Series #6 - Q&A

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. Comparison Test: Let the series be ∑ 𝑎 𝑛 ∑a n ​ , then: 𝑎 𝑛 = 𝑛 2 ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) . a n ​ = (n+1) 2 (n+2) n 2 ​ . Let 𝑏 𝑛 = 1 𝑛 2 b n ​ = n 2 1 ​ . Using the Comparison Test: lim ⁡ 𝑛 → ∞ 𝑎 𝑛 𝑏 𝑛 = lim ⁡ 𝑛 → ∞ 𝑛 2 ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) 1 𝑛 2 = lim ⁡ 𝑛 → ∞ 𝑛 4 ( 𝑛 + 1 ) 2 ( 𝑛 + 2 ) . n→∞ lim ​ b n ​ a n ​ ​ = n→∞ lim ​ n 2 1 ​ (n+1) 2 (n+2) n 2 ​ ​ = n→∞ lim ​ (n+1) 2 (n+2) n 4 ​ . Simplify: lim ⁡ 𝑛 → ∞ 𝑛 4 𝑛 4 = 1. n→∞ lim ​ n 4 n 4 ​ =1. Thus, ∑ 𝑎 𝑛 ∑a n ​ converges by the Comparison Test. Question 2 Find the sum of the convergent series: ∑ 𝑛 = 1 ∞ 4 𝑛 ( 𝑛 + 2 ) . n=1 ∑ ∞ ​ n(n+2) 4 ​ . Answer 1 Given series: ∑ 𝑛 = 1 ∞ 4 𝑛 ( 𝑛 + 2 ) . n=1 ∑ ∞ ​ n(n+2) 4 ​ . Rewrite: 4 𝑛 ( 𝑛 + 2 ) = 2 𝑛 − 2 𝑛 + 2 . n(n+2) 4 ​ = n 2 ​ − n+2 2 ​ . The series can be written as: ∑ 𝑛 = 1 ∞ ( 2 𝑛 − 2 𝑛 + 2 ) = ( 2 − 2 3 ) + ( 1 − 2 4 ) + ( 2 3 − 2 5 ) + …   . n=1 ∑ ∞ ​ ( n 2 ​ − n+2 2 ​ )=(2− 3 2 ​ )+(1− 4 2 ​ )+( 3 2 ​ − 5 2 ​ )+…. This simplifies to: 2 + 1 − 2 4 − 2 5 + …   . 2+1− 4 2 ​ − 5 2 ​ +…. Telescoping: ∑ 𝑛 = 1 ∞ 4 𝑛 ( 𝑛 + 2 ) = 3. n=1 ∑ ∞ ​ n(n+2) 4 ​ =3. Thus, the sum is 3 3. Question 3 Find the sum of the infinite geometric series: 3 + 2 + 4 3 + 8 9 + …   . 3+2+ 3 4 ​ + 9 8 ​ +…. Answer 1 Given: 𝑆 = 3 + 2 + 4 3 + 8 9 + …   . S=3+2+ 3 4 ​ + 9 8 ​ +…. We know that the sum of an infinite geometric series is given by: 𝑆 = 𝑎 1 − 𝑟 , S= 1−r a ​ , where 𝑎 a is the first term and 𝑟 r is the common ratio. Here, 𝑎 = 3 a=3, 𝑟 = 2 3 r= 3 2 ​ . Substitute values: 𝑆 = 3 1 − 2 3 = 3 1 3 = 9. S= 1− 3 2 ​ 3 ​ = 3 1 ​ 3 ​ =9. Thus, the sum of the given infinite geometric series is 9 9. Question 4 Find the Maclaurin series for the function: 𝑓 ( 𝑥 ) = 𝑒 𝑥 4 . f(x)=e x 4 . Answer 1 The Maclaurin Series is shown below: 𝑓 ( 𝑥 ) = 𝑓 ( 0 ) + 𝑓 ′ ( 0 ) 𝑥 + 𝑓 ′ ′ ( 0 ) 2 ! 𝑥 2 + 𝑓 ′ ′ ′ ( 0 ) 3 ! 𝑥 3 + …   . f(x)=f(0)+f ′ (0)x+ 2! f ′′ (0) ​ x 2 + 3! f ′′′ (0) ​ x 3 +…. Here: 𝑓 ( 𝑥 ) = 𝑒 𝑥 4 . f(x)=e x 4 . Finding derivatives: 𝑓 ( 0 ) = 𝑒 0 = 1. f(0)=e 0 =1. 𝑓 ′ ( 𝑥 ) = 4 𝑥 3 𝑒 𝑥 4 ⇒ 𝑓 ′ ( 0 ) = 0. f ′ (x)=4x 3 e x 4 ⇒f ′ (0)=0. 𝑓 ′ ′ ( 𝑥 ) = 12 𝑥 2 𝑒 𝑥 4 + 16 𝑥 6 𝑒 𝑥 4 ⇒ 𝑓 ′ ′ ( 0 ) = 0. f ′′ (x)=12x 2 e x 4 +16x 6 e x 4 ⇒f ′′ (0)=0. Thus, the Maclaurin series is: 𝑓 ( 𝑥 ) = 1 + 0 + 0 + …   . f(x)=1+0+0+…. Question 5 Determine if the series converges or diverges. If the series converges, find its sum: ∑ 𝑛 = 1 ∞ 6 ( 4 𝑛 − 1 ) ( 4 𝑛 + 3 ) . n=1 ∑ ∞ ​ (4n−1)(4n+3) 6 ​ . Answer 1 Given: 𝑆 𝑛 = ∑ 𝑛 = 1 ∞ 6 ( 4 𝑛 − 1 ) ( 4 𝑛 + 3 ) . S n ​ = n=1 ∑ ∞ ​ (4n−1)(4n+3) 6 ​ . Rewrite using partial fractions: 6 ( 4 𝑛 − 1 ) ( 4 𝑛 + 3 ) = 𝐴 4 𝑛 − 1 + 𝐵 4 𝑛 + 3 . (4n−1)(4n+3) 6 ​ = 4n−1 A ​ + 4n+3 B ​ . Solve for 𝐴 A and 𝐵 B: 𝐴 = 1 , 𝐵 = − 1. A=1,B=−1. Thus, 6 ( 4 𝑛 − 1 ) ( 4 𝑛 + 3 ) = 1 4 𝑛 − 1 − 1 4 𝑛 + 3 . (4n−1)(4n+3) 6 ​ = 4n−1 1 ​ − 4n+3 1 ​ . The series becomes telescoping: ∑ 𝑛 = 1 ∞ ( 1 4 𝑛 − 1 − 1 4 𝑛 + 3 ) . n=1 ∑ ∞ ​ ( 4n−1 1 ​ − 4n+3 1 ​ ). The sum is: 𝑆 𝑛 = 2. S n ​ =2.