39 ■ Example 2.1.7 Let an =(−1) n for n ∈ N. Then the sequence {an} is divergent. Indeed, suppose by contradiction that lim an = ℓ. n→∞ Then every subsequence of {an} converges to a number ℓ ∈ R. From the previous theorem, it follows, in particular, that ℓ = lim a2k = 1 and ℓ = lim a2k+1 = −1. k→∞ k→∞ 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 defned only for the integers greater than or equal to k0. 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 n + 1 an = . (n− 1)(n − 2) although a1 and a2 are not defned. In all cases, the sequence must be defned from some integer onwards. Exercises 2.1.1 Prove the following directly from the defnition of limit. 2n2 + 2 (a) lim = 0. n→∞ 3n3 + 1 n + 1 1 (b) lim = . n→∞ 5n + 1 5 n − 5 1 (c) lim = . n→∞ 2n + 1 2 3 − 2 3 n 1 (d) lim = . n→∞ 4n3 + n 2 2 5− 3n 3 (e) lim = − . n→∞ 2n2 + 1 2 √ n + 5 1 (f) lim √ = . n→∞ 3 n+ 2 3 2.1.2 Prove the following directly from the defnition of limit. 2n + n − 7 1 (a) lim = . n→∞ 3n2 + 5 3 2n + 1 1 (b) lim = . n→∞ 5n2 + n+ 1 5
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