Series #8 - Q&A

Question 1 Starting with the geometric series ∑ 𝑛 = 0 ∞ 𝑎 𝑛 ∑ n=0 ∞ ​ a n , find the sum of the series ∑ 𝑛 = 1 ∞ 𝑛 𝑎 𝑛 − 1 ∑ n=1 ∞ ​ na n−1 , ∣ 𝑎 ∣ < 1 ∣a∣<1. Answer 1: Consider the geometric series, ∑ 𝑛 = 0 ∞ 𝑎 𝑛 = 1 + 𝑎 + 𝑎 2 + 𝑎 3 + …   . n=0 ∑ ∞ ​ a n =1+a+a 2 +a 3 +…. Find the sum of the series ∑ 𝑛 = 1 ∞ 𝑛 𝑎 𝑛 − 1 ∑ n=1 ∞ ​ na n−1 , ∣ 𝑎 ∣ < 1 ∣a∣<1: ∑ 𝑛 = 1 ∞ 𝑛 𝑎 𝑛 − 1 = 1 + 2 𝑎 + 3 𝑎 2 + 4 𝑎 3 + …   . n=1 ∑ ∞ ​ na n−1 =1+2a+3a 2 +4a 3 +…. Let 𝑆 = 1 + 2 𝑎 + 3 𝑎 2 + 4 𝑎 3 + …   . S=1+2a+3a 2 +4a 3 +…. Multiply by 𝑎 a: 𝑎 𝑆 = 𝑎 + 2 𝑎 2 + 3 𝑎 3 + 4 𝑎 4 + …   . aS=a+2a 2 +3a 3 +4a 4 +…. Subtract 𝑎 𝑆 aS from 𝑆 S: 𝑆 − 𝑎 𝑆 = 1 + ( 2 𝑎 − 𝑎 ) + ( 3 𝑎 2 − 2 𝑎 2 ) + ( 4 𝑎 3 − 3 𝑎 3 ) + … , S−aS=1+(2a−a)+(3a 2 −2a 2 )+(4a 3 −3a 3 )+…, 𝑆 ( 1 − 𝑎 ) = 1 + 𝑎 + 𝑎 2 + 𝑎 3 + …   . S(1−a)=1+a+a 2 +a 3 +…. The series ∑ 𝑛 = 0 ∞ 𝑎 𝑛 = 1 1 − 𝑎 ∑ n=0 ∞ ​ a n = 1−a 1 ​ , 𝑆 ( 1 − 𝑎 ) = 1 1 − 𝑎 . S(1−a)= 1−a 1 ​ . Divide through by ( 1 − 𝑎 ) (1−a): 𝑆 = 1 ( 1 − 𝑎 ) 2 . S= (1−a) 2 1 ​ . Thus, ∑ 𝑛 = 1 ∞ 𝑛 𝑎 𝑛 − 1 = 1 ( 1 − 𝑎 ) 2 . n=1 ∑ ∞ ​ na n−1 = (1−a) 2 1 ​ . Question 2 Let 𝑃 ( 𝑘 ) P(k) be a statement that 1 2 + 1 3 + ⋯ + 1 𝑘 ( 𝑘 + 1 ) = 𝑘 𝑘 + 1 2 1 ​ + 3 1 ​ +⋯+ k(k+1) 1 ​ = k+1 k ​ . For the basis step to prove 𝑃 ( 𝑘 ) P(k) is that at 𝑘 = 1 k=1, 𝑃 P is true. a. Show that 𝑃 ( 1 ) P(1) is true by completing the basis step proof. Left side of 𝑃 ( 𝑘 ) P(k) and right side of 𝑃 ( 𝑘 ) P(k). b. Identify the inductive hypothesis used to prove 𝑃 ( 𝑘 ) P(k). c. Identify the inductive step used to prove 𝑃 ( 𝑘 + 1 ) P(k+1). Answer 1: Let the property 𝑃 ( 𝑘 ) P(k) be: 1 2 + 1 3 + ⋯ + 1 𝑘 ( 𝑘 + 1 ) = 𝑘 𝑘 + 1 . 2 1 ​ + 3 1 ​ +⋯+ k(k+1) 1 ​ = k+1 k ​ . a. Basis step: Show that 𝑃 ( 1 ) P(1) is true for all integers 𝑘 ≥ 1 k≥1 using mathematical induction. Basis step: 𝑃 ( 1 ) P(1) is true: That is to show that: 1 2 = 1 1 + 1 . 2 1 ​ = 1+1 1 ​ . The left-hand side of the equation is: 1 2 , 2 1 ​ , and right-hand side is: 1 2 . 2 1 ​ . It follows that 1 2 = 1 2 2 1 ​ = 2 1 ​ . Hence 𝑃 ( 1 ) P(1) is true. b. Inductive hypothesis: Show that for all integers 𝑛 ≥ 1 n≥1, 𝑃 ( 𝑘 ) P(k) is true, then 𝑃 ( 𝑘 + 1 ) P(k+1) is also true. Suppose 𝑃 ( 𝑘 ) P(k) is true: 1 2 + 1 3 + ⋯ + 1 𝑘 ( 𝑘 + 1 ) = 𝑘 𝑘 + 1 . 2 1 ​ + 3 1 ​ +⋯+ k(k+1) 1 ​ = k+1 k ​ . The inductive hypothesis is: 1 2 + 1 3 + ⋯ + 1 𝑘 ( 𝑘 + 1 ) = 𝑘 𝑘 + 1 . 