The series does not satisfy the hypotheses of the alternating series test as stated:
[tex]$\frac{2}{3}-\frac{3}{5}+\frac{4}{7}-\frac{5}{9}+\ldots$[/tex]
State which hypothesis is not satisfied. (Select all that apply.)
A. The series is of the form [tex]$\sum_{n=1}^{\infty}(-1)^{n+1} b_n$[/tex] and [tex]$0 \leq b_{n+1} \leq b_n$[/tex] for all [tex]$n \geq 1$[/tex].
B. The series is of the form [tex]$\sum_{n=1}^{\infty}(-1)^{n+1} b_n$[/tex] and [tex]$\lim _{n \rightarrow \infty} b_n=0$[/tex]
Answer (1)
The general term of the series is b n = 2 n + 1 n + 1 .
The limit of b n as n approaches infinity is lim n → ∞ 2 n + 1 n + 1 = 2 1 .
Since the limit is not 0, the condition lim n → ∞ b n = 0 is not satisfied.
The condition 0 ≤ b n + 1 ≤ b n for all n ≥ 1 is satisfied.
The series is of the form n = 1 ∑ ∞ ( − 1 ) n + 1 b n and n → ∞ lim b n = 0
Explanation
Problem Analysis We are given the series 3 2 − 5 3 + 7 4 − 9 5 + … and asked to determine which hypothesis of the alternating series test is not satisfied. The alternating series test states that for a series of the form ∑ n = 1 ∞ ( − 1 ) n + 1 b n , the series converges if: 1) b n ≥ 0 for all n , 2) b n + 1 ≤ b n for all n ≥ 1 , and 3) lim n → ∞ b n = 0 .
Finding the General Term First, we need to express the general term of the series. We can write the series as ∑ n = 1 ∞ ( − 1 ) n + 1 2 n + 1 n + 1 . Thus, b n = 2 n + 1 n + 1 .
Checking the Limit Now, let's check the limit of b n as n approaches infinity: lim n → ∞ 2 n + 1 n + 1 = lim n → ∞ 2 + 1/ n 1 + 1/ n = 2 1 . Since the limit is 2 1 and not 0, the condition lim n → ∞ b n = 0 is not satisfied.
Checking if the Sequence is Decreasing Next, we need to check if b n + 1 ≤ b n for all n ≥ 1 . This is equivalent to checking if 2 n + 3 n + 2 ≤ 2 n + 1 n + 1 . Cross-multiplying, we get ( n + 2 ) ( 2 n + 1 ) ≤ ( n + 1 ) ( 2 n + 3 ) , which simplifies to 2 n 2 + 5 n + 2 ≤ 2 n 2 + 5 n + 3 . This inequality is equivalent to 2 ≤ 3 , which is always true. Thus, b n + 1 ≤ b n for all n ≥ 1 .
Conclusion Since lim n → ∞ b n = 2 1 = 0 , the condition lim n → ∞ b n = 0 is not satisfied. The other condition, 0 ≤ b n + 1 ≤ b n for all n ≥ 1 , is satisfied.
Examples
Consider designing a system where the efficiency of each component decreases with each subsequent component. If the efficiency decrease approaches a non-zero constant, like in this series, the overall system's efficiency will not converge to zero, indicating a fundamental limitation in the system's design. Understanding series convergence helps engineers identify and address such limitations.
