cauchy and montone sequences

Definitions Theorems

Cauchy and Montone Sequences

Dr. Lance Nielsen

Creighton University Department of Mathematics

January 31, 2007


Outline

1 DefinitionsCauchy SequencesMonotonic Sequences

2 TheoremsComments and CautionA Corollary - The Contraction Principle


Definition of Cauchy Sequence

DefinitionLet {an}∞n=1 be a sequence of real numbers. We say that{an}∞n=1 is a Cauchy sequence if, for every ε > 0, there is anN ∈ N such that |am − an| < ε whenever m, n > N.

There are sequences that are not Cauchy. For example,an =

∑nk=1

1k is not a Cauchy sequence.

The idea of Cauchy sequence is very closely related to theidea of completeness.


Outline




Definition of Monotonic Sequence

DefinitionLet {an}∞n=1 be a sequence of real numbers. We say that{an}∞n=1 is a monotonic increasing sequence if an < an+1 for alln ∈ N. (It is possible for a sequence to be monotonic increasingfor n ≥ N0 for some N0; i.e. it becomes monotonic for largeenough n.) We say that {an}∞n=1 is monotonic decreasing ifan > an+1 for all n ∈ N. (As before, it is possible for a sequenceto become monotonic decreasing for large enough n.)If we replace the strict inequalities above with a "≤" or "≥", wethen say that the sequence is nondecreasing or nonincreasing,respectively.


Examples of Monotonic Sequences

an = nn+1 : We show that this sequence is monotonic

increasing. Since nn+1 < n+1

n+2 is equivalent to 0 < 1(n+1)(n+2)

and this inequality is certainly true for all n ∈ N.bn = n+1

2n+1 : We show that this sequence is monotonicdecreasing. Since n+1

2n+1 > n+12n+2 is equivalent to n

2n+2 > 0 andthis last inequality is true for all n, the sequence ismonotonic decreasing.

cn = n7−n2+1n+e2n : This sequence is not monotonic initially but

is eventually monotonic. Can you figure out if it’s eventuallyincreasing or decreasing? How?


A Theorem on Monotonic Sequences

TheoremSuppose that {an}∞n=1 is a monotonic increasing sequence thatis bounded above. Then there is a real number L such thatlimn→∞ an = L. Similarly, if {bn}∞n=1 is a monotonic decreasingsequence that is bounded below, then there is a real number Msuch that limn→∞ bn = M.

Proof.We assume, without loss of generality, that {an} is monotonicdecreasing and bounded below. Let a = inf {an : n ∈ N}. (Whydoes a exist?) Given ε > 0, note that a + ε is not a lower boundfor the sequence. Hence there is an n0 such that an0 < a + ε.But, for any n ≥ n0, an ≤ an0 < a + ε. Moreover, a is a lowerbound so a− ε < a ≤ an < a + ε which implies that |an − a| < εfor all n ≥ n0.


Example

Define an recursively by an+1 = 12

(a2

n + 1), a1 = 0.

Claim: {an} is an increasing sequence.Proof: We calculate as follows.

an+1 − an =12(a2

n + 1)− an =12

(an − 1)2 ≥ 0

Claim: {an} is bounded.Proof: We proceed by induction. If |an| ≤ 1, then|an+1| ≤ 1

2 · 2 = 1 and also, |a1| ≤ 1.We have proved that an is an increasing and boundedsequence. Therefore it has a limit. What is this limit?


A Theorem on Cauchy Sequences

TheoremLet {an}∞n=1 be a sequence of real numbers. The sequence isCauchy if and only if it converges to some limit a ∈ R.

Proof:(Convergent implies Cauchy) Assume that the sequenceconverges to a limit a. Take any ε > 0. Then there is a positiveinteger N such that if n ≥ N, then |an − a| < ε/2. Hence

|aj − ak | ≤ |aj − a|+ |a− ak | < ε/2 + ε/2 = ε

if j , k ≥ N. It follows that the sequence is Cauchy. We addressthe implication Cauchy implies convergence on the next slides.


Proof, Continued

Conversely, suppose that the sequence is Cauchy. Define theset S = {x ∈ R : x < aj except for finitely many j}.Since thesequence is bounded (why?), say by M > 0, we know thatevery term in the sequence is bigger than −M. Therefore−M ∈ S. Also, every term of the sequence is smaller than M,so that S is bounded by M. Hence we see that S is a boundedand nonempty subset of real numbers and so it has a leastupper bound, say a. We claim that an → a as n →∞. Let ε > 0be given. Choose N so that

|aj − ak | <ε

2

if j , k ≥ N. Then, in particular, |aj − aN+1| < ε2 . It follows that,

for j ≥ N,aj > aN+1 −

ε

2for j ≥ N.


Proof, Continued

Thus, aN+1 − ε2 ∈ S, and we have that a ≥ aN+1 − ε

2 . We alsohave that

aj < aN+1 +ε

2for j ≥ N and therefore aN+1 + ε

2 /∈ S. Thus a ≤ aN+1 + ε2 . But

now, looking at the inequalities above, we have that|a− aN+1| < ε

2 . We finally obtain∣∣a− aj∣∣ ≤ |a− aN+1|+

∣∣aN+1 − aj∣∣ <

ε

2+

ε

2

for j ≥ N and the proof is finished.


Outline




Further Comments About Cauchy Sequences

1 It is often difficult to tell whether a sequence is Cauchy.The fact that the convergent sequences are the Cauchysequences is usually used when trying to prove generaltheorems about sequences.

2 Keep in mind that for a sequence to be Cauchy we musthave, for arbitrary ε > 0, |ak − al | < ε for all k , l > N. It isnot sufficient that the difference in successive terms besmall. That is to say, even if |an+1 − an| < ε for all n > N,the sequence may still diverge. Suppose an =

√n. It is

clear that this sequence diverges. However,

|an+1 − an| =(√

n + 1−√

n)(√

n + 1 +√

n)√n + 1 +

√n

<1

2√

n

and so the difference in successive terms can be madearbitrarily small.


Outline




Contraction Principle

There is, however, a result that does give a test forconvergence stated in terms of the size of successivedifferences of successive terms of the sequence.

Corollary

Let {an}∞n=1 be a sequence of real numbers. If r is any realconstant such that 0 < r < 1 and |an+2 − an+1| ≤ r |an+1 − an|,then {an}∞n=1 converges.


Proof of the Contraction Principle

We show that the sequence is Cauchy. It will then follow at oncethat it converges. Note that if n > m, the difference an − am canbe written as a telsecoping sum of successive differences:

an − am =n−1∑k=m

(ak+1 − ak ). (1)

If we apply the hypothesis k − 1 times, we can approximate|ak+1 − ak | in terms of r and the initial difference |a2 − a1|.

|ak+1 − ak | ≤ r |ak − ak−1| ≤ r2|ak−1 − ak−2| ≤ · · ·≤ r k−1|a2 − a1|

(2)


Proof Continued

n−1∑k=m

r k−1|a2 − a1| = rm−1|a2 − a1|n−m−1∑

k=0

r k−1

= |a2 − a1|rm−1 · 1− rn−m

1− r

< |a2 − a1|rm−1 · 11− r

,

(4)

where the inequality follows from Example 4.2.2. But rn → 0(problem 4.2.4) and therefore we can, given ε > 0, find an Nsuch that if m − 1 > N, then rm−1 < ε(1− r)/|a2 − a1|. Usingthis bound for rm−1, it follows that if m − 1 > N, then|an − am| < ε and the proof is done.

cauchy and montone sequences

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