Problem 53
Question
Determine whether the sequence \(\left\\{a_{n}\right\\}\) is monotonic. Is the sequence bounded? \(a_{n}=3-\frac{1}{n}\)
Step-by-Step Solution
Verified Answer
The sequence \(a_n = 3-\frac{1}{n}\) is monotonic since the difference between consecutive terms, \(a_{n+1} - a_n = \frac{1}{n} - \frac{1}{n+1}\), is positive for all \(n \in \mathbb{N}\). The sequence is also bounded since all terms are greater than \(2\) (lower bound) and converging to \(3\) (upper bound).
1Step 1: Define the sequence
The given sequence is defined as \(a_{n}=3-\frac{1}{n}\), where \(n \in \mathbb{N}\).
2Step 2: Find the difference between consecutive terms
To determine if the sequence is monotonic, we need to find the difference between consecutive terms, i.e., \(a_{n+1} - a_n\). First, we express \(a_{n+1}\):
\(a_{n+1} = 3-\frac{1}{n+1}\)
Now, let us calculate the difference between consecutive terms, \(a_{n+1} - a_n\):
\(a_{n+1} - a_n = \left(3-\frac{1}{n+1}\right) - \left(3-\frac{1}{n}\right)\)
3Step 3: Simplify the difference expression
Next, we need to simplify the expression for the difference we obtained in step 2:
\(a_{n+1} - a_n = 3 - \frac{1}{n+1} - 3 + \frac{1}{n}\)
\(a_{n+1} - a_n = \frac{1}{n} - \frac{1}{n+1}\)
4Step 4: Analyze the sign of the difference expression
Now, we analyze the sign of the simplified difference expression:
Since \(n \in \mathbb{N}\), \(n\) will always be a positive integer. Hence, the terms \(\frac{1}{n}\) and \(\frac{1}{n+1}\) are both positive. Since \(n+1 > n\), we have \(\frac{1}{n} > \frac{1}{n+1}\), which implies that their difference is positive:
\(\frac{1}{n} - \frac{1}{n+1} > 0\)
This indicates that \(a_{n+1} - a_n > 0\) for all \(n \in \mathbb{N}\). The sequence is increasing, so it is monotonic.
5Step 5: Determine whether the sequence is bounded
Finally, we need to check if the sequence is bounded.
Since \(a_n = 3-\frac{1}{n}\), as \(n\) goes to infinity, the term \(\frac{1}{n}\) goes to zero, and hence, \(a_n\) approaches \(3\). Therefore, the sequence is increasing but converging to \(3\). It is clear that all terms are greater than \(2\) (i.e., \(a_n > 2\) when \(n \geq 1\)).
Thus, we have a lower bound \(2\) and an upper bound \(3\), and we conclude that the sequence is monotonic and bounded.
Key Concepts
Bounded SequencesConvergenceNatural Numbers
Bounded Sequences
A sequence is said to be bounded if there exists a real number that serves as an upper bound and another real number that serves as a lower bound for all terms in the sequence. For the sequence given as \(a_n = 3 - \frac{1}{n}\), we can see:
- The sequence approaches 3 as \(n\) becomes very large, indicating that 3 acts as an upper boundary.
- For every term in this sequence when \(n \geq 1\), we have \(a_n = 3 - \frac{1}{n}\) where each term is always greater than 2, establishing 2 as a lower boundary.
Convergence
In mathematical terms, a sequence is said to converge if its terms approach a specific number as \(n\), the term number, goes infinitely large. The concept of convergence is critically important when considering the long-term behavior of sequences.
- The given sequence \(a_n = 3 - \frac{1}{n}\) converges because as \(n\) gets larger, \(\frac{1}{n}\) becomes smaller and smaller, nearing zero.
- Thus, the value of \(a_n\) becomes increasingly close to 3.
Natural Numbers
The sequence \(a_n = 3 - \frac{1}{n}\) is defined for natural numbers \(n\). Natural numbers \(\mathbb{N}\) are essentially the set of positive integers starting from 1 upwards (1, 2, 3, ...). These numbers are foundational in mathematics, providing the building blocks for more complex ideas.
- Because the sequence is based on \(n\) being a natural number, \(n\) is always a positive integer, which influences the behavior of the sequence.
- Natural numbers ensure that \(\frac{1}{n}\) is well-defined and remains a positive fraction, allowing \(a_n\) to consistently approximate the limit of 3 as \(n\) grows.
Other exercises in this chapter
Problem 53
Determine whether the given series converges or diverges. If it converges, find its sum. \(\sum_{n=2}^{\infty} \frac{n}{\ln n}\)
View solution Problem 53
Use the Integral Test to show that \(\sum_{n=3}^{\infty} \frac{1}{n(\ln n)[\ln (\ln n)]^{p}}\) converges if \(p>1\) and diverges if \(p \leq 1\).
View solution Problem 54
Use a power series to obtain an approximation of the definite integral to four decimal places of accuracy. \(\int_{0}^{1} \sin x^{2} d x\)
View solution Problem 54
Prove that (a) \(\int_{1}^{\infty} \frac{1}{\sqrt{x(x+1)(x+2)}} d x\) converges and (b) \(\int_{1}^{\infty} \frac{1}{\sqrt{x(x+1)}} d x\) diverges.
View solution