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Section 1 A sequence(of real numbers) is
a list of infinitely many real numbers arranged in such a manner that it has a beginning but no end such as:
S1: 1, 2, 4, 8, …S2 : 1, ½, 1/3, ¼, 1/5, …S3 : 1, -1, 1, -1, …S4 : 1, -1/2, 1/3, -1/4, …S5 : 2, 2, 2, …S6 : -20, -10, -5, -4, 2, 2, 2, 2, …
Increasing sequenceIncreasing sequence
Decreasing sequenceDecreasing sequence
Oscillating sequenceOscillating sequence
Oscillating sequenceOscillating sequence
Constant sequenceConstant sequence
Constant sequence!Constant sequence!
Among them, which are convergent sequences?Among them, which are convergent sequences?
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Convergent sequences
A sequence xn is said to be convergent
iff
(or we say that xn converges)
Otherwise, it is said to be divergent. (or we say that xn diverges)
Lxlim nn 0n0nn
Nn ,|Lx|NN0 iff Lxlim
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nnsin)1(
x:1 Sequence2n
n
-0.5
-0.3
-0.1
0.1
0.3
0.5
0 50 100 150 200 250
Sequence1
Is xn convergent? Is xn convergent?
L= 0 ! L= 0 !
Discussion:p.293 Ex.7.1, Q.2Discussion:p.293 Ex.7.1, Q.2
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Sequence 2:n2
xn
n
Xn=2̂ n/n
0
2E+74
4E+74
6E+74
0 100 200 300n
Does it converge?
Does it converge?Discussion:Ex.7.1, Q.3Discussion:Ex.7.1, Q.3
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Two important sequences
1) Let q be a fixed real number and |q| < 1, then
2) Let a be a fixed positive number,
then
?qlim n
n
00
?alim n
n
11
Xn=2̂ (1/n)
0
1
2
3
0 100 200 300n
Xn=0.5̂ n
0
0.2
0.4
0.6
0 100 200 300n
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Two important theorems
0. x1
lim then
N,n0, xand xlim(1)If
nn
nnn
. x1
lim then
N,n0, xand 0xlim(2)If
nn
nnn
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Section 2 Infinity
Which of the following sequences are divergent? How many categories are there?
a) 0, 1, 0, 2, 0, 3, 0, 4, …
b) 2/1, 4/2, 8/3, 16/4, …
c) 1, -1, 2, -2, 3, -3, 4, -4,…
d) xn= n2 + 1
OscillatingOscillating
Tends to infinityTends to infinity
Tends to infinityTends to infinity
Tends to infinityTends to infinity
Discussion : p.298 Ex.7.2, 4-6Discussion : p.298 Ex.7.2, 4-6
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Section 3 Bounded and unbounded sequences
Xn=(-1)̂ n/n
-2
0
2
0 100 200 300
n
We say that Xn is bounded by 1.5 since |xn|< 1.5 for any natural no. n.
We say that Xn is bounded by 1.5 since |xn|< 1.5 for any natural no. n.
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Is bounded?nsinnxn
Xn=nsin n
-400
-200
0
200
400
0 100 200 300n
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Bounded above and below
is bounded below by 0.
is bounded above by 9.
is both bounded above and below. i.e. it is bounded.
nxn
n9xn
nnsin
xn
xn> 0 for all n.xn> 0 for all n.
xn< 9 for all n.xn< 9 for all n.
|xn|<1 for all n.|xn|<1 for all n.
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An important theorem
Can a convergent sequence be unbounded?
If it approaches to L as n tends to infinity, then it can’t go too far from L.
Therefore every convergent sequence must be bounded.
Discussion : Ex.7.3 Q.2
Discussion : Ex.7.3 Q.2
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Section 4 Properties of a sequence
Theorem 4.1
N.nany for bathen
,Nnany for yxsuch that
Ninteger positive a exists there
and bylim , axlim If
nn
nn
nn
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The uniqueness of limit
The limit of a convergent sequence is unique.
Reason: The sequence can’t have two ‘continuous’ tails.
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Sandwich Theorem
0
0.1
0.2
0.3
0.4
0.5
0 100 200 300
1數列2數列3數列
n1
z
n2n1
y
n1
x
n
2n
2n
Can you state the theorem?Can you state the theorem?(Principle of Squeezing, or Squeezing Theorem)
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Statement of Sandwich Theorem
.aylim then ,azlimxlim and
N,n allfor zyxsuch that
Ninteger positive a is thereIf
nnnnnn
nnn
Example 4.1 Example 4.1
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Example1: ?convergent 54 Is n
1 nn
4.98
5.02
5.06
5.1
5.14
5.18
1 24 47 70 93 116139162185208231254
Zn=5.21/nZn=5.21/n
yn=(4n+5n)1/nyn=(4n+5n)1/n
Xn =5Xn =5
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Prove that 5)54(lim n
1nn
n
554lim
Theorem,Sandwich theby
5152lim52.5lim and 55lim
2.5)5.2(54)5(5
5.255545
n
1nn
n
n
1
n
n
1
nn
n
1
n
1nn
1nnn
1n
nnnnnn
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Prove thatis convergent.
