Review 4'18 Monotone Sequences and Cauchy Sequences

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Review 4.18 Monotone Sequences and Cauchy
Sequences

• Monotone Sequences
• 1) Definition
• A sequence (sn) is increasing if sn sn1 for
  all n.
• A sequence (sn) is decreasing if sn sn1 for
  all n.
• A sequence is monotone if it is increasing or
  decreasing.

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• 2) Examples
• (a) sn 2n1
• (b) sn 1-1/n
• (c) sn (0.9)n
• (d) sn -4n
• (e) sn (-1)n
• (a), (b) are increasing, (c) , (d) are
  decreasing.
• So (a), (b), (c) , (d) are monotone.
• (e) is neither increasing nor decreasing. So it
  is not monotone.

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Monotone Convergence Theorem

• 3) Theorem A monotone sequence is convergent iff
  it is bounded.
• Proof Assume that (sn) is a monotone Sequence.
• Suppose (sn) is convergent. Then it follows from
  Theorem 16.13 that (sn) is bounded.
• Conversely, suppose that (sn) is bounded.
• Without loss of generality, assume (sn) is
  decreasing.

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Proof

• Let S denote the set sn n is a natural number.
• Then S is bounded. In particular, S is bounded
  below. So S has a greatest lower bound. Let sinf
  S.
• We prove that lim sns.
• Given any There exists an integer K
  such that
• sKlts because s is the greatest lower
  bound.
• Since sn is decreasing and s is an lower bound,
  then
• ssnsKlts , that is sn-slt
  , for all ngtK.
• So lim sns.

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More Examples

• Let (sn) be defined by s11 and sn1
• Prove that (sn) is a bounded increasing sequence.
  And find its limit.
• Proof

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• We use the induction to prove it is bounded above
  by 3.
• Clearly, s11lt3. Suppose that sklt3. Then
• By induction, snlt3 for all n. Moreover, sngt0.
• So it is bounded.
• We also use the induction to prove it is
  increasing.
• s11lt . Assume that sklt sk1.
  Now we have
• So by the induction, (sn) is increasing.
  Therefore it is

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• convergent.
• Assume lim sns. Then lim sn1s.
• Thus
• Since sn is positive, s can not be negative. So

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More Theorem

• Theorem (a) If (sn) is an unbounded increasing
  sequence, then lim sn .
• (b) If (sn) is an unbounded
  decreasing sequence, then lim sn
• Proof (a) Since (sn) is increasing, then (sn)
  is bounded below by s1. Thus (sn) is not bounded
  above.
• Given any M, there exists an integer K, such that
  
• sKgtM. Since (sn) is increasing, snsKgtM for all
  ngtK.
• So lim sn

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Cauchy Sequences

• 2. Cauchy Sequences
• 1) Definition
• A sequence (sn) is said to be a Cauchy sequence
  if for each there exists a number K
  such that
• sn-smlt
• 2) Lemma Every convergent sequence is a Cauchy
  sequence.

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Proof

• Proof Suppose that (sn) converges to s. Given
  any
• there exists a number K such that
• for all ngtK.
• So for m, ngtk, we have
• This proves that (sn) is a Cauchy sequence.

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More Lemma

• Lemma Every Cauchy sequence is bounded.
• Proof (Exercise 18.11)
• (Hint similar to the proof of the
  theorem that every convergent sequence is
  bounded.)

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Cauchy Convergent Criterion

• Theorem A sequence is convergent iff it is a
  Cauchy sequence.
• Proof Assume (sn) is convergent. Then it follows
  from the lemma that it is a Cauchy sequence.
• Conversely, assume (sn) is a Cauchy sequence.
• Let S sn n is a natural number.
• 1) If S contains only one number s, then sn s
  for

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• all n, therefore it is convergent.
• 2) If S is finite and contains more than one
  number, let d be the minimum distance between two
  distinct points of S. Then dgt0.
• Since (sn) is a Cauchy sequence, there exists a
  number K, such that sn- sm ltd. Let n0 be an
  integer greater than K. Then sn- s n0ltd, for
  all ngtK. Since d is the minimal distance, then
• sn- s n00, that is sn s n0 for all ngtK. It
  follows that lim sn s n0 .

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• 3) If S infinite. By the lemma, it is bounded. It
  follows from the Bolzano-Weierstrass Theorem that
  there exists a number s that is an accumulation
  pint of S.
• We prove that (sn) converges to s.
• Given any there exists a number K
  such that
• for all m, ngtK.
• Since s is an accumulation point of S, there
  exists an integer pgtk such that

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• So for ngtK, we have
• Hence lim sn s .

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Example

• Prove that
• is divergent.
• Proof If mgtn, then

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• In particular, when m2n we have
• Thus the sequence sn can not be Cauchy and hence
  it is divergent.

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Discussion

• 18.1 a), b)
• 18.2 b), c)

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Homework

• 18.3 a), c), d), 4, 5 a), 6 a), 7, 10 a), 13
• 3 c) and 13 are due Wednesday (11/16)
• Exam 2 Monday (11/21)
• Coverage 3.13, 3.14, 4.16, 4.17, 4.18, 4.19