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Magazine Chapter 9: Problem 121
Determine whether the sequence is monotonic and whether it is bounded. $$a_{n}=\frac{3 n+1}{n+1}$$
Short Answer
Expert verified
The sequence is monotonic (increasing) and bounded.
Step by step solution
01
Understand the Sequence
The given sequence is \( a_n = \frac{3n+1}{n+1} \). We need to determine if this sequence is monotonic and bounded. A sequence is monotonic if it is either entirely non-increasing or non-decreasing. It is bounded if there exist real numbers \( M \) and \( m \) such that \( m \leq a_n \leq M \) for every \( n \).
02
Check for Monotonicity
To check if the sequence is monotonic, consider the difference between consecutive terms:\[a_{n+1} - a_n = \frac{3(n+1) + 1}{(n+1) + 1} - \frac{3n + 1}{n+1} = \frac{3n+4}{n+2} - \frac{3n+1}{n+1}\]Simplify:\[a_{n+1} - a_n = \frac{(3n+4)(n+1) - (3n+1)(n+2)}{(n+2)(n+1)}\]Expanding and simplifying the numerator:\( (3n+4)(n+1) = 3n^2 + 3n + 4n + 4 = 3n^2 + 7n + 4 \)\( (3n+1)(n+2) = 3n^2 + 6n + n + 2 = 3n^2 + 7n + 2 \)Thus:\[a_{n+1} - a_n = \frac{3n^2 + 7n + 4 - (3n^2 + 7n + 2)}{(n+2)(n+1)} = \frac{2}{(n+2)(n+1)}\]Since \( \frac{2}{(n+2)(n+1)} > 0 \) for all positive integers \( n \), the sequence is increasing and hence monotonic.
03
Check if the Sequence is Bounded
To determine if the sequence is bounded, evaluate its limits as \( n \to \infty \):\[\lim_{{n \to \infty}} a_n = \lim_{{n \to \infty}} \frac{3n+1}{n+1}\]Apply L'H么pital's Rule or divide the numerator and the denominator by \( n \):\[\lim_{{n \to \infty}} \frac{3n+1}{n+1} = \lim_{{n \to \infty}} \frac{3 + \frac{1}{n}}{1 + \frac{1}{n}} = \frac{3}{1} = 3\]This implies the sequence approaches 3 as \( n \to \infty \). Since the sequence is increasing and the first term \( a_1 = 2 \) is less than future terms, the sequence is bounded below by 2 and above by a value approaching 3.
04
Conclusion
The sequence \( a_n = \frac{3n+1}{n+1} \) is increasing and hence monotonic. It is bounded below by its first term \( a_1 = 2 \) and it approaches 3 as \( n \to \infty \), so it can be considered bounded above by any number greater than 3, like 3.1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Monotonic Sequence
A monotonic sequence is one that moves in a consistent direction. It either increases or decreases steadily without any deviation. To determine if a sequence is monotonic, we examine the difference between each consecutive term. For the given sequence \(a_n = \frac{3n+1}{n+1}\), we calculate this difference as:
- \(a_{n+1} - a_n = \frac{3n^2 + 7n + 4 - (3n^2 + 7n + 2)}{(n+2)(n+1)} = \frac{2}{(n+2)(n+1)}\).
Bounded Sequence
A sequence is considered bounded if there are specific limits it doesn't surpass, both above and below. For a sequence to be bounded, there should exist real numbers \(M\) and \(m\) such that \(m \leq a_n \leq M\) for every \(n\). Looking at \(a_n = \frac{3n+1}{n+1}\),we found the first term to be \(a_1 = 2\), which acts as a lower boundary since the sequence increases.
- The sequence approaches 3 as shown by the limit calculation:
- \(\lim_{{n \to \infty}} a_n = 3\).
Limit of a Sequence
The limit of a sequence defines the value that the terms of a sequence approach as the sequence progresses indefinitely. For the sequence \(a_n = \frac{3n+1}{n+1}\), we seek to find how close the values get as \(n\) becomes extremely large. To achieve this, we calculate:
- Use L'H么pital's Rule or simplify:\(\lim_{{n \to \infty}} \frac{3n+1}{n+1} = \lim_{{n \to \infty}} \frac{3 + \frac{1}{n}}{1 + \frac{1}{n}} = 3\).
