160 Solutions and Hints for Selected Exercises Thus, p √ an+1 = anbn ≥ anan = an for all n ∈ N, an + bn bn + bn bn+1 = ≤ = bn for all n ∈ N. 2 2 It follows that {an} is monotone increasing and bounded above by b1, and {bn} is decreasing and bounded below by a1. Let x = limn →∞ an and y = limn →∞ bn. Then √ x + y x = xy and y = . 2 Therefore, x = y. SECTION 2.4 Exercise 2.4.1. Here we use the fact that in R a 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) Defne αn = sup(an + bn), βn = supak, γn = supbk. k≥n k≥n k≥n By the defnition, limsup(an + bn)= lim αn, limsupan = lim βn, limsupbn = lim γn. n→∞ n→∞ n→∞ n →∞ n→∞ n→∞ By Exercise 2.5.2, αn ≤ βn + γn for all n ∈ N . This implies lim αn ≤ lim βn + lim γn for all n ∈ N. n→∞ n→∞ n→∞ Therefore, limsup (an + bn) ≤ limsup an + limsup bn. n→∞ n→∞ n→∞ This conclusion remains valid for unbounded sequences provided that the right-hand side is welldefned. Note that the right-hand side is not well-defned, for example, when limsupn →∞ an = ∞ and limsupn →∞ bn = −∞. (b) Defne αn = inf(an + bn), βn = inf ak, γn = inf bk. k≥n k≥n k≥n Proceed as in part (a), but use part (b) of Exercise 2.5.2. (c) Consider an =(−1) n and bn =(−1) n+1.
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