Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use any test developed so far, including any from Section \(9.2\), to decide about the convergence or divergence of the series. Give a reason for your conclusion. $$ \sum_{k=1}^{\infty}\left[\left(\frac{1}{2}\right)^{k}+\frac{k-1}{2 k+1}\right] $$

Step-by-Step Solution

Answer

The series diverges because the second component diverges.

Step 1: Identify the Series Components

The given series can be separated into two parts: \( \sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{k} \) and \( \sum_{k=1}^{\infty}\frac{k-1}{2k+1} \). We will analyze each part separately to determine convergence or divergence.

Step 2: Analyze the First Series Component

The first part of the series, \( \sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{k} \), is a geometric series with a common ratio \( r = \frac{1}{2} \). Since \(|r| < 1\), this geometric series converges.

Step 3: Analyze the Second Series Component

Consider the second part of the series \( \sum_{k=1}^{\infty}\frac{k-1}{2k+1} \). We can use the Comparison Test to examine its convergence or divergence. Compare it to \( \sum_{k=1}^{\infty}\frac{k}{2k}\), which simplifies to \( \frac{1}{2} \sum_{k=1}^{\infty}1 \), a divergent harmonic series.

Step 4: Conclusion for the Second Series

Since \( \frac{k-1}{2k+1} \approx \frac{k}{2k} = \frac{1}{2} \) for large \(k\), \( \sum_{k=1}^{\infty}\frac{k-1}{2k+1} \) diverges by the Comparison Test with the harmonic series \( \sum_{k=1}^{\infty}1 \).

Step 5: Conclude the Original Series

The original series \( \sum_{k=1}^{\infty}\left[ \left(\frac{1}{2}\right)^{k} + \frac{k-1}{2k+1} \right] \) diverges. Although the geometric series converges, the divergence of any component (in this case the second series) leads to the divergence of the entire series.

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term is derived by multiplying the previous one by a fixed, non-zero number called the common ratio. The general form of a geometric series can be written as \(a + ar + ar^2 + ar^3 + \ldots \), where \(a\) represents the first term and \(r\) is the common ratio. To determine convergence, it is important to check the value of \(r\).
- If \(|r| < 1\), the series converges to \(\frac{a}{1-r}\).
- If \(|r| \geq 1\), the series diverges.
This rule makes it easy to identify the behavior of geometric series. For example, in the series \(\sum_{k=1}^{\infty} \left(\frac{1}{2}\right)^k\), the common ratio is \(\frac{1}{2}\), which is less than 1. Thus, this series converges, allowing us to focus our attention on other parts of a complex series when analyzing convergence or divergence.

Harmonic Series

The harmonic series is a well-known divergent series that is represented as \(\sum_{k=1}^{\infty} \frac{1}{k}\). Even though the terms approach zero as \(k\) increases, the sum of the series tends to infinity, leading to divergence. This is a crucial aspect when evaluating series by comparison, as it provides a clear benchmark.
Unlike geometric series, the divergence of a harmonic series does not depend on a common ratio. Instead, it stems from the behavior where the accumulation of steadily smaller terms still manages to grow unbounded over time. In many exercises, including the one at hand, the harmonic series or its variations such as \(\sum_{k=1}^{\infty} 1\), play a pivotal role in the application of the comparison test when determining the divergence of another series.

Comparison Test

The comparison test is a valuable tool for analyzing series convergence by comparing a series of interest to a second series whose convergence properties are already known. To apply the test, you generally need to:
- Identify a well-understood series, such as a geometric or harmonic series, which is similar in form to the series under consideration.
- Compare the terms of the two series. If the terms of your series are always less than or equal to the terms of a known converging series, your series converges. Conversely, if they are always greater than or equal to a known diverging series, your series diverges.
In our exercise, the series \(\sum_{k=1}^{\infty} \frac{k-1}{2k+1}\) was compared to a divergent harmonic series using the comparison test. Observing algebraic transformations helps align the series for comparison: \(\frac{k-1}{2k+1} \approx \frac{k}{2k} = \frac{1}{2}\). This approximation highlighted its divergence, reinforcing the decision to look at known series for fruitful conclusions about behavior.