Problem 5
Question
a) Let \(x_{n}:=\frac{(-1)^{n}}{n} .\) Find \(\limsup x_{n}\) and \(\liminf x_{n}\). b) Let \(x_{n}:=\frac{(n-1)(-1)^{n}}{n} .\) Find \(\limsup x_{n}\) and \(\liminf x_{n}\).
Step-by-Step Solution
Verified Answer
a) Both \(\limsup\) and \(\liminf\) are 0. b) \(\limsup = 1\), \(\liminf = -1\).
1Step 1: Identifying the Sequence for Part (a)
The sequence given is \(x_n = \frac{(-1)^n}{n}\). This sequence is composed of terms: \(\frac{1}{1}, -\frac{1}{2}, \frac{1}{3}, -\frac{1}{4}, \dots\). It alternates between positive and negative and approaches 0 as \(n\) increases.
2Step 2: Finding \(\limsup\) for Part (a)
The \(\limsup\) of the sequence is the largest limit point. For the given sequence \(\frac{1}{n}\), the positive terms \(\left(\frac{1}{1}, \frac{1}{3}, \frac{1}{5}, \dots\right)\) converge to 0. Therefore, \(\limsup x_n = 0\).
3Step 3: Finding \(\liminf\) for Part (a)
The \(\liminf\) of the sequence is the smallest limit point. For the given sequence \(-\frac{1}{2}, -\frac{1}{4}, -\frac{1}{6}, \dots\), since these are negative and also converge to 0, \(\liminf x_n = 0\).
4Step 4: Conclusion for Part (a)
Both the \(\limsup\) and \(\liminf\) of \(x_n = \frac{(-1)^n}{n}\) are 0, as both positive and negative subsequences converge to 0.
5Step 5: Identifying the Sequence for Part (b)
The sequence given is \(x_n = \frac{(n-1)(-1)^n}{n}\). This simplifies to \(\frac{n-1}{n}\) for positive \(n\) and \(-\frac{n-1}{n}\) for negative, giving terms like \(0, -\frac{1}{2}, \frac{2}{3}, -\frac{3}{4}, \dots\).
6Step 6: Finding \(\limsup\) for Part (b)
The sequence's positive terms are \(0, \frac{2}{3}, \frac{4}{5}, \dots\), which increase monotonically towards 1. Therefore, \(\limsup x_n = 1\).
7Step 7: Finding \(\liminf\) for Part (b)
The sequence's negative terms \(-\frac{1}{2}, -\frac{3}{4}, -\frac{5}{6}, \dots\) also approach \(-1\). Thus, \(\liminf x_n = -1\).
8Step 8: Conclusion for Part (b)
For \(x_n = \frac{(n-1)(-1)^n}{n}\), the \(\limsup\) is 1 and the \(\liminf\) is \(-1\).
Key Concepts
Sequence AnalysisConvergence of SeriesAlternating Sequences
Sequence Analysis
Sequence analysis involves studying the behavior of the terms in a sequence as they progress. Consider sequences like \(x_n = \frac{(-1)^n}{n}\) and \(x_n = \frac{(n-1)(-1)^n}{n}\). These sequences alternate terms, switching between positive and negative values as \(n\) increases. Analyzing sequences often involves looking at how the values change, finding limit points, and understanding overall trends.
In the first sequence, \(x_n\), the terms diminish in size, converging towards zero. The terms alternate due to the \((-1)^n\) factor, causing changes in sign. This creates a repeating, zig-zag pattern of values that becomes less extreme as \(n\) grows. Similarly, the second sequence \(x_n\) not only alternates signs, but also approaches bounds determined by the fractions \(\frac{n-1}{n}\), which hover between values within increasing proximity to one (for positive) and negative one (for negative).
Through sequence analysis, you learn to identify these trends and predict what happens as sequences extend. This is crucial for solving problems involving limit superior (lim sup) and limit inferior (lim inf) as these concepts pinpoint the upper and lower bounds that sequences approach.
In the first sequence, \(x_n\), the terms diminish in size, converging towards zero. The terms alternate due to the \((-1)^n\) factor, causing changes in sign. This creates a repeating, zig-zag pattern of values that becomes less extreme as \(n\) grows. Similarly, the second sequence \(x_n\) not only alternates signs, but also approaches bounds determined by the fractions \(\frac{n-1}{n}\), which hover between values within increasing proximity to one (for positive) and negative one (for negative).
