Problem 47 - Calculus: One and Several Variables

Question

Complete the limit comparison test. Let: \(\sum a_{k}\) and \(\sum b_{z}\) be scrics with positive terms. Suppose that \(a_{k} / b_{k} \rightarrow 0\) (a) Show that if \(\sum b_{k}\) converges, then \(\sum a_{k}\) converges. (b) Show that if \(\sum a_{k}\) diverges, then \(\sum b_{i}\) diverges. (c) Show by cxanple that if \(\sum a_{k}\) converges, iben \(\sum b_{k}\) may converge or diverge. (d) Show by example that if \(\sum b_{k}\) diverges, fhen \(\sum a_{k}\) may converge or diverge. [Parts (c) and (d) explain why we slipulatcd \(L \Rightarrow 0\) in Theorem \(12.3 .7 .]\)

Step-by-Step Solution

Answer

(a) Since \(\lim_{k\to\infty} \frac{a_k}{b_k} \rightarrow 0\), there exists a positive integer \(N\) such that for all \(k > N\), \(a_k < b_k\). If \(\sum b_k\) converges, then using the comparison test for convergence, \(\sum a_k\) converges. (b) If \(\sum a_k\) diverges and \(0 < \lim_{k\to\infty} \frac{b_k}{a_k} = L\), then for all \(k > N\), \(a_k > \frac{1}{L}b_k\). Since \(\sum a_k\) diverges, \(\sum b_k\) must also diverge. (c) Consider \(a_k = \frac{1}{k^2}\) and \(b_k = \frac{1}{k}\). \(\sum a_k\) converges while \(\sum b_k\) diverges. (d) Consider \(a_k = \frac{1}{k^2}\) and \(b_k = \frac{1}{k}\). \(\sum b_k\) diverges while \(\sum a_k\) converges.

Step 1: Define terms and conditions

Given that \(\sum a_{k}\) and \(\sum b_{z}\) are series with positive terms and that \(\lim_{k\to\infty} \frac{a_k}{b_k} \rightarrow 0\), we want to show that if \(\sum b_k\) converges, then \(\sum a_k\) converges.

Step 2: Apply Limit Comparison Test

Since \(\lim_{k\to\infty} \frac{a_k}{b_k} \rightarrow 0\), we can say that there exists a positive integer \(N\) such that for all \(k > N\), \(\frac{a_k}{b_k} < 1\). Therefore, \(a_k < b_k\) for all \(k > N\). We know that \(\sum b_k\) converges, so using the comparison test for convergence, since \(\sum a_k < \sum b_k\), it must be that \(\sum a_k\) converges. (b) Prove that if \(\sum a_k\) diverges, then \(\sum b_k\) diverges

Step 1: Re-write the given condition with the limit

The given condition states that \(\lim_{k\to\infty} \frac{a_k}{b_k} \rightarrow 0\). We rewrite this as \(0 < \lim_{k\to\infty} \frac{b_k}{a_k} = L\) for some real number L.

Step 2: Show divergence using the comparison test

We know that if \(\sum a_k\) diverges and \(0 < \lim_{k\to\infty} \frac{b_k}{a_k} = L\), then for some positive integer \(N\), for all \(k > N\), \(0 < \frac{b_k}{a_k} < L \Rightarrow a_k > \frac{1}{L}b_k\). As \(\sum a_k\) is a divergent series, this means that \(\sum \frac{1}{L}b_k\) also diverges, so \(\sum b_k\) must diverge. (c) Provide an example where if \(\sum a_k\) converges, \(\sum b_k\) may converge or diverge

Step 1: Choose example series

Consider \(a_k = \frac{1}{k^2}\) and \(b_k = \frac{1}{k}\).

Step 2: Evaluate series

We know that \(\sum \frac{1}{k^2}\) converges (as it is a p-series with \(p>1\)) and \(\sum \frac{1}{k}\) diverges (as it is a harmonic series). So this example serves as proof that if \(\sum a_k\) converges, \(\sum b_k\) may either converge or diverge. (d) Provide an example where if \(\sum b_k\) diverges, \(\sum a_k\) may converge or diverge

Step 1: Choose example series

Consider \(a_k = \frac{1}{k^2}\) and \(b_k = \frac{1}{k}\).

Step 2: Evaluate series

We already know that \(\sum \frac{1}{k}\) (harmonic series) diverges and \(\sum \frac{1}{k^2}\) converges. In this case, \(\lim_{k\to\infty} \frac{a_k}{b_k}=\lim_{k\to\infty} \frac{\frac{1}{k^2}}{\frac{1}{k}}=\lim_{k\to\infty} \frac{1}{k} = 0\), which satisfies the given condition. This example serves as proof that if \(\sum b_k\) diverges, \(\sum a_k\) may either converge or diverge.

Key Concepts

Convergence and Divergence of Series

Understanding the behavior of series is crucial in calculus and analysis. A series is a sum of an infinite sequence of terms. Whether a series converges or diverges tells us about its stability:

• Convergence: A series converges if the sum of its terms approaches a finite limit. This means the terms are getting smaller and smaller, eventually adding up to a specific value.
• Divergence: On the other hand, a series diverges if the sum increases indefinitely. This happens when the terms do not decrease sufficiently rapidly.

To determine convergence or divergence, we use several criteria and tests. One popular method is the Limit Comparison Test, especially for series with positive terms.

Comparison Test

The Comparison Test is an effective tool for analyzing series. It helps us conclude about the convergence or divergence of a target series by comparing it to another series with known behavior. Here's how it generally works:

• If a series \( \sum a_k \) is being compared to a known converging series \( \sum b_k \), and \( a_k \leq b_k \) for all terms, then \( \sum a_k \) must also converge.
• Conversely, if \( \sum a_k \) is compared to a diverging series \( \sum b_k \), and \( a_k \geq b_k \), then \( \sum a_k \) must also diverge.

This test simplifies determining series behavior when direct analysis might be challenging. By using a series with known characteristics, we create a benchmark for comparison.

Series with Positive Terms

Series consisting solely of positive terms, such as \( \sum a_k > 0 \), possess unique characteristics that make them easier to analyze compared to alternating series. These series are particularly amenable to comparison tests because:

• If all terms are positive, the partial sums are always increasing.
• This makes it easier to use inequalities, as shown in the Limit Comparison Test, where ratios of terms guide us to conclusions.

The behavior of a series can often be inferred by comparing its terms to similar series like known converging or diverging series. Understanding the properties of series with positive terms lays a strong foundation for mastering many convergence tests.