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

2024
Views:

48

Pages:

3
Author:

Reynaldo Wong

Question 1 Use the Direct Comparison Test to determine the convergence or divergence of the series. ∑ 𝑛 = 1 ∞ sin ⁡ 2 ( 𝑛 ) 𝑛 2 n=1 ∑ ∞ n 2 sin 2 (n) Answer 1: Given: The series: ∑ 𝑛 = 1 ∞ sin ⁡ 2 ( 𝑛 ) 𝑛 2 . n=1 ∑ ∞ n 2 sin 2 (n) . To determine the convergence or divergence of the series using the Direct Comparison Test. Direct Comparison Test: Let 𝑎 𝑛 ≥ 0 , 𝑏 𝑛 ≥ 0 a n ≥0,b n ≥0 for all 𝑛 ∈ 𝑁 n∈N and 𝑎 𝑛 ≤ 𝑏 𝑛 a n ≤b n . Then: If ∑ 𝑏 𝑛 ∑b n is convergent, then ∑ 𝑎 𝑛 ∑a n is convergent. If ∑ 𝑎 𝑛 ∑a n is divergent, then ∑ 𝑏 𝑛 ∑b n is divergent. Let: 𝑧 𝑛 = sin ⁡ 2 ( 𝑛 ) 𝑛 2 and 𝑦 𝑛 = 1 𝑛 2 . z n = n 2 sin 2 (n) andy n = n 2 1 . Here, consider 𝑧 𝑛 = sin ⁡ 2 ( 𝑛 ) 𝑛 2 z n = n 2 sin 2 (n) and 𝑦 𝑛 = 1 𝑛 2 y n = n 2 1 : 0 ≤ 𝑧 𝑛 = sin ⁡ 2 ( 𝑛 ) 𝑛 2 ≤ 1 𝑛 2 . 0≤z n = n 2 sin 2 (n) ≤ n 2 1 . ∑ 1 𝑛 2 ∑ n 2 1 is a convergent 𝑝 p-series with 𝑝 = 2 p=2. That is, sin ⁡ 2 ( 𝑛 ) 𝑛 2 ≤ 1 𝑛 2 . n 2 sin 2 (n) ≤ n 2 1 . By 𝑝 p-series test, the series ∑ 1 𝑛 2 ∑ n 2 1 is convergent. Therefore, by the Direct Comparison Test, ∑ 𝑧 𝑛 ∑z n is convergent. Thus, by using the Direct Comparison Test, the series: ∑ 𝑛 = 1 ∞ sin ⁡ 2 ( 𝑛 ) 𝑛 2 n=1 ∑ ∞ n 2 sin 2 (n) is absolutely convergent. Question 2 Find the sum of the infinite geometric series. ∑ 𝑛 = 1 ∞ 12 ( − 0.7 ) 𝑛 − 1 . n=1 ∑ ∞ 12(−0.7) n−1 . Answer 1: To find: The sum of the infinite series ∑ 𝑛 = 1 ∞ 12 ( − 0.7 ) 𝑛 − 1 ∑ n=1 ∞ 12(−0.7) n−1 . Concept used: Sum of infinite geometric series can be obtained by the formula: 𝑆 ∞ = 𝑎 1 − 𝑟 ( ∣ 𝑟 ∣ < 1 ) , S ∞ = 1−r a (∣r∣<1), where 𝑎 a is first term, 𝑟 r is common ratio, and 𝑆 ∞ S ∞ is sum of infinite geometric series. Calculation: Expand the summation as follows: ∑ 𝑛 = 1 ∞ 12 ( − 0.7 ) 𝑛 − 1 = 12 ( 1 ) + 12 ( − 0.7 ) + 12 ( − 0.7 ) 2 + 12 ( − 0.7 ) 3 + … . n=1 ∑ ∞ 12(−0.7) n−1 =12(1)+12(−0.7)+12(−0.7) 2 +12(−0.7) 3 +…. It is observed from the series that the first term is 12 12 and the common ratio is ( − 0.7 ) (−0.7). Common ratio ∣ 𝑟 ∣ < 1 ∣r∣<1. Apply the formula: 𝑆 ∞ = 𝑎 1 − 𝑟 . S ∞ = 1−r a . Plug in the values: 𝑆 ∞ = 12 1 − ( − 0.7 ) = 12 1 + 0.7 = 12 1.7 . S ∞ = 1−(−0.7) 12 = 1+0.7 12 = 1.7 12 . Simplify: 𝑆 ∞ = 7.0588. S ∞ =7.0588. Thus, the sum of the series is: 𝑆 ∞ = 7.0588. S ∞ =7.0588. Question 3 Use the Limit Comparison Test to determine the convergence or divergence of the series. ∑ 𝑛 = 1 ∞ 2 𝑛 2 − 1 𝑛 3 + 1 . n=1 ∑ ∞ n 3 +1 2n 2 −1 . Answer 1: We have to check the given series: ∑ 𝑛 = 1 ∞ 2 𝑛 2 − 1 𝑛 3 + 1 . n=1 ∑ ∞ n 3 +1 2n 2 −1 . is convergent or divergent using the Limit Comparison Test. According to the Limit Comparison Test: If two series ∑ 𝑎 𝑛 ∑a n and ∑ 𝑏 𝑛 ∑b n hold: lim ⁡ 𝑛 → ∞ 𝑎 𝑛 𝑏 𝑛 = 𝑐 , n→∞ lim b n a n =c, where 0 < 𝑐 < ∞ 0 1 L>1, the series diverges. If 𝐿 = 1 L=1, the Root Test is inconclusive. Calculation: 𝐿 = lim ⁡ 𝑛 → ∞ ( 𝑛 500 ) 𝑛 𝑛 . L= n→∞ lim n ( 500 n ) n . Simplify: 𝐿 = lim ⁡ 𝑛 → ∞ 𝑛 500 . L= n→∞ lim 500 n . As 𝑛 → ∞ n→∞: 𝐿 → ∞ . L→∞. Since 𝐿 > 1 L>1, the series diverges. Question 5 Use the formula for the sum of a geometric series to find the sum. ∑ 𝑛 = 0 ∞ ( − 4 9 ) 𝑛 . n=0 ∑ ∞ (− 9 4 ) n . Answer 1: Given Series: ∑ 𝑛 = 0 ∞ ( − 4 9 ) 𝑛 . n=0 ∑ ∞ (− 9 4 ) n . Concept Used: The sum of an infinite geometric series is given by: 𝑆 = 𝑎 1 − 𝑟 , for ∣ 𝑟 ∣ < 1 , S= 1−r a ,for ∣r∣<1, where 𝑎 a is the first term and 𝑟 r is the common ratio. Step 1: Identify 𝑎 a and 𝑟 r: The first term 𝑎 = 1 a=1 (since 𝑛 = 0 n=0 results in ( − 4 9 ) 0 = 1 (− 9 4 ) 0 =1). The common ratio 𝑟 = − 4 9 r=− 9 4 . Since ∣ 𝑟 ∣ = 4 9 < 1 ∣r∣= 9 4 <1, the series converges. Step 2: Calculate the Sum: 𝑆 = 1 1 − ( − 4 9 ) . S= 1−(− 9 4 ) 1 . Simplify the denominator: 𝑆 = 1 1 + 4 9 = 1 9 + 4 9 = 1 13 9 = 9 13 . S= 1+ 9 4 1 = 9 9+4 1 = 9 13 1 = 13 9 . Thus, the sum of the series is: 𝑆 = 9 13 . S= 13 9 .

Series #11 - Q&A

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