36 2.1 Convergence Remark 2.1.3 Note that in the definition of convergence of a sequence (Definition 2.1.1), the inequality|an−a| <ε only needs to be satisfied for “sufficiently large” values of the indexn(n≥N). Therefore, it is sometimes convenient to assume that nis already, say, larger than 20, if that helps simplify a calculation. In this case, the ultimate value of Nchosen would also need to be greater than 20. The following example illustrates this technique. ■ Example 2.1.5 Consider the sequence given by an = n2 +5 4n2 +n . We prove directly from the definition that {an}converges to 1/4. Let ε >0 and consider the expression |an −a| given by n2 +5 4n2 +n− 1 4 = 4n2 +20−4n2 −n 4(4n2 +n) = | 20−n| 4(4n2 +n) . If n≥20, then |20−n| =n−20. Therefore, for such nwe have |20−n| 4(4n2 +n) = n−20 4(4n2 +n) = n−20 4n(4n+1) ≤ n 16n2 = 1 16n . Now if n≥N, then 1 16n ≤ 1 16N. We also observe that 1 16N <ε if N≥ 1 16ε . However, we have also required another condition on n, that is n≥20. To account for both conditions, we choose a positive integer N>max 1 16ε ,20 . Then, for n≥Nwe get n2 +5 4n2 +n− 1 4 = n−20 4n(4n+1) ≤ 1 16n ≤ 1 16N <ε. Hence, lim n→∞ n2 +5 4n2 +n = 1 4 . Remark 2.1.4 The proof of Theorem 2.1.1 below uses a new technique which is convenient when combining two or more inequalities to achieve an estimate. The definition of convergence states that for any arbitrary positive number (called ε in Definition 2.1.1) there exists an index (N) for which a certain inequality is satisfied (|an−a| <ε, for all n≥N). Now, if ε >0, then so is ε/2 (or ε/3, ε/7, 5ε or evencε for anyc>0). So, if an →athere also exists anN1 (possibly different fromN) so that |an −a| <ε/2 for all n≥N1. This technique will be used several times in subsequent proofs. As anticipated in Definition 2.1.1, we now prove that a convergent sequence can have only one limit. This justifies referring tothe limit of a sequence (rather than to alimit of a sequence). Theorem 2.1.1 A convergent sequence {an}has at most one limit.
RkJQdWJsaXNoZXIy NTc4NTAz