2 1 ​ + 3 1 ​ +⋯+ k(k+1) 1 ​ = k+1 k ​ . c. Inductive step: Show that 𝑃 ( 𝑘 + 1 ) P(k+1) is true. That is to show that: 1 2 + 1 3 + ⋯ + 1 𝑘 ( 𝑘 + 1 ) + 1 ( 𝑘 + 1 ) ( 𝑘 + 2 ) = 𝑘 + 1 𝑘 + 2 . 2 1 ​ + 3 1 ​ +⋯+ k(k+1) 1 ​ + (k+1)(k+2) 1 ​ = k+2 k+1 ​ . Question 3 Give examples of: a. Harmonic series b. Alternating harmonic series c. Geometric series Answer 1: a. Harmonic series is the divergent infinite series: ∑ 𝑛 = 1 ∞ 1 𝑛 = 1 + 1 2 + 1 3 + 1 4 + …   . n=1 ∑ ∞ ​ n 1 ​ =1+ 2 1 ​ + 3 1 ​ + 4 1 ​ +…. b. Alternating harmonic series It's a converging series, which converges to ln ⁡ 2 ln2: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 − 1 𝑛 = 1 − 1 2 + 1 3 − 1 4 + …   . n=1 ∑ ∞ ​ n (−1) n−1 ​ =1− 2 1 ​ + 3 1 ​ − 4 1 ​ +…. c. Geometric series is a series with a constant ratio between successive terms: ∑ 𝑛 = 0 ∞ 𝑎 𝑟 𝑛 = 𝑎 + 𝑎 𝑟 + 𝑎 𝑟 2 + 𝑎 𝑟 3 + …   . n=0 ∑ ∞ ​ ar n =a+ar+ar 2 +ar 3 +…. Question 4 Using only the definition of Riemann sum and your knowledge of limits, compute the exact area under the curve 2 𝑥 2 2x 2 between 𝑥 = − 2 x=−2 and 𝑥 = 1 x=1. Answer 1: Divide the area in 𝑛 n strips of rectangles with breadth: Δ 𝑥 = 1 − ( − 2 ) 𝑛 . Δx= n 1−(−2) ​ . Δ 𝑥 = 3 𝑛 . Δx= n 3 ​ . Now using right-hand Riemann sums, the area under the curve 2 𝑥 2 2x 2 is: 𝑆 𝑛 = Δ 𝑥 ∑ 𝑖 = 1 𝑛 2 ( − 2 + 𝑖 Δ 𝑥 ) 2 . S n ​ =Δx i=1 ∑ n ​ 2(−2+iΔx) 2 . 𝑆 𝑛 = 3 𝑛 ∑ 𝑖 = 1 𝑛 2 ( − 2 + 3 𝑖 𝑛 ) 2 . S n ​ = n 3 ​ i=1 ∑ n ​ 2(−2+ n 3i ​ ) 2 . Expanding: 𝑆 𝑛 = 6 𝑛 ∑ 𝑖 = 1 𝑛 ( 4 − 12 𝑖 𝑛 + 9 𝑖 2 𝑛 2 ) , S n ​ = n 6 ​ i=1 ∑ n ​ (4− n 12i ​ + n 2 9i 2 ​ ), 𝑆 𝑛 = 6 𝑛 [ 4 𝑛 − 6 𝑛 + 9 𝑛 ( 𝑛 + 1 ) ( 2 𝑛 + 1 ) 6 𝑛 2 ] . S n ​ = n 6 ​ [4n−6n+ 6n 2 9n(n+1)(2n+1) ​ ]. Simplify and take the limit as 𝑛 → ∞ n→∞: lim ⁡ 𝑛 → ∞ 𝑆 𝑛 = 6. n→∞ lim ​ S n ​ =6. Thus, the area under the curve 2 𝑥 2 2x 2 is 6. Question 5 Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum: 10 − 2 + 0.4 − 0.08 + …   . 10−2+0.4−0.08+…. Answer 1: Geometric series: ∑ 𝑛 = 0 ∞ 𝑎 𝑟 𝑛 = 𝑎 1 − 𝑟 , n=0 ∑ ∞ ​ ar n = 1−r a ​ , where 𝑎 a and 𝑟 r are constants. If ∣ 𝑟 ∣ < 1 ∣r∣<1, then the series converges to: 𝑎 1 − 𝑟 . 1−r a ​ . Here: 𝑎 = 10 , 𝑟 = − 1 5 . a=10,r=− 5 1 ​ . Substitute: 𝑆 = 10 1 − ( − 1 5 ) = 10 1 + 1 5 = 10 6 5 = 50 6 = 25 3 . S= 1−(− 5 1 ​ ) 10 ​ = 1+ 5 1 ​ 10 ​ = 5 6 ​ 10 ​ = 6 50 ​ = 3 25 ​ . The series converges to: 25 3 . 3 25 ​ .