Introduction to Mathematical Analysis I - Second Edition

34 2.1 CONVERGENCE Lemma 2.1.4 Given real numbers a , b , then a ≤ b if and only if a < b + ε for all ε > 0. Proof: Suppose a < b + ε for all ε > 0. And suppose, by way of contradiction, that a > b . then set ε 0 = a − b . Then ε 0 > 0. By assumption, we should have a < b + ε 0 = b + a − b = a , which is a contradiction. It follows that a ≤ b . The other direction follows immediately from the order axioms. The following comparison theorem shows that (non-strict) inequalities are preserved “in the limit”. Theorem 2.1.5 — Comparison Theorem. Suppose { a n } and { b n } converge to a and b , respec- tively, and a n ≤ b n for all n ∈ N . Then a ≤ b . Proof: For any ε > 0, there exist N 1 , N 2 ∈ N such that a − ε 2 < a n < a + ε 2 , for n ≥ N 1 , b − ε 2 < b n < b + ε 2 , for n ≥ N 2 . Choose N > max { N 1 , N 2 } . Then a − ε 2 < a N ≤ b N < b + ε 2 . Thus, a < b + ε for any ε > 0. Using Lemma 2.1.4 we conclude a ≤ b . Theorem 2.1.6 — The Squeeze Theorem. Suppose the sequences { a n } , { b n } , and { c n } satisfy a n ≤ b n ≤ c n for all n ∈ N , and lim n → ∞ a n = lim n → ∞ c n = ` . Then lim n → ∞ b n = ` . Proof: Fix any ε > 0. Since lim n → ∞ a n = ` , there exists N 1 ∈ N such that ` − ε < a n < ` + ε for all n ≥ N 1 . Similarly, since lim n → ∞ c n = ` , there exists N 2 ∈ N such that ` − ε < c n < ` + ε for all n ≥ N 2 . Let N = max { N 1 , N 2 } . Then, for n ≥ N , we have ` − ε < a n ≤ b n ≤ c n < ` + ε , which implies | b n − ` | < ε . Therefore, lim n → ∞ b n = ` . Definition 2.1.2 A sequence { a n } is bounded above if the set { a n : n ∈ N } is bounded above. Similarly, the sequence { a n } is bounded below if the set { a n : n ∈ N } is bounded below. We say that the sequence { a n } is bounded if the set { a n : n ∈ N } is bounded, that is, if it is both bounded above and bounded below. It follows from the observation after Definition 1.5.1 that the sequence { a n } is bounded if and only if there is M ∈ R such that | a n | ≤ M for all n ∈ N .