. Get closed expressions for the following subsequences and decide if they converge. (1) a n+1 = (2) a 2n = (3) a 2n+1 = (4) a n 2 = (5) b n+1 =

Transcription

1 Math 316, Intro to Analysis subsequences. Recall one of our arguments about why a n = ( 1) n diverges. Consider the subsequences a n = ( 1) n = +1. It converges to 1. On the other hand, the subsequences a n+1 = ( 1) n+1 = 1 converges to 1. If a n had a limit L then the subsequences a n and a n+1 would have the same limits: 1 = L and 1 = L. But this is impossible since Today we make this explicit. Definition 1. Let f : N N be a strictly increasing sequence on natural numbers. (That is, if n < m then f(n) f(m)). Let a : N R be a sequence of real numbers. Then a f is called a subsequence of a. The idea is to build a new sequence by skipping some (possibly many) of the entries of the original sequence, but respecting the order of the original sequence, and never repeating the same term in the sequence. Notation: We often write a n instead of a(n) similarly we will write our sequence of natural numbers as n k instead of f(k). Thus, the standard notation for a subsequence if a nk where n 1 < n < n 3 <... is an increasing sequence. Example: Let a n = ( 1) n and b n = 1 n. Get closed expressions for the following subsequences and decide if they converge. (1) a n+1 = () a n = (3) a n+1 = (4) a n = (5) b n+1 = (6) b n = (7) b n+1 = (8) b n = Theorem. Let a n be any sequence and a R be any number. a n a if and only if every subsequence of a n converges to a 1

2 Proof. ( =) Suppose that for every subsequence a nk of a n a nk a. Notice that a n is a subsequence of itself. Thus, a nk = a k converges of a, as we claimed. (= ) Now suppose that a n a then for all ɛ > 0 there is an N such that if n > N then a n a < ɛ. Now, let n k be any strictly increasing sequence of natural numbers. Claim 1: n k k for all k. We will prove this claim soon. Suppose that k > N, but then n k k > N so that a nk a < ɛ. This completes the proof. Proof of Claim 1. The proof is by induction. Base Case. Show that n 1 1 inductive step Let k N. Assume that n k k. Thus, by the principle of mathematical induction n k k for all k N Use this theorem to prove the following (which we have previously proven directly) Corollary 3. a n = ( 1) n does not converge. Exercise 4. Show that there exists a sequence which diverges but for which there exists at least one convergent subsequence. Hint: Use a n = ( 1) n Cool. We ve have some divergent sequences with convergent subsequences. In order to study this phenomenon more we make some notation: Definition 5. If a n is a sequence and a R then a is a sub-sequential limit of a n if there is a subsequence a nk which converges to a. Exercise 6. Make a guess for what the set of all sub-sequential limits of a n = ( 1) n is.

3 Theorem 7. For a sequence a n and a R a is a sub-sequential limit of a n if and only if for all ɛ > 0 there exists infinitely many n N with a n a < ɛ. Proof. ( =) Suppose that a is a sub-sequential limit. Then there is a subsequence a nk with a nk a. Consider any ɛ > 0. Since a nk a there is a K such that if k > K then 3 How large is the set {n k : k > K}? Say a few words completing the proof. (= ) Suppose that for every ɛ there are infinitely many n N with a n a < ɛ. We need to build a subsequence with a nk a. We will do so using the following inductive argument: Claim : There exists a strictly increasing sequence n 1 < n <... of natural numbers such that a nk a < 1 k for all k N. Proof of Claim. Base case: By assumption there exist infinitely many n such that a n a < 1. Let n 1 be one of these. I ll include a first inductive step even though it is not strictly necessary in order to provide motivation. First inductive step: By assumption there are infinitely many n with a n a < 1. Since {1,,... n 1 } is a finite set, there are infinitely many n N with n n 1 and a n a 1. Let n be one such number. inductive step: Fix k N. Assume that that we have found n 1 < n < < n k with a nj a < 1 j for j = 1,,..., k. By assumption, there are infinitely many n N such that a n a < 1 k+1. The set {1,,... n k} is finite, so there are infinitely many natural numbers n with Let n k+1 be one such number. Thus, we have a sub-sequence a nk with a nk a < 1 k. Show that a n k a.

