159 Solving this quadratic equation yields ℓ = −1 or ℓ = 2. Therefore, limn →∞ an = 2. Defne a more general sequence as follows: a1 = c > 0, √ an+1 = c + an for n ∈ N. √ 1+ 1+ 4c We can prove that {an} is monotone increasing and bounded above by . In fact, {an} √ 2 1 + 1 + 4c √ converges to this limit. The number is obtained by solving the equation ℓ = c+ ℓ, 2 where ℓ> 0. Exercise 2.3.2. (a) The limit is 3. (b) The limit is 3. (c) The limit is 1. (d) We use the well-known inequality √ a+ b + c 3 ≥ abc for a,b,c ≥ 0. 3 By induction, we see that an > 0 for all n ∈ N. Moreover, s 1 1 1 1 1 3 1 an+1 = (2an + )= (an + an + ) ≥ an · an · = 1. a2 a2 a2 3 3 3 n n n We also have, for n ≥ 2, 3 2 1 1 −a n + 1 −(an − 1)(an + an + 1) an+1 − an = 2an + − an = = < 0. 3 a2 3a2 3a2 n n n Thus, {an} is monotone deceasing (for n ≥ 2) and bounded below. We can show that limn →∞ an = 1. √ √ (e) Use the inequality x+y ≥ xy for x,y ≥ 0 to show that an+1 √≥ b for all n ∈ N. Then follow 2 item (c) to show that {an} is monotone decreasing. The limit is b. √ Exercise 2.3.3. (a) Let {an} be the given sequence. Observe that an+1 = 2an. Then show that {an} is monotone increasing and bounded above. The limit is 2. (b) Let {an} be the given sequence. Then 1 an+1 = . 2 + an Show that {a2n+1} is monotone decreasing and bounded below; {a2n} is monotone increasing and √ bounded above. Thus, {an} converges by Exercise 2.1.16. The limit is 2− 1. Exercise 2.3.7. Observe that p an + bn bn+1 = ≥ anbn = an+1 for all n ∈ N. 2
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