Problem 28
Question
Limits of Sequences If the sequence with the given \(n\) th term is convergent, find its limit. If it is divergent, explain why. $$a_{n}=\frac{(-1)^{n}}{n}$$
Step-by-Step Solution
Verified Answer
The sequence converges to 0 as \( n \rightarrow \infty \).
1Step 1: Understand the Sequence Behavior
Let's first look at the sequence given: \( a_n = \frac{(-1)^n}{n} \). The presence of \((-1)^n\) implies that the sequence alternates in sign. This means for even \(n\), \(a_n\) is positive and for odd \(n\), \(a_n\) is negative.
2Step 2: Analyze Absolute Values
Consider the absolute value of the sequence: \( |a_n| = \frac{1}{n} \). As \(n\) approaches infinity, \(\frac{1}{n} \) approaches 0. This indicates that the magnitude of terms becomes smaller and trends towards zero.
3Step 3: Investigate Behavior at Infinity
Although the sequence alternates between positive and negative values, we see that each term \( a_n \) in absolute value becomes arbitrarily small as \( n \) increases, because \( |a_n| = \frac{1}{n} \rightarrow 0 \) as \( n \rightarrow \infty \).
4Step 4: Convergence Conclusion
Despite the alternating signs, the overall trend of the sequence \( a_n = \frac{(-1)^n}{n} \) drives it towards zero as \( n \rightarrow \infty \). Therefore, the sequence converges to 0.
Key Concepts
Limits of SequencesAlternating SequencesSequence Behavior
Limits of Sequences
The concept of limits in sequences is crucial for understanding whether a sequence converges or diverges. To find the limit of a sequence means to determine what value, if any, the terms of the sequence approach as the index, often labeled as \(n\), goes towards infinity. If a sequence tends towards a specific number, we say it converges to that number, known as the limit.
When analyzing limits, consider the sequence given in the exercise: \(a_n = \frac{(-1)^n}{n}\). At first glance, it might seem challenging due to the alternating sign from \((-1)^n\). However, by focusing on its absolute values, \(|a_n| = \frac{1}{n}\), we observe a clear path. As \(n\) increases, \(\frac{1}{n}\) approaches \(0\). Thus, we conclude that the sequence \(a_n\) converges to \(0\) because the magnitude of its terms become negligible as \(n\) grows indefinitely.
Understanding limits helps to predict sequence behavior, one of the many stepping stones in calculus that encourages precise functions analysis and deeper insight into mathematical phenomena.
When analyzing limits, consider the sequence given in the exercise: \(a_n = \frac{(-1)^n}{n}\). At first glance, it might seem challenging due to the alternating sign from \((-1)^n\). However, by focusing on its absolute values, \(|a_n| = \frac{1}{n}\), we observe a clear path. As \(n\) increases, \(\frac{1}{n}\) approaches \(0\). Thus, we conclude that the sequence \(a_n\) converges to \(0\) because the magnitude of its terms become negligible as \(n\) grows indefinitely.
Understanding limits helps to predict sequence behavior, one of the many stepping stones in calculus that encourages precise functions analysis and deeper insight into mathematical phenomena.
Alternating Sequences
Alternating sequences are marked by their terms' signs switching between positive and negative, and they are often denoted by the factor \((-1)^n\) or similar, which flips the sign for successive terms. This switch introduces an intriguing dimension to sequence analysis.
For the sequence \(a_n = \frac{(-1)^n}{n}\), we see this exact alternation:
While analyzing alternating sequences, the changes in signs do not alter the magnitude, which is crucial in determining the limits and convergence. As seen with our sequence, despite the alternation, the \(|a_n| = \frac{1}{n}\) heads towards 0, making the alternating sign merely a feature rather than an obstacle to finding convergence.
For the sequence \(a_n = \frac{(-1)^n}{n}\), we see this exact alternation:
- For even \(n\): the term \(a_n\) becomes positive.
- For odd \(n\): the term \(a_n\) becomes negative.
While analyzing alternating sequences, the changes in signs do not alter the magnitude, which is crucial in determining the limits and convergence. As seen with our sequence, despite the alternation, the \(|a_n| = \frac{1}{n}\) heads towards 0, making the alternating sign merely a feature rather than an obstacle to finding convergence.
Sequence Behavior
Understanding sequence behavior provides valuable insights into whether a series of numbers can seemingly stabilize at a particular value or not. One essential behavior is the convergence, where a sequence approaches an identifiable limit as \(n\) becomes very large.
The sequence \(a_n = \frac{(-1)^n}{n}\) illustrates fascinating behavior:
Sequences like these are incredibly useful in understanding mathematical concepts in calculus and other branches, where recognizing patterns and applying knowledge of limits and convergence guides to recognizing functions and series behavior, supporting accurate and meaningful conclusions in analysis.
The sequence \(a_n = \frac{(-1)^n}{n}\) illustrates fascinating behavior:
- It oscillates due to alternate signs but diminishes in magnitude over time.
- Each term magnitude decreases, following \(|a_n| = \frac{1}{n}\), which trends to zero as \(n\rightarrow \infty\).
Sequences like these are incredibly useful in understanding mathematical concepts in calculus and other branches, where recognizing patterns and applying knowledge of limits and convergence guides to recognizing functions and series behavior, supporting accurate and meaningful conclusions in analysis.
Other exercises in this chapter
Problem 28
Find the following for the given function \(f:\) (a) \(f^{\prime}(a),\) where \(a\) is in the domain of \(f,\) and (b) \(f^{\prime}(3)\) and \(f^{\prime}(4)\) $
View solution Problem 28
Finding Limits Evaluate the limit if it exists. $$\lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{h}$$
View solution Problem 29
One-Sided Limits Graph the piecewise-defined function and use your graph to find the values of the limits, if they exist. $$f(x)=\left\\{\begin{array}{ll} x^{2}
View solution Problem 29
Find the following for the given function \(f:\) (a) \(f^{\prime}(a),\) where \(a\) is in the domain of \(f,\) and (b) \(f^{\prime}(3)\) and \(f^{\prime}(4)\) $
View solution