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If \(\sum a_{n}\) converges and \(a_{n}>0\) for all \(n,\) can anything be said about \(\sum\left(1 / a_{n}\right) ?\) Give reasons for your answer.

Short Answer

Expert verified
\( \sum \left( \frac{1}{a_n} \right) \) diverges because \( a_n \to 0 \) implies \( \frac{1}{a_n} \to \infty.\)

Step by step solution

01

Understand the Given Information

We are given that the series \( \sum a_n \) converges and that each term \( a_n \) is positive. This means that the sequence \( a_n \rightarrow 0 \) as \( n \rightarrow \infty \).
02

Consider the Reciprocal Series

We need to evaluate the convergence of the series \( \sum \left( \frac{1}{a_n} \right) \). The task is to analyze whether this series converges, diverges, or whether no definite conclusion can be made.
03

Apply Limit Comparison Test Thoughts

If \( a_n \) approaches 0, then \( \frac{1}{a_n} \) becomes very large as \( n \rightarrow \infty \). Specifically, as \( a_n \to 0^+ \), \( \frac{1}{a_n} \to \infty \). This suggests that the terms \( \frac{1}{a_n} \) do not approach zero, indicating that the series may diverge.
04

Conclude Based on Harmonic Series Analogy

Since \( a_n \rightarrow 0 \) and \( \sum a_n \) converges, it implies that \( a_n \) converges to zero sufficiently fast. However, the harmonic series \( \sum \frac{1}{n} \) diverges even though \( \frac{1}{n} \to 0 \). Thus, \( \sum \left( \frac{1}{a_n} \right) \) diverges similarly due to terms becoming too large.

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Key Concepts

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

Positive Sequence
In the realm of arithmetic sequences and series, understanding the nature of a positive sequence is crucial. A positive sequence is a series of terms, each of which is greater than zero.

When we say the series \( \sum a_n \) converges and each term \( a_n > 0 \), it implies a specific behavior. Each term getting progressively closer to zero as \( n \rightarrow \infty \). This is vital for convergence because, if the terms remained large, the series would not graduate to a finite sum.

  • The notion of positiveness ensures that each term adds positively to the sum.
  • Convergence suggests the terms decrease rapidly.
  • Positivity implies that there are no sign changes that affect summation.
With this in mind, when analyzing the reciprocal series like \( \sum \left( \frac{1}{a_n} \right) \), we need to consider how each positive \( a_n \) affects the behavior of its inverse.
Limit Comparison Test
The Limit Comparison Test is a mathematical tool used to determine the convergence or divergence of series. It is especially useful when dealing with reciprocal transformations, as seen in the problem.

This test compares the terms of a questionable series with those of a known reference, typically a simpler harmonic or geometric series. Here's how it works:

  • Assume two positive series \( \sum a_n \) and \( \sum b_n \).
  • Compute the limit \( \lim_{{n \to \infty}} \frac{a_n}{b_n} \).
  • If this limit is a positive finite number, then both series behave similarly—in the sense of convergence or divergence.
Applying this to \( \sum \left( \frac{1}{a_n} \right) \), where \( a_n \to 0 \), the terms \( \frac{1}{a_n} \to \infty \), resembling a divergent series. By comparing \( \frac{1}{a_n} \) terms to those of a harmonic series \( \sum \frac{1}{n} \), we find that they diverge due to the limiting behaviors.
Harmonic Series
The harmonic series plays an essential role as a standard reference in assessing convergence or divergence. It is the infinite series given by \( \sum \frac{1}{n} \), which remarkably diverges even as its terms approach zero.

In comparison with a series \( \sum \left( \frac{1}{a_n} \right) \), where \( a_n \) decreases to zero, each reciprocal term \( \frac{1}{a_n} \) becomes extremely large. Such large terms could mirror those of a harmonic series, signaling potential divergence.

The divergence of the harmonic series, despite its terms shrinking to zero, offers a potent analogy:
  • Like \( \sum \frac{1}{n} \), if \( \sum \left( \frac{1}{a_n} \right) \) contains unbounded terms, it hints at divergence.
  • Even when individual terms get smaller, their sum can still grow indefinitely.
  • This teaches that achieving convergence requires more than just decreasing term size—it demands a steep decrease at a rapid pace.
Understanding this principle can clarify why certain series fail to converge despite apparent decreases in individual term value.

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Most popular questions from this chapter

It is not yet known whether the series \begin{equation}\sum_{n=1}^{\infty} \frac{1}{n^{3} \sin ^{2} n}\end{equation} converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps. \begin{equation} \begin{array}{l}{\text { a. Define the sequence of partial sums }}\end{array} \end{equation} \begin{equation} s_{k}=\sum_{n=1}^{k} \frac{1}{n^{3} \sin ^{2} n}. \end{equation} \begin{equation} \begin{array}{l}{\text { What happens when you try to find the limit of } s_{k} \text { as } k \rightarrow \infty ?} \\ {\text { Does your CAS find a closed form answer for this limit? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. Plot the first } 100 \text { points }\left(k, s_{k}\right) \text { for the sequence of partial }} \\ \quad {\text { sums. Do they appear to converge? What would you estimate }} \\ \quad {\text { the limit to be? }} \\ {\text { c. Next plot the first } 200 \text { points }\left(k, s_{k}\right) . \text { Discuss the behavior in }} \\ \quad {\text { your own words. }} \\ {\text { d. Plot the first } 400 \text { points }\left(k, s_{k}\right) \text { . What happens when } k=355 \text { ? }} \\\ \quad {\text { Calculate the number } 355 / 113 . \text { Explain from you calculation }} \\ \quad {\text { what happened at } k=355 . \text { For what values of } k \text { would you }} \\ \quad {\text { guess this behavior might occur again? }} \end{array} \end{equation}

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so do the following. $$\begin{array}{ll}{\text { a. }} & {\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. }} & {\sum_{n=1}^{\infty} k a_{n}(k \text { any number })}\end{array}$$

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{2^{n}+3^{n}}{3^{n}+4^{n}}\end{equation}

If \(\sum a_{n}\) is a convergent series of positive terms, prove that \(\sum \sin \left(a_{n}\right)\) converges.

When \(a\) and \(b\) are real, we define \(e^{(a+i b) x}\) with the equation \begin{equation} e^{(a+i b) x}=e^{a x} \cdot e^{i b x}=e^{a x}(\cos b x+i \sin b x) \end{equation} Differentiate the right-hand side of this equation to show that \begin{equation} \frac{d}{d x} e^{(a+i b) x}=(a+i b) e^{(a+i b) x} \end{equation} Thus the familiar rule \((d / d x) e^{k x}=k e^{k x}\) holds for \(k\) complex as well as real.

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