(c) If (xn) satisfies the Cauchy criterion, then there exists an α ∈ R such that 0
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[PDF] Week 3 Solutions Page 1 Exercise (241) Prove that ) is Cauchy
(n2−1 n2 ) is Cauchy using directly the definition of Cauchy sequences Proof Given ϵ > 0 Let {xn} be a sequence such that there exists a 0
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Let X = (xn) be a sequence of positive real numbers such that lim (xn+1 xn ) is Cauchy 4 Let (fn) ∈ C[0,1] be such that there exists M > 0 such that fn ∞ ≤
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A real sequence (xn) is called contractive if there exists a constant 0 0 and let N be such that x2 − x1 Cn−1 1 1−C
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Practice Problems 3 : Cauchy criterion, Subsequence
1. Show that the sequence (xn) dened below satises the Cauchy criterion.
(a)x1= 1 andxn+1= 1 +1x nfor alln1 (b)x1= 1 andxn+1=12+x2nfor alln1: (c)x1= 1 andxn+1=16 (x2n+ 8) for alln1:2. Let (xn) be a sequence of positive real numbers. Prove or disprove the following
statements. (a) Ifxn+1xn!0 then (xn) converges. (b) Ifjxn+2xn+1j3. Let (xn) be a sequence of integers such thatxn+16=xnfor alln2N. Prove or
disprove the following statements. (a) The sequence (xn) does not satisfy the Cauchy criterion. (b) The sequence (xn) cannot have a convergent subsequence.4. Suppose that 0< <1 and that (xn) is a sequence satisfying the condition:
jxn+1xnj n; n= 1;2;3;::::Show that (xn) satises the Cauchy criterion.5. Let (xn) be dened by:x1=11!
;x2=11! 12! ;:::;xn=11! 12! +:::+(1)n+1n!:Show that the sequence converges.6. Let 1x1x22 andxn+2=px
n+1xn,n2N. Show thatxn+1x n12 for all n2N,jxn+1xnj 23 jxnxn1jfor alln2Nand (xn) converges.7.(*)Show that a sequence (xn) of real numbers has no convergent subsequence if
and only ifjxnj ! 1.8.(*)Let (xn) be a sequence inRandx02R. Suppose that every subsequence of
(xn) has a convergent subsequence converging tox0. Show thatxn!x0.9.(*)Let (xn) be a sequence inR. We say that a positive integernis a peak of the
sequence ifm > nimpliesxn> xm(i.e., ifxnis greater than every subsequent term in the sequence). (a) If (xn) has innitely many peaks, show that it has a decreasing subsequence. (b) If (xn) has only nitely many peaks, show that it has an increasing subsequence. (c) From (a) and (b) conclude that every sequence inRhas a monotone subse- quence. Further, every bounded sequence inRhas a convergent subsequence (An alternate proof of Bolzano-Weierstrass Theorem).Hints/Solutions
1. (a) Note thatjxn+1xnj=j1x
n1x n1j=jxn1xnx nxn1jandjxnxn1j=j(1+1x n1)xn1j= jxn1+ 1j 2. This implies thatjxn+1xnj 12 jxnxn1j:Hence (xn) satises the contractive condition and therefore it satises the Cauchy criterion. (b) Observe thatjxn+1xnj=jx2nx2n1j(2+x2n)(2+x2n1)jxnxn1jjxn+xn1j4 24jxnxn1j: (c) We havejxn+1xnj jxnxn1jjxn+xn1j6 46
jxnxn1j: