159 Solving this quadratic equation yields ℓ =−1 or ℓ =2. Therefore, limn→∞an =2. Define a more general sequence as follows: a1 =c >0, an+1 =√c+an for n∈N. We can prove that {an}is monotone increasing and bounded above by 1+√1+4c 2 . In fact, {an} converges to this limit. The number 1+√1+4c 2 is obtained by solving the equation ℓ =√c+ℓ, 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 ≥ 3 √abcfor a,b,c ≥0. By induction, we see that an >0 for all n∈N. Moreover, an+1 = 1 3 (2an + 1 a2 n ) = 1 3 (an +an + 1 a2 n ) ≥ 1 3 3 san · an · 1 a2 n =1. We also have, for n≥2, an+1 −an = 1 3 2an + 1 a2 n − an =− a3 n +1 3a2 n =−(an −1)(a 2 n +an +1) 3a2 n <0. Thus, {an}is monotone deceasing (for n≥2) and bounded below. We can show that limn→∞an =1. (e) Use the inequality x+y 2 ≥ √xy for x,y ≥0 to show that an+1 ≥ √b for all n∈N. Then follow 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 an+1 = 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 bn+1 = an +bn 2 ≥p anbn =an+1 for all n∈N.
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