Chapter 9: Problem 9
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 \(\lim _{n \rightarrow \infty} a_{n}\). $$ a_{n}=\frac{\cos (n \pi)}{n} $$
Short Answer
Expert verified
The first five terms are -1, 1/2, -1/3, 1/4, -1/5. The sequence converges to 0.
Step by step solution
01
Understanding the sequence formula
The sequence formula given is \( a_n = \frac{\cos(n\pi)}{n} \). This sequence is defined for integer values of \( n \), where each term alternates between positive and negative because \( \cos(n\pi) = (-1)^n \).
02
Calculating the first five terms
To find the first five terms of the sequence, substitute \( n = 1, 2, 3, 4, \) and \( 5 \) into the formula:- For \( n = 1 \): \( a_1 = \frac{\cos(\pi)}{1} = -1 \)- For \( n = 2 \): \( a_2 = \frac{\cos(2\pi)}{2} = \frac{1}{2} \)- For \( n = 3 \): \( a_3 = \frac{\cos(3\pi)}{3} = -\frac{1}{3} \)- For \( n = 4 \): \( a_4 = \frac{\cos(4\pi)}{4} = \frac{1}{4} \)- For \( n = 5 \): \( a_5 = \frac{\cos(5\pi)}{5} = -\frac{1}{5} \)
03
Examining the sequence for convergence or divergence
The expression \( \cos(n\pi) = (-1)^n \) results in \( 1 \) for even \( n \) and \( -1 \) for odd \( n \). However, the denominator \( n \) increases without bound as \( n \) grows larger. Thus, \( a_n = \frac{(-1)^n}{n} \) tends to zero as \( n \) approaches infinity because \( \frac{(-1)^n}{n} \to 0 \).
04
Finding the limit of the sequence
Since \( \frac{(-1)^n}{n} \to 0 \) as \( n \to \infty \), the sequence converges to zero. Therefore, \( \lim_{n \to \infty} a_n = 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Explicit Formula for Sequences
In mathematics, a sequence is a list of numbers arranged in a specific order. An explicit formula allows us to calculate any term in the sequence without knowing the previous terms. This is very practical as it gives a formulaic method to directly find the terms. For a sequence given by an explicit formula, each term can be calculated separately by plugging in the respective value of the index.
Consider the sequence defined by the formula: \( a_{n} = \frac{\cos(n\pi)}{n} \). Here, each term in the sequence can be found by substituting the value of \( n \) as integers greater than zero.
Consider the sequence defined by the formula: \( a_{n} = \frac{\cos(n\pi)}{n} \). Here, each term in the sequence can be found by substituting the value of \( n \) as integers greater than zero.
- The numerator is controlled by the function \( \cos(n\pi) \), which switches between \( -1 \) and \( 1 \) based on whether \( n \) is odd or even.
- The denominator is simply \( n \), which increases linearly as \( n \) increases.
Exploring Limits of Sequences
The concept of limits is central to understanding sequences in calculus. The limit of a sequence \( \{a_n\} \) as \( n \) approaches infinity is the value that the terms of the sequence get arbitrarily close to as \( n \) becomes very large. If this limit exists, the sequence is said to converge; otherwise, it diverges.
Consider the sequence given by \( a_{n} = \frac{(-1)^n}{n} \). Notice here how the numerator alternates between \( -1 \) and \( 1 \), while the denominator, \( n \), grows infinitely. As \( n \) increases, \( \frac{1}{n} \) gets smaller and smaller, ultimately approaching zero.
Consider the sequence given by \( a_{n} = \frac{(-1)^n}{n} \). Notice here how the numerator alternates between \( -1 \) and \( 1 \), while the denominator, \( n \), grows infinitely. As \( n \) increases, \( \frac{1}{n} \) gets smaller and smaller, ultimately approaching zero.
- For even \( n \), \( a_{n} \) is \( \frac{1}{n} \); it becomes very small as \( n \) increases.
- For odd \( n \), \( a_{n} \) is \( -\frac{1}{n} \); it also tends toward zero from the negative side.
Understanding Alternating Sequences
Alternating sequences are sequences where terms alternate in sign. This is usually indicated by an expression like \( (-1)^n \) or \( (-1)^{n+1} \) within the formula.
For example, consider our sequence: \( a_{n} = \frac{(-1)^n}{n} \). The \( (-1)^n \) portion causes the sequence to flip between negative and positive values as \( n \) changes from odd to even.
For example, consider our sequence: \( a_{n} = \frac{(-1)^n}{n} \). The \( (-1)^n \) portion causes the sequence to flip between negative and positive values as \( n \) changes from odd to even.
- The result is a back-and-forth pattern around a central line, like zero in our example.
- This alternating pattern is significant since it affects whether the sequence has a single limit point or diverges.