Problem 26
Question
\(9-32\) n Determine whether the sequence converges or diverges. If it converges, find the limit. $$a_{n}=2^{-n} \cos n \pi$$
Step-by-Step Solution
Verified Answer
The sequence converges to 0 as \( n \to \infty \).
1Step 1: Understanding the Sequence
The given sequence is \( a_n = 2^{-n} \cos(n \pi) \). We need to determine its behavior as \( n \to \infty \).
2Step 2: Analyzing the Terms
Notice that \( \cos(n \pi) \) alternates between \( 1 \) and \( -1 \) depending on the parity of \( n \). Specifically, \( \cos(n \pi) = (-1)^n \).
3Step 3: Simplifying the Sequence
The sequence can be rewritten as \( a_n = 2^{-n}(-1)^n \). This means the sequence terms become \( a_n = \frac{(-1)^n}{2^n} \).
4Step 4: Examining Limit as n Approaches Infinity
The term \( \frac{(-1)^n}{2^n} \) gets smaller in magnitude as \( n \) increases because \( 2^n \) grows exponentially. As \( n \to \infty \), \( 2^{-n} \) approaches zero.
5Step 5: Convergence Conclusion
Since the sequence \( \frac{(-1)^n}{2^n} \) approaches zero, we conclude that the sequence converges and its limit is \( 0 \).
Key Concepts
Limit of a SequenceAlternating SequencesExponential Growth
Limit of a Sequence
The concept of the "limit of a sequence" is crucial for understanding how sequences behave as they progress towards infinity. A sequence refers to a list of numbers in a specific order, where each number is defined as a term. When we talk about the limit, we want to know what value (if any) the terms in the sequence are approaching. For a sequence \( a_n \), the limit as \( n \to \infty \) is denoted as \( \lim_{{n \to \infty}} a_n \). If the terms get closer and closer to a specific number as \( n \) increases, we say the sequence converges to that limit.
If the terms of the sequence do not approach a particular value, then the sequence diverges. For the sequence \( a_n = \frac{(-1)^n}{2^n} \), recognizing the behavior of \( \frac{1}{2^n} \) is key. As \( n \) gets large, \( 2^n \) becomes very large, making \( \frac{1}{2^n} \) close to zero. Thus, the sequence converges to a limit of 0, which means the terms approach zero as \( n \) goes to infinity.
Understanding the limit helps in predicting long-term behavior of sequences, which is widely applicable in various mathematical contexts.
If the terms of the sequence do not approach a particular value, then the sequence diverges. For the sequence \( a_n = \frac{(-1)^n}{2^n} \), recognizing the behavior of \( \frac{1}{2^n} \) is key. As \( n \) gets large, \( 2^n \) becomes very large, making \( \frac{1}{2^n} \) close to zero. Thus, the sequence converges to a limit of 0, which means the terms approach zero as \( n \) goes to infinity.
Understanding the limit helps in predicting long-term behavior of sequences, which is widely applicable in various mathematical contexts.
Alternating Sequences
Alternating sequences are sequences in which the signs of the terms change regularly from positive to negative or vice versa. This happens because of a term like \( (-1)^n \) that toggles between 1 and -1 as \( n \) increases.
For example, in our sequence \( a_n = 2^{-n} \cos(n \pi) \), the \( \cos(n \pi) \) part is equivalent to \((-1)^n\), hence the sequence term alternates signs. This component is what makes the sequence alternate between negative and positive values.
For example, in our sequence \( a_n = 2^{-n} \cos(n \pi) \), the \( \cos(n \pi) \) part is equivalent to \((-1)^n\), hence the sequence term alternates signs. This component is what makes the sequence alternate between negative and positive values.
- When \( n \) is even, \( \cos(n \pi) = 1 \), making the term positive.
- When \( n \) is odd, \( \cos(n \pi) = -1 \), making the term negative.
Exponential Growth
Exponential growth describes the rapid increase of a quantity over time, often modeled with expressions like \( 2^n \). In mathematics, this type of growth means that as \( n \) increases, \( 2^n \) becomes significantly larger. This concept is very relevant when analyzing sequences because it can determine the behavior of the sequence over time.
In the provided sequence \( a_n = \frac{(-1)^n}{2^n} \), the term \( 2^n \) grows exponentially as \( n \) becomes larger. This exponential growth causes the denominator to outgrow the numerator significantly, resulting in the entire fraction \( \frac{1}{2^n} \) getting closer to zero. Therefore, even though the sequence alternates signs, the exponential growth of \( 2^n \) ensures that the magnitudes of the terms shrink, contributing to the sequence's convergence.
Understanding exponential growth helps to anticipate how quickly terms will approach zero or diverge depending on other factors in the sequence.
In the provided sequence \( a_n = \frac{(-1)^n}{2^n} \), the term \( 2^n \) grows exponentially as \( n \) becomes larger. This exponential growth causes the denominator to outgrow the numerator significantly, resulting in the entire fraction \( \frac{1}{2^n} \) getting closer to zero. Therefore, even though the sequence alternates signs, the exponential growth of \( 2^n \) ensures that the magnitudes of the terms shrink, contributing to the sequence's convergence.
Understanding exponential growth helps to anticipate how quickly terms will approach zero or diverge depending on other factors in the sequence.
Other exercises in this chapter
Problem 26
Use the binomial series to expand the function as a power serics. State the radius of convergence. $$(1-x)^{2 / 3}$$
View solution Problem 26
\(19-40=\) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty} \frac{n !}{100^{n}} $$
View solution Problem 27
Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int x^{2} \ln (1+x) d x $$
View solution Problem 27
If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth. (a) If
View solution