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Chapter 9: Problem 23

Find an explicit formula a \(a_{n}=\) ____ for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). $$ -1, \frac{2}{3},-\frac{3}{5}, \frac{4}{7},-\frac{5}{9}, \ldots $$

Short Answer

Expert verified
The sequence is \(a_n = \frac{(-1)^n n}{2n-1}\); it diverges.

Step by step solution

01

Identify the sequence pattern

Observe the given sequence: \(-1, \frac{2}{3}, -\frac{3}{5}, \frac{4}{7}, -\frac{5}{9}, \ldots\). Notice that each term \(a_n\) can be expressed with the formula: \(-1, \frac{2}{3}, \ldots\). The pattern is an alternating sequence of fractions where the numerator increases by 1 starting from -1 and the denominator also increases by 2 starting from 1.
02

Develop a formula for the sequence

Given the pattern, express the \(n\)-th term as \(a_n = \frac{(-1)^n n}{2n-1}\). Here, \((-1)^n\) ensures that the terms alternate sign, while \(n\) and \(2n-1\) respectively position as the numerator and denominator reflecting the pattern.
03

Check if the sequence converges or diverges

To determine convergence, find \(\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{(-1)^n n}{2n-1}\). As \(n\) approaches infinity, the ratio \(\frac{n}{2n-1}\) simplifies to \(\frac{1}{2}\), but alternating sign \((-1)^n\) does not converge to a single value, indicating divergence.
04

Decide the limit of the sequence if it converges

Since we established in Step 3 that \(\lim_{n \to \infty} a_n\) does not yield a single finite value due to the alternating sign component \((-1)^n\), the sequence diverges and has no limit.

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Key Concepts

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

Understanding Sequences
A sequence is essentially a list of numbers in a specific order. Each number in the sequence is termed as a 'term'. For example, in the sequence \(-1, \frac{2}{3}, -\frac{3}{5}, \frac{4}{7}, -\frac{5}{9}, \ldots\), each expression corresponds to a term.
  • The position of a term in a sequence is often denoted by \(n\), representing its order.
  • The general term in a sequence can usually be represented by a formula involving \(n\), called the closed form or the explicit formula.
In the given problem, the sequence shows a predictable pattern of alternating positive and negative fractions, where both the numerator and denominator of each term increase incrementally. Such sequences are termed as alternating sequences. Understanding the pattern helps in deriving the formula for each term.
Convergence in Sequences
The convergence of a sequence is a fundamental concept in calculus and means that as \(n\) gets larger, the terms in the sequence become closer to a specific number, called the limit. Convergence is significant because it indicates the behavior of sequences in the long run.
For a sequence to converge, its limit \(\lim_{n \to \infty} a_n\) must exist and be a finite value. In mathematical terms, this means that as you go through the sequence, the terms approach a single finite value without fluctuating or going off to infinity.
  • You can determine convergence by calculating the limit as \(n\) approaches infinity.
  • If a limit exists and is finite for the sequence, it means the sequence converges.
In our main problem, despite finding a limiting-like behavior in the fractions themselves (\(\frac{1}{2}\)), the alternating sign prevents convergence.
Divergence Explained
Divergence is the opposite of convergence. A sequence diverges if it does not approach any specific finite value as \(n\) becomes very large. In mathematical terms, if the limit \(\lim_{n \to \infty} a_n\) does not exist or is not finite, the sequence is said to diverge.
  • Divergence can occur if the terms of the sequence increase without bound or if they keep alternating without settling on a specific value.
  • A classic indicator of divergence is when terms oscillate without nearing any single point.
In our problem, even though the non-alternating part of the terms approaches \(\frac{1}{2}\), the presence of \((-1)^n\) causes the actual terms to never settle, thus the sequence diverges. Recognizing the behavior of oscillation due to alternating signs is key to identifying divergence.
The Role of Limits
Limits help us understand the long-term behavior of sequences and functions, a pivotal concept in calculus. The limit of a sequence \(a_n\) as \(n\) approaches infinity, denoted \(\lim_{n \to \infty} a_n\), gives insights into convergence or divergence.
When we are finding a limit, we are asking: What number does the sequence get closer and closer to as the sequence gets longer?
  • If a sequence's terms settle at a specific number, the sequence converges to that limit.
  • If the sequence never settles at a single number, then it diverges and does not have a limit per se.
In our problem, the limit of the sequence terms appeared to approach a fraction, but the alternating sign shifted the sequence's terms back and forth, thwarting convergence and anchoring the sequence in divergence instead.

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Most popular questions from this chapter

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\left(1-x^{2}\right)^{2 / 3}\)

Show that \(\lim _{n \rightarrow \infty} a_{n}=0\) is not sufficient to guarantee the convergence of the alternating series \(\Sigma(-1)^{n+1} a_{n} .\) Hint: Alternate the terms of \(\sum 1 / n\) and \(\Sigma\left(-1 / n^{2}\right)\).

Test for convergence or divergence using the Root Test. (a) \(\sum_{n=2}^{\infty}\left(\frac{1}{\ln n}\right)^{n}\) (b) \(\sum_{n=1}^{\infty}\left(\frac{n}{3 n+2}\right)^{n}\) (c) \(\sum_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{n}\right)^{n}\)

Determine the order \(n\) of the Maclaurin polynomial for \(4 \tan ^{-1} x\) that is required to approximate \(\pi=4 \tan ^{-1} 1\) to five decimal places, that is, so that \(\left|R_{n}(1)\right| \leq 0.000005\).

Let \(r\) be a fixed number with \(|r|<1 .\) Then it can be shown that \(\sum_{k=1}^{\infty} k r^{k}\) converges, say with sum \(S .\) Use the properties of \(\Sigma\) to show that $$ (1-r) S=\sum_{k=1}^{\infty} r^{k} $$ and then obtain a formula for \(S\), thus generalizing Problem 47 a.
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