Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that if \(a_{n} \geq 0, b_{n}>0, \lim _{n \rightarrow \infty} a_{n} / b_{n}=0\), and \(\sum b_{n}\) converges then \(\sum a_{n}\) converges.

Step-by-Step Solution

Answer

The series \(\sum a_n\) converges by the comparison test with \(\sum \frac{1}{2}b_n\), which converges.

Step 1: Analyze Given Conditions

We know that \(a_n \geq 0\), \(b_n > 0\), and \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\). Also, the series \(\sum b_n\) converges.

Step 2: Understand Series Convergence

Since \(\sum b_n\) converges, the terms \(b_n\) must approach zero as \(n\) approaches infinity.

Step 3: Use Limit Condition

Given \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\), for any \(\epsilon > 0\), there exists an \(N\) such that for all \(n > N\), \(\frac{a_n}{b_n} < \epsilon\). Thus, \(a_n < \epsilon b_n\).

Step 4: Apply Comparison Test

Choose \(\epsilon = \frac{1}{2}\), so \(a_n < \frac{1}{2}b_n\) for \(n > N\). Since \(\sum_{n=1}^{\infty} b_n\) converges, by the comparison test, \(\sum_{n=1}^{\infty} \frac{1}{2}b_n\) also converges and is smaller than \(\sum b_n\).

Step 5: Conclude Convergence

By the comparison test, since \(a_n < \frac{1}{2}b_n\), \(\sum a_n\) converges as \(\sum \frac{1}{2}b_n\) converges.

Key Concepts

Comparison Test

The comparison test is a powerful tool in determining the convergence of series. It involves comparing the series \(\sum a_n\) with another series \(\sum b_n\), where the behavior of \(\sum b_n\) is already known. Here's how it works:

• If the terms \(a_n\) of your series are always less than the terms \(b_n\) of a known convergent series \(\sum b_n\), then \(\sum a_n\) also converges.
• Conversely, if \(a_n\) terms are always greater than those of a divergent series \(\sum b_n\), then \(\sum a_n\) also diverges.

This test helps to conclude the convergence by comparing with simpler known series. In our exercise, since we showed that \(a_n < \frac{1}{2}b_n\), and \(\sum b_n\) converges, it follows by the comparison test that \(\sum a_n\) converges.

Limit of Sequences

Understanding the limit of sequences is crucial when dealing with series convergence. A sequence \(a_n\) approaches a limit \(L\) as \(n o \infty\) if for every small \(\epsilon > 0\), there exists a point \(N\) such that for all \(n > N\), the terms of the sequence are closer to \(L\) than \(\epsilon\): \(|a_n - L| < \epsilon\).In our exercise, we have the condition:

• \(\lim_{n \to \infty} \frac{a_n}{b_n} = 0\).

This implies that the sequence \(\frac{a_n}{b_n}\) becomes arbitrarily small as \(n\) becomes very large. Therefore, for large \(n\), we can find that \(a_n < \epsilon b_n\). This observation is crucial for applying the comparison test to establish the convergence of \(\sum a_n\).

Series Convergence

When discussing series convergence, we are interested in whether the sum of an infinite sequence of terms will add up to a finite number. A series \(\sum a_n\) is said to converge if the sequence of partial sums \(S_N = \sum_{n=1}^{N} a_n\) approaches a finite limit as \(N o \infty\).There are several tests to determine series convergence, including:

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

In this exercise, we used the comparison test to show the convergence of \(\sum a_n\) by comparison to the convergent series \(\sum b_n\).Understanding these concepts helps in identifying the nature of series, ensuring we don't spend unnecessary time on series that don't converge.

Positive Series

Series formed from non-negative terms have interesting properties. In positive series, denoted typically as \(\sum a_n\) where each \(a_n \geq 0\), convergence implies that the partial sums do not grow indefinitely.Let's explore this a bit:

• If a series of positive terms converges, it implies that the contribution from each additional term eventually becomes negligible.
• The behavior of these series is easier to analyze using tests like the comparison test, since negative terms can't cancel out growth.

For our exercise, both \(a_n\) and \(b_n\) are positive, and thus, easier to handle. Knowing \(a_n < \frac{1}{2}b_n\), with \(\sum b_n\) converging, simplifies our understanding that \(\sum a_n\) must also converge by the comparison test.