Series #10 - Q&A

Question 1 a. Find an upper bound for the remainder in terms of 𝑛 n. b. Find how many terms are needed to ensure that the remainder is less than 1 0 − 3 10 −3 . c. Find lower and upper bounds ( 𝑙 𝑛 ln and 𝑈 𝑛 Un, respectively) on the exact value of the series ∑ 𝑘 = 1 ∞ 1 𝑘 ( 3 ) 𝑘 . k=1 ∑ ∞ ​ k(3) k 1 ​ . Answer 1: (a): We know: 𝑅 𝑛 < ∫ 𝑛 ∞ 1 3 𝑥 𝑑 𝑥 . R n ​ <∫ n ∞ ​ 3 x 1 ​ dx. ∫ 𝑛 ∞ 1 3 𝑥 𝑑 𝑥 = lim ⁡ 𝑎 → ∞ ∫ 𝑛 𝑎 1 3 𝑥 𝑑 𝑥 = lim ⁡ 𝑎 → ∞ 1 ln ⁡ ( 3 ) [ − 1 𝑥 ] 𝑛 𝑎 . ∫ n ∞ ​ 3 x 1 ​ dx= a→∞ lim ​ ∫ n a ​ 3 x 1 ​ dx= a→∞ lim ​ ln(3) 1 ​ [− x 1 ​ ] n a ​ . = 1 ln ⁡ ( 3 ) [ 0 + 1 𝑛 ( 3 ) 𝑛 ] . = ln(3) 1 ​ [0+ n(3) n 1 ​ ]. An upper bound for the remainder in terms of 𝑛 n is: 𝑅 𝑛 = 1 𝑛 ( 3 ) 𝑛 . R n ​ = n(3) n 1 ​ . (b): We have 𝑅 𝑛 < 1 0 − 3 R n ​ <10 −3 : 1 𝑛 ( 3 ) 𝑛 < 1 0 − 3 . n(3) n 1 ​ <10 −3 . ln ⁡ ( 3 ) 𝑛 > ln ⁡ ( 1000 ) − ln ⁡ ( 𝑛 ( 3 ) ) . ln(3)n>ln(1000)−ln(n(3)). Now simplify: 𝑛 ( 3 𝑛 ) ( 1000 ) − ln ⁡ ( 𝑙 𝑛 ( 3 ) ) ≈ 𝑛 > 2.645. n(3 n )(1000)−ln(ln(3))≈n>2.645. Question 2 Answer: Question 2 For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. ∑ 𝑛 = 1 ∞ 1 𝑛 ( 𝑛 + 1 ) . n=1 ∑ ∞ ​ n(n+1) 1 ​ . Answer 1: Given: The series: ∑ 𝑛 = 1 ∞ 1 𝑛 ( 𝑛 + 1 ) . n=1 ∑ ∞ ​ n(n+1) 1 ​ . To determine: The explicit formula for the sequence of partial sums and convergence of the series. Explanation: The general term ( 𝑘 k-th term) of the sequence of partial sums is of a series is the sum of the first 𝑘 k terms. Here, the given series is: ∑ 𝑛 = 1 ∞ 1 𝑛 ( 𝑛 + 1 ) . n=1 ∑ ∞ ​ n(n+1) 1 ​ . Simplify using partial fraction decomposition: 1 𝑛 ( 𝑛 + 1 ) = 1 𝑛 − 1 𝑛 + 1 . n(n+1) 1 ​ = n 1 ​ − n+1 1 ​ . Thus, ∑ 𝑛 = 1 ∞ 1 𝑛 ( 𝑛 + 1 ) = ∑ 𝑛 = 1 ∞ ( 1 𝑛 − 1 𝑛 + 1 ) . n=1 ∑ ∞ ​ n(n+1) 1 ​ = n=1 ∑ ∞ ​ ( n 1 ​ − n+1 1 ​ ). This forms a telescoping series: ( 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + …   . (1− 2 1 ​ )+( 2 1 ​ − 3 1 ​ )+( 3 1 ​ − 4 1 ​ )+…. The general term of sequence of partial sums is: 𝑆 𝑘 = ( 1 − 1 𝑘 + 1 ) . S k ​ =(1− k+1 1 ​ ). Conclusion: As 𝑘 → ∞ k→∞, 𝑆 𝑘 = 1 − 1 𝑘 + 1 → 1. S k ​ =1− k+1 1 ​ →1. Thus, the sequence of partial sums converges to 1 1, and the series converges. Question 3 For the following infinite series, find the first four terms of the sequence of partial sums. Then make a conjecture about the value of the infinite series or state that the series diverges. ∑ 𝑘 = 0 ∞ 1 0 𝑘 . k=0 ∑ ∞ ​ 10 k . Answer 1: Given: The series is: ∑ 𝑘 = 0 ∞ 1 0 𝑘 . k=0 ∑ ∞ ​ 10 k . We have to find the first four terms of the sequence of partial sums, make a conjecture, and