Introduction to Mathematical Analysis I - Second Edition

36 2.1 CONVERGENCE Then every subsequence of { a n } converges to a number ` ∈ R . From the previous theorem, it follows, in particular, that ` = lim k → ∞ a 2 k = 1 and ` = lim k → ∞ a 2 k + 1 = − 1 . This contradiction shows that the sequence is divergent. Since the sequence { a n } is bounded but not convergent, this example illustrates the fact that the converse of theorem 2.1.7 is not true. Remark 2.1.10 Given a positive integer k 0 , it will be convenient to also talk about the sequence { a n } n ≥ k 0 , that is, a function defined only for the integers greater than or equal to k 0 . For simplicity of notation, we may also denote this sequence by { a n } whenever the integer k 0 is clear from the context. For instance, we talk of the sequence { a n } given by a n = n + 1 ( n − 1 )( n − 2 ) . although a 1 and a 2 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 → ∞ 2 n 2 + 2 3 n 3 + 1 = 0. (b) lim n → ∞ n 2 + 1 5 n 2 + n + 1 = 1 5 . (c) lim n → ∞ 2 n 3 + 1 4 n 3 − n = 1 2 . (d) lim n → ∞ 3 n 2 + 5 6 n 2 + n = 1 2 . (e) lim n → ∞ 4 n 2 − 1 n 2 − n = 4. 2.1.2 Prove that if { a n } is a convergent sequence, then {| a n |} is a convergent sequence. Is the converse true? 2.1.3 Let { a n } be a sequence. Prove that if the sequence {| a n |} converges to 0, then { a n } also converges to 0. 2.1.4 Prove that lim n → ∞ sin n n = 0. 2.1.5 Let { x n } be a bounded sequence and let { y n } be a sequence that converges to 0. Prove that the sequence { x n y n } converges to 0. 2.1.6 Prove that the following limits are 0. ( Hint: use Theorem 2.1.6 . ) (a) lim n → ∞ n + cos ( n 2 − 3 ) 2 n 2 + 1 .