Problem 37 Prove that if \(a_{n} \geq 0, b_... [FREE SOLUTION]

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Calculus Early Transcendentals: Pearson New International Edition

Dale Varberg, Edwin J. Purcell, Steve E. Rigdon

$Math Studyset 91影视 Explanations$ Math

1 Edition

Chapter 9: Problem 37

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.

Short Answer

Expert verified

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

Step by step solution

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.

Understand Series Convergence

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

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, $a_n < \epsilon b_n$.

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 $\sum b_n$.

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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91影视

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.

Short Answer

Step by step solution

Analyze Given Conditions

Understand Series Convergence

Use Limit Condition

Apply Comparison Test

Conclude Convergence

Key Concepts

Comparison Test

Limit of Sequences

Series Convergence

Positive Series

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