Convergence of the Sequence rⁿ
For what values of r is the sequence rⁿ convergent?
Answer:
The sequence rⁿ is convergent if and only if |r| < 1. Let's break down the cases to understand why:
- For |r| < 1:
- When the absolute value of r is less than 1, each term is multiplied by a fraction, making the terms get closer and closer to 0.
- Example with r = 0.5: 0.5 × 0.5 = 0.25, 0.25 × 0.5 = 0.125, …
- Example with r = -0.5: (-0.5) × (-0.5) = 0.25, 0.25 × (-0.5) = -0.125, …
- For |r| = 1:
- When the absolute value of r is exactly 1, the terms will either be constant or oscillate between two values.
- Example with r = 1: 1 × 1 = 1, 1 × 1 = 1, …
- Example with r = -1: (-1) × (-1) = 1, 1 × (-1) = -1, …
- For |r| > 1:
- When the absolute value of r is greater than 1, each term is multiplied by a number greater than 1, making the terms grow without bound.
- Example with r = 2: 2 × 2 = 4, 4 × 2 = 8, 8 × 2 = 16, …
- Example with r = -2: (-2) × (-2) = 4, 4 × (-2) = -8, 16 × (-2) = 16, …
Why r = 1 is the Breakpoint:
- The value r = 1 is the natural breakpoint because it’s the boundary between the three distinct behaviors of the sequence.
- Choosing other values like r = 2 or r = 0.5 as the breakpoint would not capture the essential differences in behavior.
This analysis helps us understand why the sequence rⁿ is convergent for |r| < 1 and divergent otherwise.
Increasing Sequences
An increasing sequence is a sequence where each term is greater than or equal to the previous term. In mathematical terms, a sequence (aₙ) is increasing if aₙ₊₁ ≥ aₙ for all n.
Understanding the Subscripts:
The subscript n represents the position of a term in the sequence. For example, aₙ is the nth term, a₃ is the third term, and so on.
The subscript n+1 represents the next position after n. So if aₙ is the nth term, then aₙ₊₁ is the (n+1)th term, or the term immediately following aₙ.
Example:
Consider the sequence defined by aₙ = n². This sequence is increasing because:
- aₙ = n² (the nth term)
- aₙ₊₁ = (n+1)² (the (n+1)th term, or the term immediately after the nth term)
- Since (n+1)² > n², we have aₙ₊₁ > aₙ for all n, so the sequence is increasing.
Here’s a comparison between consecutive terms:
- a₁ (first term) = 1² = 1
- a₂ (second term) = 2² = 4, a₂ > a₁
- a₃ (third term) = 3² = 9, a₃ > a₂
- …
In this example, the subscript n represents the position of the term, and n+1 represents the next position. By comparing the terms at these positions, we can see that the sequence is increasing.
Decreasing Sequences
A decreasing sequence is a sequence where each term is less than or equal to the previous term. In mathematical terms, a sequence (aₙ) is decreasing if aₙ₊₁ ≤ aₙ for all n.
Understanding the Subscripts:
The subscript n represents the position of a term in the sequence. For example, aₙ is the nth term, a₃ is the third term, and so on.
The subscript n+1 represents the next position after n. So if aₙ is the nth term, then aₙ₊₁ is the (n+1)th term, or the term immediately following aₙ.
Example:
Consider the sequence defined by aₙ = -n². This sequence is decreasing because:
- aₙ = -n² (the nth term)
- aₙ₊₁ = -(n+1)² (the (n+1)th term, or the term immediately after the nth term)
- Since -(n+1)² < -n², we have aₙ₊₁ < aₙ for all n, so the sequence is decreasing.
Here’s a comparison between consecutive terms:
- a₁ (first term) = -1² = -1
- a₂ (second term) = -2² = -4, a₂ < a₁
- a₃ (third term) = -3² = -9, a₃ < a₂
- …
In this example, the subscript n represents the position of the term, and n+1 represents the next position. By comparing the terms at these positions, we can see that the sequence is decreasing.
Example: Decreasing Sequence \( \left\{ \frac{2}{n + 1} \right\} \)
The sequence defined by \( a_n = \frac{2}{n + 1} \) is decreasing. We can prove this by comparing consecutive terms:
- \( a_n = \frac{2}{n + 1} \) (the nth term)
- \( a_{n+1} = \frac{2}{(n + 1) + 1} = \frac{2}{n + 2} \) (the (n+1)th term, or the term immediately after the nth term)
- Since \( \frac{2}{n + 1} > \frac{2}{n + 2} \), we have \( a_{n+1} < a_n \) for all \( n \geq 0 \), so the sequence is decreasing.
Explanation:
- The denominator \( n + 1 \) in \( a_n \) represents the nth term’s denominator.
- The denominator \( n + 2 \) in \( a_{n+1} \) represents the (n+1)th term’s denominator, which is one more than the nth term’s denominator.
- Since the numerators are the same (2) and the denominators are increasing, the fractions are decreasing.
This example shows that the sequence \( \left\{ \frac{2}{n + 1} \right\} \) is decreasing, and it provides a clear understanding of how the terms were compared to reach this conclusion.
Example: Decreasing Sequence \( \left\{ \frac{n}{n^2 + 1} \right\} \)
We want to show that the sequence is decreasing by comparing consecutive terms.
Given: \( a_n = \frac{n}{n^2 + 1} \) and \( a_{n+1} = \frac{n+1}{(n+1)^2 + 1} \)
Step 1: Start by comparing consecutive terms: \( \frac{n+1}{(n+1)^2 + 1} < \frac{n}{n^2 + 1} \)
Step 2: Cross-multiply to eliminate fractions: \( n(n+1)^2 + n < n(n^2 + 1) + n^2 + 1 \)
Step 3: Expand both sides to simplify: \( n^3 + 2n^2 + n + n < n^3 + n^2 + n + n^2 + 1 \)
Step 4: Rearrange terms to isolate the inequality: \( 1 < n^2 + n \)
Step 5: This inequality \( 1 < n^2 + n \) has the same meaning as our original inequality but is simpler to understand. It clearly shows that the terms of the sequence are decreasing.
Since \( n > 1 \), we conclude that \( 1 < n^2 + n \), and the sequence is decreasing.
Conclusion: The sequence \( \left\{ \frac{n}{n^2 + 1} \right\} \) is decreasing as we have shown that consecutive terms are less than the previous ones. The simplified inequality \( 1 < n^2 + n \) helped us understand this more clearly.
Example: Decreasing Sequence \( \left\{ \frac{n}{n^2 + 1} \right\} \) Using the Function Version
We can analyze the sequence by considering the corresponding function and its derivative.
Define the Function
Consider the function \( f(x) = \frac{x}{x^2 + 1} \).
Find the Derivative
We’ll use the quotient rule to find the derivative:
- \( f'(x) = \frac{(x^2 + 1) \cdot 1 – x \cdot 2x}{(x^2 + 1)^2} \)
- \( f'(x) = \frac{x^2 + 1 – 2x^2}{(x^2 + 1)^2} \)
- \( f'(x) = \frac{1 – x^2}{(x^2 + 1)^2} \)
Analyze the Derivative
The function is decreasing when \( f'(x) < 0 \), which occurs when \( x^2 > 1 \), i.e., \( x > 1 \) or \( x < -1 \).
The function is increasing when \( f'(x) > 0 \), which occurs when \( -1 < x < 1 \).
Conclusion for the Sequence
Since the derivative is negative for \( x > 1 \), the function is decreasing for those values of \( x \), and thus the sequence \( \left\{ \frac{n}{n^2 + 1} \right\} \) is decreasing for \( n > 1 \).