160 Solutions and Hints for Selected Exercises Thus, an+1 =panbn ≥ √anan =an for all n∈N, bn+1 = an +bn 2 ≤ bn +bn 2 =bn for all n∈N. It follows that {an} is monotone increasing and bounded above byb1, and{bn} is decreasing and bounded below bya1. Let x =limn →∞ an andy =limn →∞ bn. Then x =√xy andy = x+y 2 . Therefore, x =y. SECTION 2.4 Exercise 2.4.1. Here we use the fact that inRa sequence is a Cauchy sequence if and only if it is convergent. (a) Not a Cauchy sequence. See Example 2.1.7. (b) A Cauchy sequence. This sequence converges to 0. (c) A Cauchy sequence. This sequence converges to 1. (d) A Cauchy sequence. This sequence converges to 0 (see Exercise 2.1.6). SECTION 2.5 Exercise 2.5.3. (a)Define αn =sup k≥n (an +bn), βn =sup k≥n ak, γn =sup k≥n bk. By the definition, limsup n→∞ (an +bn)=lim n→∞ αn, limsup n→∞ an =lim n→∞ βn, limsup n→∞ bn =lim n→∞ γn. By Exercise 2.5.2, αn ≤βn +γn for all n∈N. This implies lim n→∞ αn ≤lim n→∞ βn +lim n→∞ γn for all n∈N. Therefore, limsup n→∞ (an +bn) ≤limsup n→∞ an +limsup n→∞ bn. This conclusion remains valid for unbounded sequences provided that the right-hand side is welldefined. Note that the right-hand side is not well-defined, for example, when lim supn →∞ an =∞and limsupn →∞ bn =−∞. (b)Define αn =inf k≥n (an +bn), βn =inf k≥n ak, γn =inf k≥n bk. Proceed as in part (a), but use part (b) of Exercise 2.5.2. (c) Consider an =(−1) n andbn =(−1) n+1.
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