Through sequence analysis, you learn to identify these trends and predict what happens as sequences extend. This is crucial for solving problems involving limit superior (lim sup) and limit inferior (lim inf) as these concepts pinpoint the upper and lower bounds that sequences approach.
Convergence of Series
Convergence of series is a foundational concept in calculus and sequence analysis. When a sequence converges, its terms approach a single, specific value as \(n\) becomes very large. This value is known as the limit of the sequence. For the sequences in our exercise, we used terms like \(\limsup\) and \(\liminf\) to define their convergence.
The sequence \(x_n = \frac{(-1)^n}{n}\) converges to zero in both its limit superior and inferior as both the positive and negative terms approach zero. In contrast, the sequence \(x_n = \frac{(n-1)(-1)^n}{n}\) is more complex. Its limit superior, or largest accumulation point, reaches one for positive terms, while its limit inferior, or smallest accumulation point, dips to negative one for negative terms. Convergence analysis for sequences like these involves carefully checking if limit points exist and how they relate to each other.
Understanding convergence helps in predicting how the sequence behaves eventually, which is vital in determining whether functions describing these sequences remain bounded, grow indefinitely, or stabilize.
The sequence \(x_n = \frac{(-1)^n}{n}\) converges to zero in both its limit superior and inferior as both the positive and negative terms approach zero. In contrast, the sequence \(x_n = \frac{(n-1)(-1)^n}{n}\) is more complex. Its limit superior, or largest accumulation point, reaches one for positive terms, while its limit inferior, or smallest accumulation point, dips to negative one for negative terms. Convergence analysis for sequences like these involves carefully checking if limit points exist and how they relate to each other.
Understanding convergence helps in predicting how the sequence behaves eventually, which is vital in determining whether functions describing these sequences remain bounded, grow indefinitely, or stabilize.
Alternating Sequences
Alternating sequences are those that switch signs between consecutive terms, typically influenced by a component like \((-1)^n\). Such sequences are intriguing due to their oscillating nature, making them useful for demonstrating the behavior of more complex mathematical systems.
For \(x_n = \frac{(-1)^n}{n}\), the sequence alternates between positive and negative fractions, each time reducing in absolute value as \(\frac{1}{n}\) becomes smaller. Despite the changes in sign, the sequence approaches zero steadily. Similarly, the sequence \(x_n = \frac{(n-1)(-1)^n}{n}\) creates a pattern where positive terms increase toward one, and negative terms decrease toward negative one. This zig-zag movement allows for distinguishing clear limits like \(\limsup\) and \(\liminf\).
Such alternating patterns are not just academically interesting; they demonstrate principles of balance and recurring adjustments. Analyzing alternating sequences helps grasp the extremes they can reach and understand their periodic behavior in depth.
For \(x_n = \frac{(-1)^n}{n}\), the sequence alternates between positive and negative fractions, each time reducing in absolute value as \(\frac{1}{n}\) becomes smaller. Despite the changes in sign, the sequence approaches zero steadily. Similarly, the sequence \(x_n = \frac{(n-1)(-1)^n}{n}\) creates a pattern where positive terms increase toward one, and negative terms decrease toward negative one. This zig-zag movement allows for distinguishing clear limits like \(\limsup\) and \(\liminf\).
Such alternating patterns are not just academically interesting; they demonstrate principles of balance and recurring adjustments. Analyzing alternating sequences helps grasp the extremes they can reach and understand their periodic behavior in depth.
Other exercises in this chapter
Problem 5
For the following power series, find if they are convergent or not, and if so find their radius of convergence. a) \(\sum_{n=0}^{\infty} 2^{n} x^{n}\) b) \(\sum
View solution Problem 5
Suppose a Cauchy sequence \(\left\\{x_{n}\right\\}\) is such that for every \(M \in \mathbb{N},\) there exists a \(k \geq M\) and an \(n \geq M\) such that \(x_
View solution Problem 5
For \(j=1,2, \ldots, n\), let \(\left\\{x_{j k}\right\\}_{k=1}^{\infty}\) denote n sequences. Suppose that for each \(j\) $$ \sum_{k=1}^{\infty} x_{j, k} $$ is
View solution Problem 5
In the following exercises, feel free to use what you know from calculus to find the limit, if it exists. But you must prove that you found the correct limit, o
View solution