(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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[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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Week 3 Solutions Page 1
Exercise(2.4.1).Prove that
n21n 2 is Cauchy using directly the denition ofCauchy sequences.
Proof.Given >0, letM2Nbe such thatr2
< M.Then, for anym;nM,
jxmxnj=m21m 2n21n 2 1n 21m2 1n 2+1m 2 1M 2+1M 2 2M 2
Therefore,
n21n 2 is a Cauchy sequence.Exercise(2.4.2).Letfxngbe a sequence such that there exists a0< C <1 such that jxn+1xnj Cjxnxn1j: Prove thatfxngis Cauchy. Hint: You can freely use the formula (forC6= 1)1 +C+C2++Cn=1Cn+11C:
Proof.Let >0 be given. Note that
jx3x2j Cjx2x1j jx4x3j Cjx3x2j CCjx2x1j=C2jx2x1j and in general, one could prove that jxn+1xnj Cjxnxn1j C2jxn1xn2j Cn1jx2x1j:Week 3 Solutions Page 2
Now, form > n, we can evaluate the quantity
jxmxnj jxmxm1j+jxm1xm2j++jxn+1xnjCm2jx2x1j+Cm3jx2x1j++Cn1jx2x1j
= (Cm2++Cn1)jx2x1j =Cn1(1 +C++Cmn1)jx2x1j =Cn11Cmn1C jx2x1jCn111C
jx2x1j Now, since 0< C <1,Cn1!0 asn! 1. Therefore, there existsN2N such that whenevernN, jCn10j< 11C jx2x1j:For this sameN, wheneverm > nN
jxmxnj Cn111C jx2x1j 11C jx2x1j11C jx2x1jTherefore,fxngis a Cauchy sequence.Exercise.Prove the following statement using Bolzano-Weierstrass theorem.
Assume that(xn)n2Nis a bounded sequence inRand that there existsx2Rsuch that any convergent subsequence(xni)i2Nconverges tox. Thenlimn!1xn=x. Proof.Assume for contradiction thatxn6!x. Then9 >0 such that jxnxj for innitely manyn. From this, we can create a subsequencefxnjg such thatjxnjxj for allj2N. Since our original sequence is bounded, this subsequence is bounded, and so, by Bolzano-Weierstrass, there is a convergent subsequence of this subsequence, fxnjkg. By assumption,fxnjkgconverges tox. However, this is a contradiction sincejxnjkxj for allk2N.