Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

\(\sum_{n=1}^{\infty} \frac{5}{3^{n}+1}\)

Step-by-Step Solution

Answer

The series \(\sum_{n=1}^{\infty} \frac{5}{3^{n}+1}\) is convergent.

Step 1: Determine if the series is convergent

To determine if the series \(\sum_{n=1}^{\infty} \frac{5}{3^{n}+1}\) converges, we can apply convergence tests. Recognize that the series \(\frac{5}{3^n+1}\) can be compared with the series \(\frac{1}{3^n}\), which is a geometric series with ratio \(r = \frac{1}{3}\). Since \(\frac{1}{3^n}\) is a convergent geometric series, we compare the original series.

Step 2: Use the Limit Comparison Test

Apply the Limit Comparison Test with the comparand series \(b_n = \frac{1}{3^n}\). Compute the limit:\[\lim_{n \to \infty} \frac{\frac{5}{3^n+1}}{\frac{1}{3^n}} = \lim_{n \to \infty} \frac{5 \cdot 3^n}{3^n + 1} = \lim_{n \to \infty} \frac{5}{1 + \frac{1}{3^n}} = 5.\]Since the limit is a finite positive number, by the Limit Comparison Test, both series converge.

Step 3: Conclude Convergence and Note the Series Behavior

Since the series \(\sum_{n=1}^{\infty} \frac{1}{3^n}\) is convergent and the limit comparison test shows convergence between \(\frac{5}{3^n+1}\) and \(\frac{1}{3^n}\), we conclude that the original series \(\sum_{n=1}^{\infty} \frac{5}{3^n+1}\) also converges. This means the sum of the infinite series exists as a finite number.

Key Concepts

Limit Comparison Test

The Limit Comparison Test is a valuable tool when determining the convergence of an infinite series. To use this test, we compare the series of interest with a second series, whose convergence is already understood. The key steps involve:

• Choosing a comparand series that is similar in form to the original series.
• Calculating the limit of the ratio of the terms from the two series as the term number approaches infinity.

If this limit is a positive finite number, the Limit Comparison Test tells us that both series either converge or diverge together.
In the context of the given exercise, we compared the series \( \sum_{n=1}^{\infty} \frac{5}{3^n+1} \) with \( \sum_{n=1}^{\infty} \frac{1}{3^n} \). By finding the limit of their term ratio as described, we confirmed that they both converge since the limit result was a positive finite number.

Geometric Series

Geometric series are one of the simplest types of infinite series, characterized by each term being a constant multiple of the preceding one.
The general form of a geometric series is \( \sum_{n=0}^{\infty} a r^n \), where \( a \) is the first term and \( r \) is the common ratio.
Geometric series play an important role in convergence tests because:

• We know that a geometric series converges if \( |r| < 1 \).
• It diverges otherwise.

In the original exercise, \( \frac{1}{3^n} \) is identified as a simple convergent geometric series with ratio \( r = \frac{1}{3} \), which falls under the convergent criteria \( |r| < 1 \). This makes it an appropriate comparand for applying convergence tests like the Limit Comparison Test.

Convergence Tests

Convergence tests are crucial in understanding whether an infinite series adds up to a finite value, which means it converges, or if it continues indefinitely, meaning it diverges.
There are various tests to determine the convergence of series, including:

• The Ratio Test
• The Root Test
• The Comparison Test
• The Limit Comparison Test
• The Integral Test

For this exercise, we focused on the Limit Comparison Test, which is useful when you can easily find a similar series known to converge or diverge. Applying convergence tests is essential because it helps filter out which series can be summed up to a finite value, allowing for deeper mathematical and practical applications. Understanding and selecting the appropriate convergence test requires practice, but once mastered, it provides a concrete understanding of the behavior of infinite series.