4 4 The Bolzano Weierstrass Theorem. Exercise 8. (1) The sequence a n = ( 1) n does not converge. Find a subsequence of a n which does converge. () The sequence b n = n diverges to infinity. ( i.e. For all M there exists and N such that if n > N then b n > M) Show that every subsequence of b n also diverges to infinity. What is different about these two sequences? Today we prove the following important theorem. Its proof is an application of Theorem 7 and of the nested intervals theorem. Theorem 9. If x n is a bounded sequence, then there is a subsequence x nk which converges. Proof. In order to prove the theorem we find a sub sequential limit. The subsequence which converges to that limit is a convergent subsequence. We will first make use of the Nested intervals theorem. Claim 3: If u is an upper bound for x n and l is lower bound then there is a sequence of nested intervals [a 1, b 1 ] [a, b ] [a 3, b 3 ] with b n a n = u l. and such that for each k n there are infinitely many n such that x n I k. We now use the claim to prove the theorem. By the nested intervals theorem there is an element of the intersection p a limit point of x n.. We will use Theorem 7 to show that p is Consider any ɛ > 0. Since 1 k, there is a k > 0 such that 1 <. k Then u l k = (u l) 1 k < ɛ Consider any q [a k, b k ]. a k p b k and a k q b k If p q then since q b k and a k p it follows that q p = q p = < ɛ For you: If q < p then show that q p < ɛ

5 5 Thus, for all q [a k, b k ] q is within ɛ of p. But by Claim 3 there are infinitely many x n [a k, b k ]. Thus there are infinitely many n with x n p < ɛ and p is a sub-sequential limit of x n. All that remains is to prove Claim 3: If u is an upper bound for x n and l is lower bound then there is a sequence of nested intervals [a 1, b 1 ] [a, b ] [a 3, b 3 ] with b n a n = u l. n and such that for each k there are infinitely many n such that x n I k. Proof of Claim 3. We will use an inductive construction to build the intervals [a n, b n ]. Let m = u l. Since x n [l, u] for all n (and in particularly infinitely often) it follows that at least one of the following hold: (Case 1) There are infinitely many n with x n [l, m] or (Case ) There are infinitely many n with x n [m, u] If (case 1) holds then set a 1 = l and b 1 = m. If (case 1) does not hold, then (case ) does. Set a 1 = m and b 1 = u. Either way we see that x n [a 1, b 1 ] for infinitely many n, and either (Case 1 ) b 1 a 1 = This completes the base case for our induction. Assume that there exists a sequence of k nested intervals = u l or (Case ) b 1 a 1 = = u l. [a 1, b 1 ] [a, b ] [a k, b k ] such that for all j = 1,,..., k two conditions hold: We need only build a subinterval [a k+1, b k+1 ] [a k, b k ] such that Let m k = b k+a k be the midpoint of [a k, b k ]. Since x n [l, u] for infinitely many n it follows that at least one of the following hold: (Case 1) There are infinitely many n with x n [a k, m k ] or (Case ) There are infinitely many n with x n [m k, b k ] If (case 1) holds then set a k+1 = a k and b k+1 = m k. If (case 1) does not hold, then (case ) does. Set a k+1 = m k and b k=1 = b k. Either way we see that x n [a k+1, b k+1 ] for infinitely many n.

6 6 In (case 1) In (case ) b k+1 a k+1 = = = by the inductive assumption b k+1 a k+1 = = = by the inductive assumption Thus, by the principle of mathematical induction we have a sequence of nested intervals [a 1, b 1 ] [a, b ]... such that for all k x n [a k, b k ] for infinitely many n and such that [b k a k ] u l. Thus, every sequence has a subsequential limit and so has a convergent subsequence. How many might it have? Exercise 10. If a n is a convergent sequence then show that a n has exactly one subsequential limit. If a n is a bounded sequence but does not converge then show that a n has at least two subsequential limits.