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Chapter 8: Problem 15

Determine whether the series is convergent or divergent. If it is convergent, find its sum. $$\sum_{n=1}^{\infty} \frac{n-1}{3 n-1}$$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Recognize the Series Type

This is an infinite series given by \(\sum_{n=1}^{\infty} \frac{n-1}{3n-1}\). To determine its convergence or divergence, we need to analyze its behavior.
02

Use the Limit Comparison Test

Consider the terms \(a_n = \frac{n-1}{3n-1}\). For large \(n\), \(a_n\) behaves approximately like \(\frac{n}{3n} = \frac{1}{3}\). Compare this to the harmonic series \(b_n = \frac{1}{n}\). Since \(\sum \frac{1}{n}\) diverges, we can use the Limit Comparison Test to analyze \(a_n\) with \(b_n\).
03

Compute the Limit

Find the limit \(\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n-1}{3n-1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n^2-n}{3n-1} = \lim_{n \to \infty} \frac{n^2-n}{3n^2} = \lim_{n \to \infty} \frac{1-\frac{1}{n}}{3} = \frac{1}{3}\).
04

Conclusion from the Limit Comparison Test

Since the limit \(\frac{1}{3} eq 0\) and the comparison series \(\sum \frac{1}{n}\) diverges, the original series \(\sum \frac{n-1}{3n-1}\) also diverges.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is a powerful method to determine the convergence or divergence of an infinite series. It works by comparing a complex series with a simpler, more well-known series.

Here's how it usually works:
  • Start with two series that have terms \(a_n\) and \(b_n\).
  • Choose \(b_n\) such that you already know its convergence or divergence behavior, often a basic series like the harmonic series.
  • Calculate the limit \(\lim_{n \to \infty} \frac{a_n}{b_n}\).
  • If this limit is a positive finite number, both series \(\sum a_n\) and \(\sum b_n\) will either converge or diverge together.
In this particular exercise, the original series \(a_n = \frac{n-1}{3n-1}\) was compared to the harmonic series \(b_n = \frac{1}{n}\). Hence, once we found the limit to be \(\frac{1}{3}\), which is a positive finite number, this informed us that the behavior of both series on convergence or divergence is the same.
Harmonic Series
The harmonic series is one of the most fundamental infinite series in mathematics. It is defined as \(\sum_{n=1}^{\infty} \frac{1}{n}\). Despite its simple form, it has remarkable divergent behavior.

Key points about the harmonic series include:
  • The harmonic series diverges, meaning its sum grows indefinitely as more and more terms are added.
  • It is often used as a benchmark in tests for convergence or divergence of other series, such as the Limit Comparison Test.
  • Even though each individual term \(\frac{1}{n}\) gets smaller as \(n\) increases, the sum of all these terms grows without bound.
In the exercise, the harmonic series was used as a comparison to deduce the divergence of the original series \(\sum_{n=1}^{\infty} \frac{n-1}{3n-1}\). Its divergence helps in validating the behavior of other series through comparison.
Infinite Series Analysis
Analyzing an infinite series to determine if it converges or diverges is a crucial aspect of calculus. Here are some steps and tests often used in the analysis process:

  • Identify the type of series: Before applying tests, recognize the series type. Is it geometric, harmonic, or something else?
  • Use comparison tests: Both direct and limit comparison tests use known series to deduce the behavior of the target series.
  • Apply other criteria as needed: Sometimes Ratio Test, Root Test, or Integral Test might be required based on the series at hand.
In the current problem, we first approximated the series behavior by comparing \(a_n\) with \(\frac{n}{3n}\). This insight allowed us to employ the Limit Comparison Test knowledgeably.

Ultimately, knowing these strategies helps tackle a broader range of convergence and divergence questions. Understanding the nature of series within the mathematical context is essential, whether for theoretical investigations or practical applications.

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Most popular questions from this chapter

(a) Starting with the geometric series \(\sum_{n=0}^{\infty} x^{n},\) find the sum of the series $$ \sum_{n=1}^{\infty} n x^{n-1} \quad|x|<1 $$ (b) Find the sum of each of the following series. $$ \text { (i) }\sum_{n=1}^{\infty} n x^{n}, \quad|x|<1 \quad \text { (ii) } \sum_{n=1}^{\infty} \frac{n}{2^{n}} $$ (c) Find the sum of each of the following series. $$ \text { (i) }\sum_{n=2}^{\infty} n(n-1) x^{n}, \quad|x|<1 $$ $$ \text { (ii) }\sum_{n=2}^{\infty} \frac{n^{2}-n}{2^{n}} \quad \text { (iii) } \sum_{n=1}^{\infty} \frac{n^{2}}{2^{n}} $$

Show that if \(a_{n}>0\) and \(\lim _{n \rightarrow \infty} n a_{n} \neq 0,\) then \(\Sigma a_{n}\) is divergent.

\(\begin{array}{l}{\text { (a) Use the binomial series to expand } 1 / \sqrt{1-x^{2}}}. \\ {\text { (b) Use part (a) to find the Maclaurin series for } \sin ^{-1} x}.\end{array}\)

Find a power series representation for \(f,\) and graph \(f\) and several partial sums \(s_{a}(x)\) on the same screen. What happens as \(n\) increases? $$ f(x)=\tan ^{-1}(2 x) $$

Suppose that a series \(\Sigma a_{n}\) has positive terms and its partial sums \(s_{n}\) satisfy the inequality \(s_{n} \leqslant 1000\) for all \(n .\) Explain why \(\Sigma a_{n}\) must be convergent.
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