Introduction to Mathematical Analysis I - Second Edition

45 Theorem 2.3.6 Let { a n } , { b n } , and { c n } be sequences of real numbers and let k be a constant. Suppose lim n → ∞ a n = ∞ , lim n → ∞ b n = ∞ , and lim n → ∞ c n = − ∞ Then (a) lim n → ∞ ( a n + b n ) = ∞ ; (b) lim n → ∞ ( a n b n ) = ∞ ; (c) lim n → ∞ ( a n c n ) = − ∞ ; (d) lim n → ∞ ka n = ∞ if k > 0, and lim n → ∞ ka n = − ∞ if k < 0; (e) lim n → ∞ 1 a n = 0. (Here we assume a n 6 = 0 for all n .) Proof: We provide proofs for (a) and (e) and leave the others as exercises. (a) Fix any M ∈ R . Since lim n → ∞ a n = ∞ , there exists N 1 ∈ N such that a n ≥ M 2 for all n ≥ N 1 . Similarly, there exists N 2 ∈ N such that b n ≥ M 2 for all n ≥ N 1 . Let N = max { N 1 , N 2 } . Then it is clear that a n + b n ≥ M for all n ≥ N . This implies (a) . (e) For any ε > 0, let M = 1 ε . Since lim n → ∞ a n = ∞ , there exists N ∈ N such that a n > 1 ε for all n ≥ N . This implies that for n ≥ N , 1 a n − 0 = 1 a n < ε . Thus, (e) holds. The proof of the comparison theorem below follows directly from Definition 2.3.2 (see also Theorem 2.1.5 ) . Theorem 2.3.7 Suppose a n ≤ b n for all n ∈ N . (a) If lim n → ∞ a n = ∞ , then lim n → ∞ b n = ∞ . (b) If lim n → ∞ b n = − ∞ , then lim n → ∞ a n = − ∞ .