Lecture 7: Convergent Sequences of Real Numbers

18.100A: Complete Lecture Notes Lecture 7: Convergent Sequences of Real Numbers We will do another example of limits that converge: Example 1 limn→∞ 1 n2 +2n+100 = 0. Proof : Let  > 0 and choose M ∈ N such that M > n2 −1 2 . Then, ∀n ≥ M , 1 1 1 1 −0 ≤ 2 ≤ ≤ < . + 2m + 100 n + 2m + 100 2n 2M  The fact that we can go from a complicated rational function to one that works for our purposes (namely to prove the sequence converges to 0) is awesome. Example 2 Consider the sequence xn = (−1)n . This sequence is divergent. Proof : Let x ∈ R. We claim limn→∞ (−1)n 6= x. To prove this, we simply need find an epsilon that stops the sequence from converging. For instance, consider 0 = 12 . Then, for M ∈ N, 1 = |(−1)M − (−1)M +1 | ≤ |(−1)M − x| + |(−1)M +1 − x|. Thus, either |(−1)M − x| ≥ x. 1 2 or |(−1)M +1 − x| ≥ 21 . In either case, this shows that the limit cannot converge to  Theorem 3 If {xn } is convergent, then {xn } is bounded. Before we start the proof, let’s first talk about the idea of the proof. Let  = 1 such that |xn − x| < 1 for all n ≤ M for some M ∈ N. Then, there are finitely many elements not in the interval (x − 1, x + 1). We use this to our advantage. Proof : Suppose that limn→∞ xn = x. Thus, there exists an M ∈ N such that |xn − x| < 1 for all n ≥ M . Let B = max{|x1 |, |x2 |, . . . , |xM −1 |, |x| + 1}. If n < M , then |xn | ≤ B by construction. If n ≥ M , then |xn | ≤ |xn − x| + |x| < 1 + |x| ≤ B. 1 Definition 4 (Monotone) A sequence {xn } is monotone increasing if ∀n ∈ N, xn ≤ xn+1 . A sequence {xn } is monotone decreasing if ∀n ∈ N, xn ≥ xn+1 . If {xn } is either monotone increasing or monotone decreasing, we say {xn } is monotone or monotonic. Example 5 For example, xn = 1 n is monotone, yn = − n1 is monotone increasing, and (−1)n is neither. Theorem 6 Let {xn } be a monotone increasing sequence. Then, {xn } is convergent if and only if {xn } is bounded. Moreover, limn→∞ xn = sup{xn | n ∈ N}. Proof : Firstly, we know that if {xn } is convergent then it is bounded by the previous theorem, Now assume that {xn } is bounded. Then, x := sup{xn | n ∈ N} exists in R by the lowest upper bound property of R. We now prove that lim xn = x. n→∞ Let  > 0. Then, ∃M0 ∈ N such that x −  < xM0 < x since x is the supremum of the set. Let M = M0 . Then, ∀n ≥ M , we have x −  < xM0 = xM ≤ xn ≤ x < x + . Therefore, |xn | < |x + |. Therefore, xn → x. Theorem 7 Let {xn } be a monotone decreasing function. Then, {xn } is convergent if and only if {xn } is bounded. Moreover, lim xn = inf{xn | n ∈ N} n→∞ The proof of this is similar to the previous theorem and is thus omitted. Definition 8 (Subsequence) Informally, a subsequence is a sequence with entries coming from another given sequence. In other words, let {xn } be a sequence and let {nk } be a strictly increasing sequence of natural numbers. Then the sequence {xnk }∞ k=1 is called a subsequence of {xn }. Consider the sequence {xn } = n – in other words, the sequence 1, 2, 3, 4, . . . . Then, the following are subse- 2 quences of xn : 1, 3, 5, 7, 9, 11, . . . 2, 4, 6, 8, 10, . . . 2, 3, 5, 7, 11, 13, . . . . The first two are described by xnk = x2k and xnk = x2k−1 respectively. Question 9. How would we describe the third? Continuing to let {xn = n}n , the following are not subsequences: 1, 1, 1, 1, 1, 1, . . . 1, 1, 3, 3, 5, 5, . . . . Now consider the sequence {(−1)n }. Then we have the subsequences xnk = x2k−1 → −1, −1, −1, . . . ynk = x2k → 1, 1, 1, . . . . Theorem 10 If {xn } converges to x, then any subsequence of xn will converge to x. Proof : Suppose limn→∞ xn = x. Let  > 0. Then, ∃M0 ∈ N such that ∀n ≥ M0 , |xn − x| < . Choose M = M0 . If k ≥ M , then nk ≥ k ≥ M = M0 . Hence, for all  > 0 there exists an M such that for all nk > M , |xn−k − x| < . Remark 11. Notice that this also implies that the sequence {(−1)n }n is divergent. Notation 12 (DNC) We can denote the statement "a sequence does not converge"/"a sequence is divergent" as "the sequence DNC". 3