state if the series diverges. Sequence of Partial Sums: 𝑎 𝑘 = 1 0 𝑘 . a k ​ =10 k . Find the partial sums: 𝑆 1 = 𝑎 1 = 1 0 1 = 10 , S 1 ​ =a 1 ​ =10 1 =10, 𝑆 2 = 𝑎 1 + 𝑎 2 = 1 0 1 + 1 0 2 = 10 + 100 = 110 , S 2 ​ =a 1 ​ +a 2 ​ =10 1 +10 2 =10+100=110, 𝑆 3 = 𝑎 1 + 𝑎 2 + 𝑎 3 = 1 0 1 + 1 0 2 + 1 0 3 = 10 + 100 + 1000 = 1110 , S 3 ​ =a 1 ​ +a 2 ​ +a 3 ​ =10 1 +10 2 +10 3 =10+100+1000=1110, 𝑆 4 = 𝑎 1 + 𝑎 2 + 𝑎 3 + 𝑎 4 = 1 0 1 + 1 0 2 + 1 0 3 + 1 0 4 = 10 + 100 + 1000 + 10000 = 11110. S 4 ​ =a 1 ​ +a 2 ​ +a 3 ​ +a 4 ​ =10 1 +10 2 +10 3 +10 4 =10+100+1000+10000=11110. Conjecture: Since the partial sums grow without bound as 𝑘 → ∞ k→∞, the series diverges. Question 4 Find the region of convergence and absolute convergence of the series: ∑ 𝑛 = 0 ∞ ln ⁡ 𝑥 𝑛 2 . n=0 ∑ ∞ ​ n 2 lnx ​ . Answer 1: Given: The series: ∑ 𝑛 = 0 ∞ ln ⁡ 𝑥 𝑛 2 . n=0 ∑ ∞ ​ n 2 lnx ​ . To find the region of convergence and absolute convergence of the given series: Solution: Apply the ratio test to determine convergence: Let 𝑎 𝑛 = ln ⁡ 𝑥 𝑛 2 a n ​ = n 2 lnx ​ . The ratio test implies: lim ⁡ 𝑛 → ∞ ∣ 𝑎 𝑛 + 1 𝑎 𝑛 ∣ < 1. n→∞ lim ​ ​ a n ​ a n+1 ​ ​ ​ <1. Simplify: 𝑎 𝑛 + 1 𝑎 𝑛 = ln ⁡ ( 𝑥 ) ( 𝑛 + 1 ) 2 ⋅ 𝑛 2 ln ⁡ ( 𝑥 ) = 𝑛 2 ( 𝑛 + 1 ) 2 . a n ​ a n+1 ​ ​ = (n+1) 2 ln(x) ​ ⋅ ln(x) n 2 ​ = (n+1) 2 n 2 ​ . Take the limit as 𝑛 → ∞ n→∞: lim ⁡ 𝑛 → ∞ 𝑛 2 ( 𝑛 + 1 ) 2 = lim ⁡ 𝑛 → ∞ 𝑛 2 𝑛 2 + 2 𝑛 + 1 = 1 1 = 1. n→∞ lim ​ (n+1) 2 n 2 ​ = n→∞ lim ​ n 2 +2n+1 n 2 ​ = 1 1 ​ =1. Thus, the series does not converge absolutely for all 𝑥 x, but converges conditionally for ∣ ln ⁡ 𝑥 ∣ < 1 ∣lnx∣<1. Question 5 Determine whether the series ∑ 𝑎 𝑛 ∑a n ​ converges or diverges: Use the Alternating Series Test. ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑛 ln ⁡ ( 𝑛 ) . n=1 ∑ ∞ ​ (−1) n ln(n) n ​ . Answer 1: Given: The series is: ∑ 𝑛 = 1 ∞ ( − 1 ) 𝑛 𝑛 ln ⁡ ( 𝑛 ) . n=1 ∑ ∞ ​ (−1) n ln(n) n ​ . Solution: Take 𝑎 𝑛 = ( − 1 ) 𝑛 𝑛 ln ⁡ ( 𝑛 ) a n ​ =(−1) n ln(n) n ​ and 𝑏 𝑛 = 𝑛 ln ⁡ ( 𝑛 ) b n ​ = ln(n) n ​ . The Alternating Series Test is stated below: Suppose the series ∑ 𝑎 𝑛 ∑a n ​ and either 𝑎 𝑛 = ( − 1 ) 𝑛 𝑏 𝑛 a n ​ =(−1) n b n ​ or 𝑎 𝑛 = ( − 1 ) 𝑛 + 1 𝑏 𝑛 a n ​ =(−1) n+1 b n ​ , where 𝑏 𝑛 ≥ 0 b n ​ ≥0 for all 𝑛 n. Then if: lim ⁡ 𝑛 → ∞ 𝑏 𝑛 = 0 , lim n→∞ ​ b n ​ =0, 𝑏 𝑛 + 1 ≤ 𝑏 𝑛 b n+1 ​ ≤b n ​ , Then, the series converges. Condition 1: Check if lim ⁡ 𝑛 → ∞ 𝑏 𝑛 = 0 : lim n→∞ ​ b n ​ =0: 𝑏 𝑛 = 𝑛 ln ⁡ ( 𝑛 ) . b n ​ = ln(n) n ​ . lim ⁡ 𝑛 → ∞ 𝑛 ln ⁡ ( 𝑛 ) = ∞ . n→∞ lim ​ ln(n) n ​ =∞. Since lim ⁡ 𝑛 → ∞ 𝑏 𝑛 ≠ 0 lim n→∞ ​ b n ​  =0, the series diverges.