Calculus Solution of Section 11.1 7. List the first five terms of the sequence. a1 = 3, an+1 = 2an − 1. Sol. The sequence is {3, 5, 9, 17, 33, ...}. 13. Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. {1, − 32 , 49 − 8 27 , ...}. Sol. an = (− 23 )n−1 . 25. Determine whether the sequence { (2n−1)! (2n+1)! } converges or diverges. If it converges, find the limit. Sol. an = (2n−1)! (2n+1)! = (2n−1)! (2n+1)(2n)(2n−1)! = 1 (2n+1)(2n) → 0 as n → ∞. Converges. 1 35. Determine whether the sequence an = (1 + n2 ) n converges or diverges. If it converges, find the limit. 1 Sol. an = (1 + n2 ) n ⇒ ln an = 1 n ln(1 + n2 ). As n → ∞, ln an → 0. Thus, an → e0 = 1 as n → ∞. Converges. 39. Determine whether the sequence an = Sol. an = n! 2n = 1 2 · 2 2 · 3 2 · ··· · (n−1) 2 · n 2 ≥ 1 2 n! 2n · n 2 converges or diverges. If it converges, find the limit. [ for n > 1 ] = n4 → ∞ as n → ∞, so {an } diverges. 51. For what values of r is the sequence {nrn } convergent? x Sol. If |r| ≥ 1, then it is clear that {nrn } diverges. If |r| < 1 then lim xrx = lim −x = x→∞ x→∞ r 1 rx lim = lim = 0, so lim nrn = 0, and hence {nrn } converges whenever |r| < 1. −x x→∞ (− ln r)r x→∞ − ln r x→∞ 59. Determine whether the sequence an = n2n+1 is increasing, decreasing, or not monotonic. Is the sequence bounded? Sol. an = n n2 +1 defines a decreasing sequence since for f (x) = x x2 +1 1 − x2 ≤ 0 for x ≥ 1. The sequence is bounded since 0 < an ≤ (x2 + 1)2 q p √ p √ √ 61. Find the limit of the sequence { 2, 2 2, 2 2 2, ...}. Sol. an = 2 2n −1 2n , f 0 (x) = 1 2 (x2 + 1)(1) − x(2x) = (x2 + 1)2 for all n ≥ 1. 1 . lim an = lim 21− 2n = 2. n→∞ n→∞ 63. Show that the sequence defined by a1 = 1, an+1 = 3 − 1 an is increasing and an < 3 for all n. Deduce that {an } is convergent and find its limit. Sol. We show by induction that {an } is increasing and bounded above by 3. Let Pn be the proposition that an+1 > an and 0 < an < 3. Clearly P1 is true. Assume that Pn is true. Then 1 an+1 > an ⇒ − an+1 > − a1n . Now an+2 = 3 − 1 an+1 > 3− 1 an = an+1 . Since an+1 = 3 − 1 an and 0 < an < 3, we have 0 < an+1 < 3. Thus Pn+1 is ture. This proves that {an } is increasing and bounded above by 3, and 1 = a1 < an < 3, that is, {an } is bounded, and hence convergent by the Monotonic Sequence Theorem. If L = lim an , then lim an+1 = L, so L must satisfy L = 3 − ⇒ L2 − 3L + 1 = 0 ⇒ L = √ 3± 5 2 . n→∞ But L > 1, so L = 1 n→∞ √ 3+ 5 2 . 1 L

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