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?

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

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

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

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

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

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

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.

Is bounded?nsinnxn

Xn=nsin n

-400

-200

0

200

400

0 100 200 300n

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.

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

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

The uniqueness of limit

The limit of a convergent sequence is unique.

Reason: The sequence can’t have two ‘continuous’ tails.

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)

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

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

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

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!

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

.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

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

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

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?

Theorem 5.9

0|x|lim ifonly and if 0xlim nnnn

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.

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

Proof of example 8.1

say. L, to,convergent isit thusand 0xn

0

L0L

xlim1n

3limxlim

thatimplies x1n

3xthen

nnn

1nn

n1n

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

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

.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

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