(c) Show that the sequence xn is bounded below by 1 and above by 2 (d) Use (e) Use (d) in a proof to show that Sn is Cauchy and thus converges In the first case, if k = 0, then we wish to prove that k · f(x) = 0, the zero function, has limit 0
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[PDF] Prove that ) is Cauchy using directly the definition of Cauchy
Assume that (xn)n∈N is a bounded sequence in R and that there exists x ∈ R such that any convergent subsequence (xni )i∈N converges to x Then limn→∞ xn
[PDF] 14 Cauchy Sequence in R
A sequence xn ∈ R is said to converge to a limit x if • ∀ϵ > 0, ∃N s t n > N ⇒ xn − x < ϵ A sequence xn ∈ R is called Cauchy sequence if • ∀ϵ, ∃N s t n > N m > N ⇒ xn − xm < ϵ Every convergent sequence is a Cauchy sequence Proof
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Show that xn → x as well; i e to prove that a Cauchy sequence is convergent, we only need i e any bounded sequence has a convergent subsequence Show
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A sequence {an}is called a Cauchy sequence if for any given ϵ > 0, there exists N ∈ N Proof Since {an}forms a Cauchy sequence, for ϵ = 1 there exists N ∈ N such that (i) lima1/n = 1, if a > 0 (ii) limnαxn = 0, if x < 1 and α ∈ IR Solution:
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If a sequence of points in X is U-convergent in X then it fulfills U-Cauchy condition Proof Let U − lim n→∞ xn = x and ε > 0 Thus A(ε/2) = {n ∈ N : ρ(xn, x) ≥
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Definition A sequence (an) is said to be a Cauchy sequence iff for any ϵ > 0 there exists N such prove (over the course of 2 + ϵ lectures) the following theorem: If a subsequence of a Cauchy sequence converges to x, then the sequence
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Every complex Cauchy sequence is convergent Proof Put zn = x + iy Then xn is Cauchy: xx − xm ⩽ zn − zm (as
[PDF] Math 431 - Real Analysis I
(c) Show that the sequence xn is bounded below by 1 and above by 2 (d) Use (e) Use (d) in a proof to show that Sn is Cauchy and thus converges In the first case, if k = 0, then we wish to prove that k · f(x) = 0, the zero function, has limit 0
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If (X, d) is a complete metric space and Y is a closed subspace of X, then (Y,d) is complete Proof Let (xn) be a Cauchy sequence of points in Y Then (xn) also
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Math 431 - Real Analysis I
Solutions to Homework due October 22
Question 1.Sequences are frequently givenrecursively, where a beginning termx1is specied and subse- quent terms can be found using a recursive relation. One such example is the sequence dened byx1= 1 and x n+1=p2 +xn: (a) Forn= 1;2;:::;10, computexn. A calculator may be helpful. (b) Show thatxnis a monotone increasing sequence. A proof by induction might be easiest. (c) Show that the sequencexnis bounded below by 1 and above by 2. (d) Use (b) and (c) to conclude thatxnconverges.Solution 1.
(a)nx n1121:4142131:8477641:9615751:9903661:9975971:9993981:9998591:99996101:99999(b) LetA(n) be the statement thatxn+1xn. We will show thatA(n) is true for alln1. First, note
thatx1= 1SinceA(k) is true, thenxk2. Thus,
x k+1=p2 +xkp2 + 2 = 2: Thus,A(k+ 1) is true. So, by induction,xkis bounded above by 2.(d) By (b) and (c),xnis a bounded, monotone sequence; thus, by a theorem in class, it converges.Question 2.One very important class of sequences areseries, in which we add up the terms of a given
sequence. One such example is the following sequence: S n=nX k=01k!=10! +11! +12! +13! ++1n!: 1 (a) Forn= 0;1;2;:::;6, computeSn. Again, a calculator may be helpful; be sure to use several digits. (b) Show thatSnis monotone increasing. (c) Use induction to show that for alln1,n!2n1. (d) Use (c) to show that S n1 +nX k=112 k1: (e) Use well-known facts from Calculus II and the geometric series to show that 1 + nX k=112 k1<3: (f) Use (b), (d), and (e) to conclude thatSnconverges.Solution 2.
(a)nS n01 1222:532:6666642:70833352:716666662:7180555(b) Since 1=n!>0, then
S n+1=Sn+1(n+ 1)!Sn:Thus,Snis monotone increasing.
(c) LetA(n) be the statement thatn!>2n1. We will show thatA(n) is true for alln1. For the base case, notice that 1! = 11 = 211:Thus,A(1) is true. Assume thatA(k) holds. Thus,k!>2k1. We will show thatA(k+ 1) is true by showing that (k+ 1)!<2k. Notice that (k+ 1)! = (k+ 1)k!(k+ 1)2k1: Sincek1, thenk+ 12. So, (k+ 1)2k122k1= 2k:Thus,A(k+ 1) is true. So,n!2n1for alln1. (d) Sincen!2n1, then we have that12 n11n!for alln1. Comparing term-by-term, we have that n X k=11k!nX k=112 k1: Since 10! = 1, we have that S n=nX k=01k!1 +nX k=112 k1: 2 (e) From Calculus II, we recognize nX k=112 k1=n1X j=012 j=n1X j=0 12 j as a geometric series. Thus, n1X j=0 12 j =112 n112 <1112 = 2: Thus, 1 +nX k=112 k1<1 + 2 = 3: (f) By (b),Snis a monotone sequence. By (d) and (e), we have thatSn<3 and is thus bounded. Thus, since every bounded monotone sequence converges,Snconverges. (g)Question 3.In class, we learned that a sequence inRkis convergent if and only if it is Cauchy. We have
previously proven using the denition of convergence that the sequence x n=1nconverges (to 0). Thus, it should also be Cauchy. In this problem, we will prove directly that it is Cauchy.
(a) Letn;m2Z+. Show that1n 1m <1n +1m (b) Use (a) to show thatxn=1n is a Cauchy sequence. To do so, given an" >0, nd anNsuch that for alln;m > N,1n 1m <1n +1m