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

2024
Views:

429

Pages:

4
Author:

Reynaldo Wong

Question 1 Radius and interval of convergence: Determine the radius and interval of convergence of the following power series. 𝑥 = − 1 4 + 𝑥 9 − 𝑥 2 16 + … x=− 4 1 + 9 x − 16 x 2 +… Answer 1 First write the given series in formula: 𝑆 = 𝑥 − 𝑥 2 3 2 + 𝑥 3 4 2 − … . S=x− 3 2 x 2 + 4 2 x 3 −…. This can be expressed as: 𝑆 = ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 + 1 𝑥 𝑛 − 1 𝑛 2 . S= n=1 ∑ ∞ (−1) n+1 n 2 x n−1 . Use the ratio test to determine whether the series converges: For the given series: 𝑎 𝑘 = ( − 1 ) 𝑘 𝑥 𝑘 + 1 ( 𝑘 + 1 ) 2 . a k =(−1) k (k+1) 2 x k+1 . The ratio test gives: 𝐿 = lim ⁡ 𝑘 → ∞ ∣ 𝑎 𝑘 + 1 𝑎 𝑘 ∣ = ∣ 𝑥 ∣ ⋅ lim ⁡ 𝑘 → ∞ 𝑘 2 ( 𝑘 + 1 ) 2 = ∣ 𝑥 ∣ . L= k→∞ lim a k a k+1 =∣x∣⋅ k→∞ lim (k+1) 2 k 2 =∣x∣. For convergence, 𝐿 < 1 L<1: ∣ 𝑥 ∣ < 1. ∣x∣<1. Radius of convergence, 𝑅 = 1 R=1. Interval of convergence: 𝑥 ∈ [ − 1 , 1 ] . x∈[−1,1]. Question 2 Evaluate the following infinite series or state that the series diverges: ∑ 𝑘 = 0 ∞ 4 ⋅ 4 − 𝑘 . k=0 ∑ ∞ 4⋅4 −k . Answer 1 The given series can be written as: ∑ 𝑘 = 0 ∞ 4 ⋅ ( 4 − 1 ) 𝑘 . k=0 ∑ ∞ 4⋅(4 −1 ) k . This forms a geometric series with first term 𝑎 = 4 a=4 and common ratio 𝑟 = 1 / 4 r=1/4. The formula for the sum of a geometric series: 𝑆 = 𝑎 1 − 𝑟 . S= 1−r a . Substitute the values: 𝑆 = 4 1 − 1 4 = 4 3 4 = 16 / 3. S= 1− 4 1 4 = 4 3 4 =16/3. Thus, the series converges and its sum is 16 / 3 16/3. Question 3 Use the Root Test to determine the convergence or divergence of the series: ∑ 𝑛 = 1 ∞ ( 3 𝑛 + 2 𝑛 + 3 ) 𝑛 . n=1 ∑ ∞ ( n+3 3n+2 ) n . Answer 1 The Root Test states that the series ∑ 𝑎 𝑛 ∑a n converges if: lim ⁡ 𝑛 → ∞ ∣ 𝑎 𝑛 ∣ 𝑛 < 1 , n→∞ lim n ∣a n ∣ <1, and diverges if: lim ⁡ 𝑛 → ∞ ∣ 𝑎 𝑛 ∣ 𝑛 > 1. n→∞ lim n ∣a n ∣ >1. For the given series: 𝑎 𝑛 = ( 3 𝑛 + 2 𝑛 + 3 ) 𝑛 . a n =( n+3 3n+2 ) n . Compute the limit: lim ⁡ 𝑛 → ∞ ∣ 𝑎 𝑛 ∣ 𝑛 = lim ⁡ 𝑛 → ∞ 3 𝑛 + 2 𝑛 + 3 . n→∞ lim n ∣a n ∣ = n→∞ lim n+3 3n+2 . Simplify: lim ⁡ 𝑛 → ∞ 3 𝑛 + 2 𝑛 + 3 = lim ⁡ 𝑛 → ∞ 3 + 2 𝑛 1 + 3 𝑛 = 3. n→∞ lim n+3 3n+2 = n→∞ lim 1+ n 3 3+ n 2 =3. Since 3 > 1 3>1, the series diverges by the Root Test. Question 4 Determine the radius of convergence and the interval of convergence for each power series: ∑ 𝑛 = 0 ∞ 𝑛 ( 𝑧 − 1 ) 𝑛 . n=0 ∑ ∞ n (z−1) n . Answer 1 1) Radius of convergence: The formula for radius of convergence is: 1 𝑅 = lim ⁡ 𝑛 → ∞ ∣ 𝑐 𝑛 ∣ 𝑛 , R 1 = n→∞ lim n ∣c n ∣ , where 𝑐 𝑛 = 𝑛 c n = n . Compute: lim ⁡ 𝑛 → ∞ 𝑛 + 1 𝑛 = 1. n→∞ lim n n+1 =1. Radius of convergence: 𝑅 = 1 R=1. 2) Interval of convergence: If 𝑅 R is finite and non-zero, the power series converges inside the interval ∣ 𝑧 − 1 ∣ < 𝑅 ∣z−1∣