solution

Answer

The series converges by the Limit Comparison Test with a geometric series.

Step 1: Identify the Series Type

We need to determine if the series \( \sum_{n=1}^{\infty} \frac{5}{3^{n}+1} \) converges or diverges. This series has positive terms, so we can apply convergence tests suitable for positive term series.

Step 2: Apply the Limit Comparison Test

The Limit Comparison Test suggests that if two series \( \sum a_n \) and \( \sum b_n \) are both positive term series, and \( \lim_{n \to \infty} \frac{a_n}{b_n} = c \) where \( 0 < c < \infty \), then both series either converge or diverge. Let's compare our series with \( \sum b_n = \sum_{n=1}^{\infty} \frac{5}{3^n} \), a geometric series.

Step 3: Compare with a Geometric Series

Now, compute the limit: \[ \lim_{n \to \infty} \frac{5/(3^n+1)}{5/3^n} = \lim_{n \to \infty} \frac{3^n}{3^n+1}. \]Simplifying gives:\[ \lim_{n \to \infty} \frac{3^n}{3^n+1} = \lim_{n \to \infty} \frac{1}{1+1/3^n} = 1. \]Since the limit is 1 and finite (and non-zero), the Limit Comparison Test suggests that \( \sum_{n=1}^{\infty} \frac{5}{3^n+1} \) behaves like \( \sum_{n=1}^{\infty} \frac{5}{3^n} \).

Step 4: Analyze the Comparison Series

The series \( \sum_{n=1}^{\infty} \frac{5}{3^n} \) is a geometric series with a common ratio \( r = \frac{1}{3} \). Since \( |r| < 1 \), this geometric series converges.

Step 5: Conclude the Convergence

Since \( \sum_{n=1}^{\infty} \frac{5}{3^n+1} \) and \( \sum_{n=1}^{\infty} \frac{5}{3^n} \) behave similarly, and the latter converges, the given series also converges.

Key Concepts

Understanding Geometric Series

A geometric series is a special type of series where each term is a constant multiple of the previous one. This constant multiplier is known as the "common ratio" and is a key characteristic of geometric series. Geometric series can take the following basic form:

• Standard Form: \( a + ar + ar^2 + ar^3 + \ldots \)
• Summation Form: \( \sum_{n=0}^{\infty} ar^n \)

In these expressions, "\(a\)" is the first term and "\(r\)" is the common ratio. Understanding if a geometric series converges or diverges is straightforward thanks to the value of "\(r\)":

• If \(|r| < 1\), the series converges.
• If \(|r| \geq 1\), the series diverges.

To determine the convergence specifically, the infinite sum of a convergent geometric series is given by the formula \( S = \frac{a}{1-r} \). This information makes geometric series one of the easiest convergent series to analyze, provided you correctly identify the first term and common ratio.

Exploring Convergence Tests

Series convergence is a crucial concept in calculus and analysis, and understanding whether a series converges or diverges can often be determined using a variety of tests. These sequences of tests are powerful tools allowing mathematicians to understand the behavior of series effectively:

• Limit Comparison Test: This is helpful when you can compare your series directly with a series that is easier to interpret. The test uses the limit \( \lim_{n \to \infty} \frac{a_n}{b_n} \), where "\(a_n\)" is your series term and "\(b_n\)" is from a known series.
• Direct Comparison Test: Compare the original series with a series known to converge or diverge using inequality. If your series is consistently less than a known convergent series, it converges too.
• Root and Ratio Tests: Particularly useful for series involving factorials or exponential functions. They are powerful because they often simplify the process using growth rates of terms.

Selecting the right test often depends on the series form and available comparative series. As seen in our problem, the Limit Comparison Test proved to be advantageous for drawing comparisons with a geometric series.

Demystifying the Limit Comparison Test

The Limit Comparison Test is a handy method for determining the convergence or divergence of a series, particularly useful when direct assessment isn't straightforward due to complicated term structures. Here's how it effectively works:

• First, identify your series \( \sum a_n \) and a known comparison series \( \sum b_n \).
• Compute the limit: \( \lim_{n \to \infty} \frac{a_n}{b_n} \). If this limit \( c \) is finite and non-zero \((0 < c < \infty)\), then both series either converge or diverge together.

In practice, this test is robust because:

• It allows you to leverage known series (like a simple geometric series) to deduce the behavior of more complex series.
• It minimizes the need for additional computations beyond identifying suitable comparison series and the limit itself.

In applying the Limit Comparison Test to our series \( \sum_{n=1}^{\infty} \frac{5}{3^n+1} \), we compared it with \( \sum_{n=1}^{\infty} \frac{5}{3^n} \), a geometric series. The limit calculation confirmed its convergence by drawing parallels to the known behavior of geometric series.