39 ■ Example 2.1.7 Let an = (−1) n for n∈N. Then the sequence {an}is divergent. Indeed, suppose by contradiction that lim n→∞ an =ℓ. Then every subsequence of {an}converges to a number ℓ∈R. From the previous theorem, it follows, in particular, that ℓ =lim k→∞ a2k =1 and ℓ =lim k→∞ a2k+1 =−1. This contradiction shows that the sequence is divergent. Since the sequence {an}is bounded but not convergent, this example illustrates the fact that the converse of Theorem 2.1.4 is not true. Remark 2.1.5 Given a positive integer k0, it will be convenient to also talk about the sequence {an}n≥k0, that is, a function defined only for the integers greater than or equal tok0. For simplicity of notation, we may also denote this sequence by {an} whenever the integer k0 is clear from the context. For instance, we talk of the sequence {an}given by an = n+1 (n−1)(n−2) . although a1 and a2 are not defined. In all cases, the sequence must be defined from some integer onwards. Exercises 2.1.1 Prove the following directly from the definition of limit. (a) lim n→∞ 2n2 +2 3n3 +1 =0. (b) lim n→∞ n+1 5n+1 = 1 5 . (c) lim n→∞ n−5 2n+1 = 1 2 . (d) lim n→∞ 2n3 −3 4n3 +n = 1 2 . (e) lim n→∞ 5−3n2 2n2 +1 =− 3 2 . (f) lim n→∞ √n+5 3√n+2 = 1 3 . 2.1.2 Prove the following directly from the definition of limit. (a) lim n→∞ n2 +n−7 3n2 +5 = 1 3 . (b) lim n→∞ n2 +1 5n2 +n+1 = 1 5 .
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