Problem 76
Question
Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? \( a_n = 2 + \frac{(-1)^n}{n} \)
Step-by-Step Solution
Verified Answer
The sequence is not monotonic and is bounded.
1Step 1: Understand the Sequence
We are given the sequence \( a_n = 2 + \frac{(-1)^n}{n} \). The term \( \frac{(-1)^n}{n} \) alternates signs and becomes smaller in magnitude as \( n \) increases, because \( n \) is in the denominator.
2Step 2: Analyze Monotonicity
To determine if the sequence is increasing or decreasing, compare consecutive terms: \( a_n = 2 + \frac{(-1)^n}{n} \) and \( a_{n+1} = 2 + \frac{(-1)^{n+1}}{n+1} \). Since the expression \( \frac{(-1)^n}{n} \) alternates in sign, the sequence oscillates instead of being entirely increasing or decreasing, making it not monotonic.
3Step 3: Check for Boundedness
The sequence involves the term \( \frac{(-1)^n}{n} \) which oscillates between values approaching 0. Thus, the values of \( a_n \) oscillate between \( 2 + \frac{1}{n} \) and \( 2 - \frac{1}{n} \). As \( n \to \infty \), these bounds converge to 2. Since sequence terms fall between these bounds, it is bounded.
Key Concepts
Bounded SequencesAlternating SequencesSequence Analysis
Bounded Sequences
Understanding bounded sequences is key to comprehending certain sequences. A sequence is considered bounded if all its terms are confined within a fixed interval on the real number line.
For example, consider the sequence given:
For example, consider the sequence given:
- The terms of this sequence comprise the expression \( a_n = 2 + \frac{(-1)^n}{n} \), where the term \( \frac{(-1)^n}{n} \) is integral in bounding the sequence. As \( n \) increases, this term alternates in sign but decreases in magnitude due to the \( n \) in the denominator.
- This results in the sequence terms oscillating between \( 2 + \frac{1}{n} \) and \( 2 - \frac{1}{n} \). Importantly, as \( n \to \infty \), the upper bound \( 2 + \frac{1}{n} \) and the lower bound \( 2 - \frac{1}{n} \) approach 2, thus pinning all sequence terms between these two limits.
Alternating Sequences
An alternating sequence is characterized by terms that change sign with each subsequent element. This often results in an oscillating pattern.
In our sequence, the term \( \frac{(-1)^n}{n} \) plays a crucial role.
In our sequence, the term \( \frac{(-1)^n}{n} \) plays a crucial role.
- The presence of \((-1)^n\) signifies that the sequence alternates signs, as \((-1)^n\) yields \(-1\) for odd \(n\) and \(+1\) for even \(n\).
- Coupled with the fact that the magnitude of \( \frac{1}{n} \) diminishes as \(n\) grows, each subsequent term becomes smaller, thus moderating the oscillation.
Sequence Analysis
When analyzing sequences, one often investigates their nature, behavior, and characteristics. Let's apply some sequence analysis techniques to our equation.
The key points of analysis include:
The key points of analysis include:
- Examining the sequence’s behavior: Does it increase, decrease, or fluctuate? The sequence \( a_n = 2 + \frac{(-1)^n}{n} \) oscillates, which we know because its terms alternate due to the \((-1)^n\) factor.
- Checking for monotonicity: A sequence is monotonic if it keeps moving in one direction, either always increasing or always decreasing. Our sequence fails this, evident from alternating terms which fluctuate above and below the value of 2.
- Determining boundedness: Here, we determined that the sequence is bounded. Its terms do not veer beyond the bounds \(2 + \frac{1}{n}\) and \(2 - \frac{1}{n}\), consistently converging towards 2 as \(n\) increases.
Other exercises in this chapter
Problem 75
Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded? \( a_n = n(-1)^n \)
View solution Problem 76
Find the sum of the series. \( \sum_{n = 0}^{\infty} \frac {3^n}{5^n n!} \)
View solution Problem 76
Find the value of c such that \( \displaystyle \sum_{n = 0}^{\infty} e^{nc} = 10 \)
View solution Problem 77
Find the sum of the series. \( \sum_{n = 0}^{\infty} \frac {(-1)^n \pi^{2n + 1}}{4^{2n + 1}(2n + 1)!} \)
View solution