Chapter Three SEQUENCES

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Chapter ThreeSEQUENCES
Oleh Mardiyana Mathematics Education Sebelas
Maret University
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Definition 3.1.1

• A sequence of real numbers is a function on the
  set N of natural numbers whose range is contained
  in the set R of real numbers.
• X N ? R
• n ? xn
• We will denote this sequence by the notations
• X or (xn) or (xn n ? N)

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Definition 3.1.3

• If X (xn) and Y (yn) are sequences of real
  numbers, then we define
• X Y (xn yn)
• X Y (xn - yn)
• X.Y (xnyn)
• cX (cxn)
• X/Y (xn/yn), if yn ? 0 for all n ? N.

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Definition 3.1.4

• Let X (xn) be a sequence of real numbers. A
  real number x is said to be a limit of (xn) if
  for every ? gt 0 there exists a natural number
  K(?) such that for all n ? K(?), the terms xn
  belong to the ?-neighborhood V?(x) (x - ?, x
  ?).
• If X has a limit, then we say that the sequence
  is convergent, if it has no limit, we say that
  the sequence is divergent.
• We will use the notation
• lim X x or lim (xn) x

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Theorem 3.1. 5 A sequence of real numbers can
have at most one limit

• Proof
• Suppose, on the contrary, that x and y are both
  limits of X and that x ? y. We choose ? gt 0 such
  that the ?-neighborhoods V?(x) and V?(y) are
  disjoint, that is, ? lt ½ x y. Now let K and
  L be natural numbers such that if n gt K then xn ?
  V?(x) and if n gt L then xn ? V?(y). However, this
  contradicts the assumption that these
  ?-neighborhoods are disjoint. Consequently, we
  must have x y.

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Theorem 3.1.6

• Let X (xn) be a sequence of real numbers, and
  let x ? R. The following statements are
  equivalent
• X convergent to x.
• For every ?-neighborhood V?(x), there is a
  natural number K(?) such that for all n ? K(?)
  the terms xn belong to V?(x).
• For every ? gt 0, there is a natural number K(?)
  such that for all n ? K(?), the terms xn satisfy
  xn x lt ?
• For every ? gt 0, there is a natural number K(?)
  such that for all n ? K(?), the terms xn satisfy
  x - ? lt xn lt x ?.

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Tails of Sequences

• Definition 3.1.8
• If X (x1, x2, , xn, ) is a sequence of real
  numbers and if m is a given natural number, then
  the m-tail of X is the sequence
• Xm (xmn n ? N) (xm1, xm2, )
• Example
• The 3-tail of the sequence X (2, 4, 6, 8, ,
  2n, ) is the sequence X3 (8, 10, 12, , 2n
  6, ).

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Theorem 3.1.9

• If X (xn) be a sequence of real numbers and let
  m ? N. Then the m-tail Xm (xmn) of X converges
  if and only if X converges.
• In this case lim Xm lim X.

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Proof

• We note that for any p ? N, the pth term of Xm is
  the (p m)th term of X. Similarly, if q gt m,
  then the qth term of X is the (q m)th term of
  Xm.
• Assume X converges to x. Then given any ? gt 0, if
  the terms of X for n ? K(?) satisfy xn x lt ?,
  then the terms of Xm for k ? K(?) m satisfy xk
  x lt ?. Thus we can take Km(?) K(?) m, so
  that Xm also converges to x.
• Conversely, if the terms of Xm for k ? Km(?)
  satisfy xk x lt ?, then the terms of X for
  n ? Km(?) m satisfy xn x lt ?. Thus we
  can take K(?) Km(?) m.
• Therefore, X converges to x if and only if Xm
  converges to x.

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Theorem 3.1.10

• Let A (an) and X (xn) be sequences of real
  numbers and let x ? R. If for some C gt 0 and some
  m ? N we have
• xn x ? Can for all n ? N such that n ?
  m,
• and if lim (an) 0 then it follows that lim
  (xn) x.

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Proof

• If ? gt 0 is given, then since lim (an) 0, it
  follows that there exists a natural number
  KA(?/C) such that if n ? KA(?/C) then
• an an 0 lt ?/C.
• Therefore it follows that if both n ? KA(?/C) and
  n ? m, then
• xn x ? Can lt C (?/C) ?.
• Since ? gt 0 is arbitrary, we conclude that x
  lim (xn).

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Examples

1. Show that if a gt 0, then lim (1/(1 na)) 0.
2. Show that lim (1/2n) 0.
3. Show that if 0 lt b lt 1, then lim (bn) 0.
4. Show that if c gt 0, then lim (c1/n) 1.
5. Show that lim (n1/n) 1.

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Limit Theorems

• Definition 3.2.1
• A sequence X (xn) of real numbers is said to be
  bounded if there exists a real number M gt 0 such
  that
• xn ? M for all n ? N.
• Thus, a sequence X (xn) is bounded if and only
  if the set xn n ? N of its values is bounded
  in R.

