Cauchy sequences

1 MATH 201, APRIL 20, 2020 The next assignment, an online test (or quiz ) due Wednesday 4/22 at5 pm CDT (hard deadline!) has now been posted on Canvas. There are14 questions. Each question has a mathematical statement. You haveto identify whether the statement is true or false. We have covered allthe material you need for the test. In particular, there are no questionson today s new is the lesson summary from last sequence{xn}n Uisconvergentif L R. >0, M N. M n U ,|xn L|< . 4 quantifiers, compares terms against some sequence{xn}n Uis aCauchy sequenceif >0, M N. M m, n U ,|xm xn|< . 3 quantifiers, compares terms against each convergent sequence is a Cauchy estimate:|xm xn|=|(xm L) + (L xn)| |xm L|+|L xn| 2+ 2= . Cauchy sequence is {xn}n U, chooseM Uso M m, n U ,|xm xn|< k U ,|xk| max{1 +|xM|,max{|xl||M > l U}}.

2 sequences 201, APRIL 20, 2020 Homework :Show directly from the definition that{n2 1n2}0<n Nis a Cauchy start by rewriting the sequence terms asxn=n2 1n2= 1 the sequence{1/n2}converges to 0, we know that for a giventolerance , there is a (positive) costMsuch that M m, n N,1n2< , M m, n N,|xm xn|= 1n2 1m2 1n2 + 1m2 < 2+ 2= ,verifying the Cauchy will not work detail: It just uses a general version ofour method , withynas the{2/n2}sequence for the challenge :Let{xn}be a Cauchy sequence such that(1) M N, k M . xk<0 and l M . xl> that limxn= a tolerance . By the Cauchy property, there is acostMsuch that M m, n ,|xm xn|< the given property (1), we then havexl>0 andxk<0, withk, l M, so|x| xk|< /2, which means in particular that|xl|< M n ,|xn|=|(xn xl) +xl| |xn xl|+|xl|< 2+ 2= ,as required for the convergence of{xn}to 201, APRIL 20, 20203 The new material: SeriesIn our 4/6 class, we started our work with sequences by consideringsuccessive approximationss0, s1, s2.

3 To Euler s number, the basee= ..of natural logarithms:Sequence element (partial sum) Numerical =10! =10!+11! =10!+11!+12! =10!+11!+12!+13! =10!+11!+12!+13!+14!..As you have probably noticed, this sequence was rather special (nowreflected in our switch to the notationsnand partial sum ). In fact,as captured by the following definition, sequences of this type are calledseries, which come with their own special notations and N, aseriesorinfinite series n=hxnor informallyxh+xh+1+xh+2+..means the sequence{k n=hxn}h k Nofpartial sumsk n=hxnof thesummandsxn. We write n=hxn=Lto express that the sequence of partial sums converges :Euler s constante= n=01n!= ..Most of the sequence terminology carries over, so we have convergentseries, bounded series, divergent series, Cauchy series, 201, APRIL 20, 2020 For today, we start working with series by explicitly finding a limit forthe sequence of partial sums.

4 This can only be done in certain series:(2) n=0rn=11 rfor a ratio rwith|r|< , consider the partial sumsk=k n=0rn= 1 +r+r2+..+rk 1+rkfor a natural numberk. Now always, (1+r+..+rk)(1 r) = 1 rk+1, asmay be checked informally by multiplying out and cancelling (try it!),or formally by induction on the natural number parameterk. Then,noting 1 r6= 0 for|r|<1, we have(3)sk=1 rk+11 r 11 rif|r|<1, and that is exactly what the equation (2) is saying, accordingto the basic that we used Proposition (i) from the book here, settingc=|r|, to obtain lim|r|k+1= 0, and then the Squeezing Lemma appliedto |r|k+1 rk+1 |r|k+1to conclude that limrk+1= 0 in (3).Telescoping series:Given a convergent sequence{yn}h n N y, n=h(yn yn+1) =yh Y ,sincek n=h(yn yn+1) = (yh yh+1) + (yh+1 yh+2) +.

5 + (yk yk+1)=yh yk+1 yh y .Example. n=11n(n+ 1)(n+ 2)= n=1[12n(n+ 1) 12(n+ 1)(n+ 2)] 201, APRIL 20, 20205 Here is the lesson N, aseriesorinfinite series n=hxnor informallyxh+xh+1+xh+2+..means the sequence{k n=hxn}h k Nofpartial sumsk n=hxnof thesummandsxn. Then write n=hxn=Lif the sequence of partial sums converges :Euler s constante= n=01n!= ..Most of the sequence terminology carries over, so have convergentseries, bounded series, divergent series, Cauchy series, series are easy to series: n=0rn=11 rfor ratio rwith|r|< series:Given a convergent sequence{yn}h n N y, n=h(yn yn+1) =yh Ysincek n=h(yn yn+1) = (yh yh+1) + (yh+1 yh+2) +..+ (yk yk+1)=yh yk+1 yh y.