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In Problems 1-20, an explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\) \(a_{n}=\frac{n \cos (n \pi)}{2 n-1}\)

Short Answer

Expert verified
The sequence \(a_n\) diverges because it does not settle to a single value as \(n\) approaches infinity.

Step by step solution

01

Identify the Sequence Formula

The given formula for the sequence is \(a_{n} = \frac{n \cos(n\pi)}{2n - 1}\). Notice that the cosine term, \(\cos(n\pi)\), will alternate between 1 and -1 depending on whether \(n\) is even or odd.
02

Calculate the First Five Terms

Let's calculate the first five terms of the sequence:- For \(n = 1\), \(a_{1} = \frac{1 \cdot \cos(1\pi)}{2\cdot1 - 1} = \frac{1 \cdot (-1)}{1} = -1\).- For \(n = 2\), \(a_{2} = \frac{2 \cdot \cos(2\pi)}{2\cdot2 - 1} = \frac{2 \cdot 1}{3} = \frac{2}{3}\).- For \(n = 3\), \(a_{3} = \frac{3 \cdot \cos(3\pi)}{2\cdot3 - 1} = \frac{3 \cdot (-1)}{5} = -\frac{3}{5}\).- For \(n = 4\), \(a_{4} = \frac{4 \cdot \cos(4\pi)}{2\cdot4 - 1} = \frac{4 \cdot 1}{7} = \frac{4}{7}\).- For \(n = 5\), \(a_{5} = \frac{5 \cdot \cos(5\pi)}{2\cdot5 - 1} = \frac{5 \cdot (-1)}{9} = -\frac{5}{9}\).The first five terms are \(-1, \frac{2}{3}, -\frac{3}{5}, \frac{4}{7}, -\frac{5}{9}\).
03

Analyze Convergence

To determine convergence, we evaluate the limit as \(n\) approaches infinity: \[\lim_{{n \to \infty}} \frac{n \cos(n\pi)}{2n - 1}\]Consider the absolute value first, ignoring signs to analyze magnitude:\[\left| \frac{n}{2n-1} \right| = \left| \frac{1}{2 - \frac{1}{n}} \right| \to \frac{1}{2} \text{ as } n \to \infty\]Since the magnitudes approach \(\frac{1}{2}\) and the sequence alternates in sign between positive and negative, the sequence does not settle to a single value and diverges.
04

Conclude on Sequence Behavior

The sequence does not converge to a single value because the terms alternate between positive and negative, and thus it diverges. You must ensure the sequence tends towards a single finite value for convergence, which is not the case here.

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Key Concepts

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

Explicit Formula
The explicit formula is used to find the terms of a sequence without needing any preceding terms. It directly expresses each term as a function of its position number (n). In our exercise, the explicit formula is \(a_{n} = \frac{n \cos(n\pi)}{2n - 1}\). This formula tells us how each term relates to \(n\) directly, which is helpful because:
  • It streamlines calculations for terms at any position \(n\).
  • Only the formula is needed to compute the terms of the sequence.
For instance, by simply substituting \(n = 1, 2, 3, 4, 5\) into the formula, you obtain the terms \(-1, \frac{2}{3}, -\frac{3}{5}, \frac{4}{7}, -\frac{5}{9}\). Each calculation is straightforward and the formula enables us to predict the behavior of the sequence further out.
Alternating Sequence
An alternating sequence is a sequence of numbers whose terms alternate in sign. Here, the sequence \(a_{n} = \frac{n \cos(n\pi)}{2n - 1}\) is an example.
  • For even \(n\), \(\cos(n\pi) = 1\), making the term positive.
  • For odd \(n\), \(\cos(n\pi) = -1\), making the term negative.
This sign change pattern creates an alternating sequence. The fluctuations between negative and positive values impact whether a sequence converges. It's crucial to understand alternating sequences when analyzing convergence because the alternating nature can heavily influence the sequence's behavior over time.
Limit Analysis
Limit analysis helps determine whether a sequence converges towards a single finite value as \(n\) approaches infinity. To examine the sequence \(a_{n} = \frac{n \cos(n\pi)}{2n - 1}\), we consider\[\lim_{{n \to \infty}} \frac{n \cos(n\pi)}{2n - 1}\]Firstly, focusing on the magnitude, ignoring sign,\[\left| \frac{n}{2n-1} \right| = \left| \frac{1}{2 - \frac{1}{n}} \right|\]as \(n\) approaches infinity, this expression approaches \(\frac{1}{2}\). Although the magnitude suggests convergence towards \(\frac{1}{2}\), the varied sign due to \(\cos(n\pi)\) causes the sequence values to alternate. Thus, the alternating signs hinder the sequence from settling on a single value, demonstrating that \(a_{n}\) diverges. Proper limit analysis considers both magnitude and sign variation to conclude convergence or divergence.

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