Introduction to Mathematical Analysis I - 3rd Edition

49 (i) limn→∞(an +bn) =∞, (ii) limn→∞(anbn) =∞, (iii) limn→∞(ancn) =−∞, (iv) limn→∞kan =∞if k >0, and limn→∞kan =−∞if k <0, (v) limn→∞ 1 an =0. (Here we assume an̸ =0 for all n.) Proof: We provide proofs for (i) and (v) and leave the others as exercises. (i) Fix anyM∈R. Since limn→∞an =∞, there exists N1 ∈Nsuch that an ≥ M 2 for all n≥N1. Similarly, there exists N2 ∈Nsuch that bn ≥ M 2 for all n≥N2. Let N=max{N1,N2}. Then it is clear that an +bn ≥Mfor all n≥N. This proves (i). (v) For anyε >0, let M=1/ε. Since limn→∞an =∞, there exists N∈Nsuch that an > 1 ε for all n≥N. This implies that for n≥N, 1 an − 0 = 1 an <ε. Thus, (v) holds. □ The proof of the comparison theorem below follows directly from Definition 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. Define an+1 =pan +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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