4.3-4* Find an example of a sequence of real numbers satisfying each set ofproperties.20
1. Cauchy, but not monotone.2. Monotone, but not Cauchy.3. Bounded, but not Cauchy.Solutions:1. ((−10nn) is Cauchy, but not monotone.2. (1,2,3,4, ...) is monotone, but not Cauchy.3. (1,0,1,0,1,0, ...) is bounded, but not Cauchy.�•4.3-6* Let(an)and(bn)be monotone sequences. Prove or give a counterex-ample,1. The sequence(cn)given bycn=kanis monotone for anyk∈R.2. The sequence(cn)given bycn=an/bnis monotone, wherebn�= 0for alln∈N.Solutions:1. Yes. (cn) = (kan) is monotone. In fact, ifk >0, we have thatkan≤kan+1ifand only ifan≤an+1; and thatkan≥kan+1if and only ifan≥an+1. Therefore,(kan) is monotone if and only if (an) is monotone. Similarly we can consider thecasek <0 andk= 0.2. No. Here is a counterexample: (an) = (1,2,2,3,3,4,4,....) and (bn) = (1,1,2,2,3,3,4,4, ...) are monotone. But the sequence�anbn�=�1,2,1,32,1,43, ...�is not monotone.•4.3-9* Supposex >0. Define a seqeunce(sn)bys1=kandsn+1=s2n+x2snforn∈N. Prove that for anyk >0,limsn=√x.Solutions:(sn) is bounded belowWe claim thatsn≥√xfor alln≥2, i.e.,s2n−x≥0 for alln≥2. In fact,21
s2k+1−x=�s2k+x2sk�2−x=s4k+ 2s2kx+x24s2k−x=s4k+ 2s2kx+x2−4s2kx4k2=(s2k−2)24k2≥0.holds for anyk≥1.(sn) is decreasing:We want to prove that (sn) is decreasing forn≥2:sn−sn+1≥0for alln≥2. In fact,sn−sn+1=sn−x2n+x2sn=s2n−x2sn≥0.Here for the last inequality, we have used the ineqyalitys2n−x≥0 proved above.