University:
Morehouse College
Course:
HMTH 105 | College Algebra 1
Academic year:

2024
Views:

232

Pages:

5
Author:

Reynaldo Wong

Question 1 Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the convergence or divergence of the series using other methods. ∑ 𝑛 = 1 ∞ 𝑛 2 ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) . n=1 ∑ ∞ (n+1) 2 (n+2) n 2 . Answer 1 Ratio Test: Let the series be ∑ 𝑎 𝑛 ∑a n , then: lim ⁡ 𝑛 → ∞ 𝑎 𝑛 + 1 𝑎 𝑛 = 𝐿 . n→∞ lim a n a n+1 =L. If 𝐿 < 1 L<1, the series converges. If 𝐿 > 1 L>1, the series diverges. If 𝐿 = 1 L=1, the test is inconclusive. For the given series: 𝑎 𝑛 = 𝑛 2 ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) . a n = (n+1) 2 (n+2) n 2 . Now apply the Ratio Test: lim ⁡ 𝑛 → ∞ 𝑎 𝑛 + 1 𝑎 𝑛 = lim ⁡ 𝑛 → ∞ ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) 2 ( 𝑛 + 3 ) ⋅ ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) 𝑛 2 . n→∞ lim a n a n+1 = n→∞ lim (n+2) 2 (n+3) (n+1) 2 ⋅ n 2 (n+1) 2 (n+2) . Simplify: lim ⁡ 𝑛 → ∞ ( 𝑛 + 1 ) 2 ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) 2 ( 𝑛 + 3 ) 𝑛 2 . n→∞ lim (n+2) 2 (n+3)n 2 (n+1) 2 (n+1) 2 . Take the limit: 𝐿 = lim ⁡ 𝑛 → ∞ 𝑛 4 𝑛 4 = 1. L= n→∞ lim n 4 n 4 =1. Since 𝐿 = 1 L=1, the Ratio Test fails. Question 2 Use Theorem Alternating Series Remainder to determine the number of terms required to approximate the sum of the series with an error of less than 0.001 0.001: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 + 1 𝑛 3 . n=1 ∑ ∞ n 3 (−1) n+1 . Answer 1 Given that: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 + 1 𝑛 3 . n=1 ∑ ∞ n 3 (−1) n+1 . We need to use Theorem Alternating Series to determine the number of terms required to approximate the sum with an error of less than 0.001 0.001. Given series is an alternating series with 𝑎 𝑛 = 1 𝑛 3 a n = n 3 1 . We know that: ∣ 𝑅 𝑁 ∣ ≤ 𝑎 𝑁 + 1 , where 𝑎 𝑁 + 1 < 0.001 must satisfy the inequality . ∣R N ∣≤a N+1 , where a N+1 <0.001 must satisfy the inequality. 1 ( 𝑁 + 1 ) 3 < 0.001. (N+1) 3 1 <0.001. Taking the cube root on both sides: 0.001 3 < 𝑁 + 1. 3 0.001