Introduction to Mathematical Analysis I - Second Edition

38 2.2 LIMIT THEOREMS Proof: (a) Fix any ε > 0. Since { a n } converges to a , there exists N 1 ∈ N such that | a n − a | < ε 2 for all n ≥ N 1 . Similarly, there exists N 2 ∈ N such that | b n − b | < ε 2 for all n ≥ N 2 . Let N = max { N 1 , N 2 } . For any n ≥ N , one has | ( a n + b n ) − ( a + b ) | ≤ | a n − a | + | b n − b | < ε 2 + ε 2 = ε . Therefore, lim n → ∞ ( a n + b n ) = a + b . This proves (a) . (b) If k = 0, then ka = 0 and ka n = 0 for all n . The conclusion follows immediately. Suppose next that k 6 = 0. Given ε > 0, let N ∈ N be such that | a n − a | < ε | k | for n ≥ N . Then for n ≥ N , | ka n − ka | = | k || a n − a | < ε . It follows that lim n → ∞ ( ka n ) = ka as desired. This proves (b) . (c) Since { a n } is convergent, it follows from Theorem 2.1.7 that it is bounded. Thus, there exists M > 0 such that | a n | ≤ M for all n ∈ N . For every n ∈ N , we have the following estimate: | a n b n − ab | = | a n b n − a n b + a n b − ab | ≤ | a n || b n − b | + | b || a n − a | . (2.1) Let ε > 0. Since { a n } converges to a , we may choose N 1 ∈ N such that | a n − a | < ε 2 ( | b | + 1 ) for all n ≥ N 1 . Similarly, since { b n } converges to b , we may choose N 2 ∈ N such that | b n − b | < ε 2 M for all n ≥ N 2 . Let N = max { N 1 , N 2 } . Then, for n ≥ N , it follows from ( 2.1 ) that | a n b n − ab | < M ε 2 M + | b | ε 2 ( | b | + 1 ) < ε for all n ≥ N . Therefore, lim n → ∞ a n b n = ab . This proves (c) . (d) Let us first show that lim n → ∞ 1 b n = 1 b .