Introduction to Mathematical Analysis I - Second Edition

33 Since n ≥ 1, we have n 2 ≥ n and 4 n > 3. Therefore we have 4 n 2 − 1 3 n 2 − n − 4 3 = 4 n − 3 3 ( 3 n 2 − n ) ≤ 4 n − 3 3 ( 3 n 2 − n 2 ) < 4 n 6 n 2 = 4 6 n . Thus, if N > 4 6 ε , we have, for n ≥ N 4 n 2 − 1 3 n 2 − n − 4 3 ≤ 4 6 n ≤ 4 6 N < ε . Example 2.1.5 Consider the sequence given by a n = n 2 + 5 4 n 2 + n . We prove directly from the definition that { a n } converges to 1 4 . Let ε > 0. Now, n 2 + 5 4 n 2 + n − 1 4 = 4 n 2 + 20 − 4 n 2 − n 4 ( 4 n 2 + n ) = | 20 − n | 4 ( 4 n 2 + n ) . If n ≥ 20, then | 20 − n | = n − 20. Therefore, for such n we have n 2 + 5 4 n 2 + n − 1 4 = n − 20 4 ( 4 n 2 + n ) ≤ n 16 n 2 = 1 16 n . Choose N > max { 1 16 ε , 20 } . Then, for n ≥ N we get n 2 + 5 4 n 2 + n − 1 4 ≤ 1 16 n ≤ 1 16 N < ε . The following result is quite useful in proving certain inequalities between numbers. Lemma 2.1.2 Let ` ≥ 0. If ` < ε for all ε > 0, then ` = 0 . Proof: This is easily proved by contraposition. If ` > 0, then there is a positive number, for example ε = `/ 2, such that ε < ` . Theorem 2.1.3 A convergent sequence { a n } has at most one limit. Proof: Suppose { a n } converges to a and b . Then given ε > 0, there exist positive integers N 1 and N 2 such that | a n − a | < ε / 2 for all n ≥ N 1 and | a n − b | < ε / 2 for all n ≥ N 2 . Let N = max { N 1 , N 2 } . Then | a − b | ≤ | a − a N | + | a N − b | < ε / 2 + ε / 2 = ε . Since ε > 0 is arbitrary, by Lemma 2.1.2 , | a − b | = 0 and, hence, a = b . The following lemma is a simple generalization of ( 2.1.2 ) .