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Theorem 3.2.2A convergent sequence of real
numbers is bounded

• Proof
• Suppose that lim (xn) x and let ? 1. By
  Theorem 3.1.6, there is a natural number K
  K(1) such that if n ? K then xn x lt 1. Hence,
  by the Triangle Inequality, we infer that if n ?
  K, then xn lt x 1. If we set
• M sup x1, x2, , xK-1, x 1,
• then it follows that
• xn ? M, for all n ? N.

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Theorem 3.2.3

• (a). Let X and Y be sequences of real numbers
  that converge to x and y, respectively, and let c
  ? R. Then the sequences X Y, X Y, X.Y and cX
  converge to x y, x y, xy and cx,
  respectively.
• (b). If X converges to x and Z is a sequence of
  nonzero real numbers that converges to z and if z
  ? 0, then the quotient sequence X/Z converges to
  x/z.

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3.3 Monotone Sequences

• Definition
• Let X (xn) be a sequence of real numbers. We
  say that X is increasing if it satisfies the
  inequalities
• x1 ? x2 ? ? xn-1 ? xn ?
• We say that X is decreasing if it satisfies the
  inequalities
• x1 ? x2 ? ? xn-1 ? xn ?

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Monotone Convergence Theorem

• A monotone sequence of real numbers is convergent
  if and only if it is bounded. Further
• a). If X (xn) is a bounded increasing sequence,
  then
• lim (xn) sup xn.
• b). If X (xn) is a bounded decreasing sequence,
  then
• lim (xn) inf xn.

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Proof
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Example

• Let

for each n ? N. a). Prove that X (xn) is an
increasing sequence. b). Prove that X (xn) is
a bounded sequence. c). Is X (xn) convergent?
Explain!
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Exercise

• Let x1 gt 1 and xn1 2 1/xn for n ? 2. Show
  that (xn) is bounded and monotone. Find the limit.

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3.4. Subsequences and The Bolzano-Weierstrass
Theorem

• Definition
• Let X (xn) be a sequence of real numbers and
  let r1 lt r2 lt lt rn lt be a strictly increasing
  sequence of natural numbers. Then the sequence Y
  in R given by

is called a subsequence of X.
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Theorem

• If a sequence X (xn) of real numbers converges
  to a real number x, then any sequence of
  subsequence of X also converges to x.
• Proof
• Let ? gt 0 be given and let K(?) be such that if n
  ? K(?), then xn x lt ?. Since r1 lt r2 lt lt rn
  lt is a strictly increasing sequence of natural
  numbers, it is easily proved that rn ? n. Hence,
  if n ? K(?) we also have rn ? n ? K(?). Therefore
  the subsequence X also converges to x.

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Divergence Criterion

• Let X (xn) be a sequence of real numbers. Then
  the following statements are equivalent
• (i). The sequence X (xn) does not converge to x
  ? R.
• (ii). There exists an ?0 gt 0 such that for any k
  ? N, there
• exists rk ? N such that rk ? k and xrk
  x ? ?0.
• (iii). There exists an ?0 gt 0 and a subsequence
  X of X such
• that xrk x ? ?0 for all k ? N.

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Monotone Subsequence Theorem

• If X (xn) is a sequence of real numbers, then
  there is a subsequence of X that is monotone.

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The Bolzano-Weierstrass Theorem

• A bounded sequence of real numbers has a
  convergent subsequence.

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3.5 The Cauchy Criterion

• Definition
• A sequence X (xn) of real numbers is said to be
  a Cauchy Sequence if for every ? gt 0 there is a
  natural number H(?) such that for all natural
  numbers n, m ? H(?), the terms xn, xm satisfy xn
  xm lt ?.

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Lemma If X (xn) is a convergent sequence of
real numbers, then X is a Cauchy sequence.

• Proof If x lim X, then given ? gt 0 there is a
  natural number K(?/2) such that if n ? K(?/2)
  then xn x lt ?/2. Thus, if H(?) K(?/2) and
  if n, m ? H(?), then we have
• xn xm (xn x) (x xm) ? xn x
  xm x
• lt ?/2 ?/2 ?.
• Since ? gt 0 is arbitrary, it follows that (xn) is
  a Cauchy sequence.

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LemmaA Cauchy sequence of real numbers is bounded

• Proof Let X (xn) be a Cauchy sequence and let
  ? 1. If H H(1) and n ? H, then xn xH ?
  1. Hence, by the triangle Inequality we have that
  xn ? xH 1 for n ? H. If we set
• M supx1, x2, , xH-1, xH 1
• then it follows that xn ? M, for all n ? N.

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Cauchy Convergence CriterionA sequence of real
numbers is convergent if and only if it is a
Cauchy sequence.

• Proof

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Definition

• We say that a sequence X (xn) of real numbers
  is contractive if there exists a constant C, 0 lt
  C lt 1, such that
• xn2 xn1 ? Cxn1 xn
• for all n ? N. The number C is called the
  constant of the contractive sequence.

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TheoremEvery contractive sequence is a Cauchy
sequence, and therefore is convergent

• Proof