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

2024
Views:

265

Pages:

2
Author:

Reynaldo Wong

Question 1 Sum of series: arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ ( 𝑛 + 2 ) when 𝑛 varies from 0 to infinity. arctan(n+1)−arctan(n+2) when n varies from 0 to infinity. Answer 1: If you are looking for the sum of the series: ∑ 𝑛 = 0 ∞ ( arctan ⁡ ( 𝑛 + 1 ) − arctan ⁡ ( 𝑛 + 2 ) ) , n=0 ∑ ∞ (arctan(n+1)−arctan(n+2)), then what you have is a telescoping series. This means that terms will cancel. The first part of the sum is: arctan ⁡ ( 1 ) − arctan ⁡ ( 2 ) + arctan ⁡ ( 2 ) − arctan ⁡ ( 3 ) + arctan ⁡ ( 3 ) − arctan ⁡ ( 4 ) + … . arctan(1)−arctan(2)+arctan(2)−arctan(3)+arctan(3)−arctan(4)+…. The terms in the middle will cancel, for instance: − arctan ⁡ ( 2 ) + arctan ⁡ ( 2 ) = 0. −arctan(2)+arctan(2)=0. Thus your 𝑛 n-th partial sum is: 𝑆 𝑛 = arctan ⁡ ( 1 ) − arctan ⁡ ( 𝑛 + 2 ) . S n =arctan(1)−arctan(n+2). Taking the limit gives: lim ⁡ 𝑛 → ∞ ( arctan ⁡ ( 1 ) − arctan ⁡ ( 𝑛 + 2 ) ) = arctan ⁡ ( 1 ) − 𝜋 2 . n→∞ lim (arctan(1)−arctan(n+2))=arctan(1)− 2 π . And arctan ⁡ ( 1 ) = 𝜋 4 arctan(1)= 4 π , so the result is: 𝜋 4 − 𝜋 2 = − 𝜋 4 . 4 π − 2 π =− 4 π . Question 2 Find the radius of convergence and interval of convergence of the series: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 1 0 𝑛 𝑥 𝑛 𝑛 4 . n=1 ∑ ∞ n 4 (−1) n 10 n x n . Answer 1: Here: 𝑎 𝑛 = ( − 1 ) 𝑛 1 0 𝑛 𝑥 𝑛 𝑛 4 , 𝑎 𝑛 + 1 = ( − 1 ) 𝑛 + 1 1 0 𝑛 + 1 𝑥 𝑛 + 1 ( 𝑛 + 1 ) 4 . a n = n 4 (−1) n 10 n x n ,a n+1 = (n+1) 4 (−1) n+1 10 n+1 x n+1 . Now: ∣ 𝑎 𝑛 + 1 𝑎 𝑛 ∣ = ∣ 1 0 𝑛 + 1 𝑥 𝑛 + 1 ( 𝑛 + 1 ) 4 ⋅ 𝑛 4 1 0 𝑛 𝑥 𝑛 ∣ = ∣ 10 ∣ 𝑥 ∣ 𝑛 4 ( 𝑛 + 1 ) 4 ∣ . a n a n+1 = (n+1) 4 10 n+1 x n+1 ⋅ 10 n x n n 4 = 10∣x∣ (n+1) 4 n 4 . This shows that: lim ⁡ 𝑛 → ∞ ∣ 𝑎 𝑛 + 1 𝑎 𝑛 ∣ = 10 ∣ 𝑥 ∣ . n→∞ lim a n a n+1 =10∣x∣. Therefore, the given series converges if: 10 ∣ 𝑥 ∣ < 1 ⟹ ∣ 𝑥 ∣ < 1 10 . 10∣x∣<1⟹∣x∣< 10 1 . And diverges if 10 ∣ 𝑥 ∣ > 1 10∣x∣>1. Therefore, the radius of convergence is: 𝑅 = 1 10 . R= 10 1 . When 𝑥 = 1 10 x= 10 1 , the series becomes: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑛 4 , n=1 ∑ ∞ n 4 (−1) n , which converges by the Leibniz test. When 𝑥 = − 1 10 x=− 10 1 , the series becomes: ∑ 𝑛 = 1 ∞ 1 𝑛 4 , n=1 ∑ ∞ n 4 1 , which converges by the 𝑝 p-series test. Conclusion: The series converges in the interval: ∣ 𝑥 ∣ ≤ 1 10 . ∣x∣≤ 10 1 .