Problem Sets – SequencesExplicitly-defined sequences1. Determine whether the following sequences converge, and if they do, find the limit.a)an=3√n√n+ 2,b)an=3 + 5n21 +n,c)an=1 +2nnSOLUTIONS:a) Divide the numerator and denominator by√n,an=3√n√n+ 2=31 + 2/√nThe numerator and the denominator are convergent sequences, and so the limit of the original sequence is aquotient of their individual limits,limn→∞3√n√n+ 2=limn→∞3limn→∞1 + 2/√n=31= 3b) Divide the numerator and the denominator byn2,an=3 + 5n21 +n=3/n2+ 51/n2+ 1/nAgain, the numerator and the denominator are convergent sequences, BUT the limit of the denominator iszero, 1/n2+ 1/n→0 andn→ ∞. Consequently, the sequence diverges (i.e.does not converge).c) To apply the limit rules, first make the change of variablen/27→k(and son7→2k), then the original sequenceis transformed toan=1 +2nn7→bk=1 +1k2k="1 +1kk#2The limit of the sequence inside the square brackets is the natural exponential basee= 2.718281828. . .. Thefull limit is then,limn→∞an= limk→∞bk= limk→∞"1 +1kk#2="limk→∞1 +1kk#2=e2= 7.389056099. . .Implicitly-defined sequences2. Determine whether the following sequence converges and, if so, determine the limit.√2,q2√2,r2q2√2,s2r2q2√2, . . .SOLUTION:Usually implicitly-defined sequences are difficult to analyze directly, but the initial term of thissequence makes our work much easier.First, from the list of terms, the rule generating the sequence seems to be:an+1=√2an(n≥0),witha0=√2Using the exponential form for the square-root,an+1= (2an)1/2(n≥0),witha0= 21/2
Writing out a few terms,a0= 21/2a1=2×21/21/2=23/21/2= 23/4a2=2×23/41/2=27/41/2= 27/8a3= 215/16...an= 22(n+1)-12(n+1)= 21-12(n+1)Each term is 2 raised to a fraction with a power of two in the denominator, and one-minus-a-power-of-two in thenumerator. Asn→ ∞, the exponent approaches 1, and solimn→∞an= 21= 23. An implicitly-defined sequence is given byan+1=√2 +an(n≥1), with initial terma1=√2.a) Show that the sequence is increasing and bounded above by 3.SOLUTION:The claim is that3≥an+1≥anIt is certainly true for the first two terms:a1=√2 = 1.41421. . .anda2=p2 +√2 = 1.84776. . ., so3≥a2≥a1XFor the ‘induction’ part of the proof,giventhat 3≥an+1≥an, we must show that 3≥an+2≥an+1is alsotrue (i.e.that the first statement implies the second). Starting from what we are given,3≥an+1≥anwe want to convertan+1→an+2andan→an+1. To that end, add 2 to each term in the inequality,2 + 3≥2 +an+1≥2 +anThis step does not change the logic of the inequalities. The square-root function is a strictly-increasing function,and so it also won’t change the logic of the inequalities. Taking the square-root of each term,√2 + 3≥p2 +an+1≥√2 +anYou will recognize the last two terms asan+2=√2 +an+1andan+1=√2 +anso we have√2 + 3≥an+2≥an+1But√2 + 3 =√5<3, so3≥√2 + 3≥an+2≥an+1Xwhich means we have proved the claim by induction.