3333
2...
321
n
n
n
n
n
n
n
n
.convergent is n
n2...
n
3n
n
2n
n
1n
0
0...000n
n2lim...
n
3nlim
n
2nlim
n
1nlim
n
n2...
n
3n
n
2n
n
1nlim
3333
3n3n3n3n
3333n
This is a very common mistake since limits can’t be evaluated by splitting into infinite many pieces though each of them is convergent!
This is a very common mistake since limits can’t be evaluated by splitting into infinite many pieces though each of them is convergent!
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Prove thatis convergent.
0)n2(
1...
)2n(
1
)1n(
1
n
1lim
Theorem,Sandwich theby
0n
1nlim and 0
n4
1nlim
)n2(
1nlim
n
1)1n(
)n2(
1...
)2n(
1
)1n(
1
n
1
)n2(
1)1n(
2222n
2n2n2n
222222
2222 )n2(
1...
)2n(
1
)1n(
1
n
1
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.or -, , represents * where,
y lim xlim)y(x lim and convergent are
yx andyx,yx, yx sequences then the
,convergentboth are y and x sequences two theIf 5.1 Theorem
nn
nn
nnn
nnnnnnnn
nn
For {xn/yn} , yn are non-zero
and lim yn are non-zero too.
For {xn/yn} , yn are non-zero
and lim yn are non-zero too.
Section 5 Operations of Limits of
Sequences
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Three important theorems
.0yxlim then ,0ylim and
bounded is }{x sequence theIf 5.3 Theorem
nnnnn
n
?n
)1(2lim 2.g.e
?n
nsinlim 1.g.e
n
n
n
00
00
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Proof:
.0)y(xlim thusand
0|yx|lim Theorem,Sandwich by the
,0|y|limM|y|Mlim and 0|y|lim
|y|m|y||x||y|
M.|x|0 i.e.
.M|x|such that 0M exists there
bounded. is x
nnn
nnn
nn
nn
nn
nnnn
n
n
n
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Theorem 5.8
Theorem 5.8
.|L||x|lim and convergent also is
|}x{| sequence then the,Lxlim If
nn
nnn
Is the converse correct?Is the converse correct?
Counter-example: {(-1)n}
Counter-example: {(-1)n}
Can the converse be true for some value(s) of L?
Can the converse be true for some value(s) of L?
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Theorem 5.9
0|x|lim ifonly and if 0xlim nnnn
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Section 8 Monotonic Sequences
Theorem 8.1
If a sequence is monotonic increasing increasing(decreasing) and is bounded above(below), then it is convergent i.e. it has a limit.
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Example 8.1
Show that the sequence
is convergent and find its limit. !n
3x
n
n
2nfor decreasing monotonic isit
2.nfor x x
2.nfor 11n
3
3
!n
)!1n(
3
x
x
:oofPr
n1n
n
1n
n
1n
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Proof of example 8.1
say. L, to,convergent isit thusand 0xn
0
L0L
xlim1n
3limxlim
thatimplies x1n
3xthen
nnn
1nn
n1n
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A problem for discussion
limit. itsdown writeHence
above, bounded is }{x (c)
,increasing monotonic is }{x (b)
)n
1k1)...(
n
21)(
n
11(
!k
1 x(a)
: thatShow
... 3, 2, 1,nfor n
11 xLet
n
n
n
0kn
n
n
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n
0k
n
0k
k
n
0k
n
0kk
n
0kk
nk
kn
0k
nk
n
)n
1k1)...(
n
21)(
n
11(
!k
1
n
1kn...
n
2n
n
1n
n
n
!k
1
n
)1kn)...(2n)(1n(n
!k
1
n!k
)1kn)...(2n)(1n(n
n
C
n
1C
n
11
:(a)part of oofPr
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.increasing is it
x
)1n
1k1)...(
1n
21)(
1n
11(
!k
1
)1n
1k1)...(
1n
21)(
1n
11(
!k
1
)n
1k1)...(
n
21)(
n
11(
!k
1
n
11x
:(b)part of oofPr
1n
1n
0k
n
0k
n
0k
n
n
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exlim
3
2
1-1
2
1
2 2
111
!k
111
!k
1
)n
1k1)...(
n
21)(
n
11(
!k
1
n
11x
:(c)part of oofPr
nn
n
1kk
n
2k
n
0k
n
0k
n
n
Example 8.3
Discussion on Ex.7.5 Q.5Discussion on Ex.7.5 Q.5