49 (i) limn →∞(an + bn)= ∞, (ii) limn →∞(anbn)= ∞, (iii) limn →∞(ancn)= −∞, (iv) limn →∞ kan = ∞ if k > 0, and limn →∞ kan = −∞ if k < 0, 1 (v) limn →∞ = 0. (Here we assume an ≠ 0 for all n.) an Proof: We provide proofs for (i) and (v) and leave the others as exercises. (i) Fix any M ∈ R. Since limn →∞ an = ∞, there exists N1 ∈ N such that M an ≥ for all n ≥ N1. 2 Similarly, there exists N2 ∈ N such that M bn ≥ for all n ≥ N2. 2 Let N = max{N1,N2}. Then it is clear that an + bn ≥ M for all n ≥ N. This proves (i). (v) For any ε > 0, let M = 1/ε. Since limn →∞ an = ∞, there exists N ∈ N such that 1 an > for all n ≥ N. ε This implies that for n ≥ N, 1 1 − 0 = < ε. an an Thus, (v) holds. □ The proof of the comparison theorem below follows directly from Defnition 2.3.2 (see also Theorem 2.1.2). Theorem 2.3.5 Suppose an ≤ bn for all n ∈ N. (a) If limn →∞ an = ∞, then limn →∞ bn = ∞. (b) If limn →∞ bn = −∞, then limn →∞ an = −∞. Exercises √ 2.3.1 ▶ Let a1 = 2. Defne p an+1 = an + 2 for n ≥ 1. (a) Prove that an < 2 for all n ∈ N. (b) Prove that {an} is an increasing sequence.
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