Sequences (Sec.11.2)

A sequence is an infinite list of numbers

a1, a2, a3, a4, …, ordered by the natural numbers (n) 1, 2, 3,…

Examples:

If {an} is a sequence, call an the nth term. A sequence is also written

For above examples

.202020 :sequence )2(

.,,,,1 :sequence )1( 41

31

21

,, , , , ,

1}{ nna

nnn a

na )1(1 (2). ;

1 ).1(

Limits

Let be a sequence. Then

means

an approaches as n becomes larger and

larger.

Convention: If sequence has limit, we say that an converges to and that the sequence is convergent – otherwise the sequence is divergent.

1}{ nna

R)( lim

nn

a

1}{ nna

Examples

exist.not does )1(limThen

1,-1, 1,- 1, 1,- i.e. ,})1{( sequenceConsider )2(

.301

03

1

3lim

1

53lim)1(

1

11

5

2

2

2

2

n

n

nn

nn

n

nn nn

n

- todiverges say we, as Then

12

1 sequenceConsider ).3(

2

1

nn

n

nn

ana

n

n

na

Theorem: If lim x f(x) = L and an = f(n), n

then lim n an = L.

Example: Evaluate

Sol:

n

nn

)ln(lim

.0)ln(

lim ,by theorem Thus

.01

lim)ln(

lim

gives rule sHopital'L' Use

.)ln(

)( so )()ln(

1

n

nx

x

x

xxfnf

n

n

n

x

xx

Limit LawsSame limit theorems hold as for functions. Thus

if an and bn are convergent sequences and

lim n an = L , lim n bn = M then

(i) lim n (an + bn ) = L + M,

(ii) lim n an bn = LM,

(iii) lim n an / bn = L/ M provided M 0.

Also the squeeze law applies:If an, bn cn for all n and

lim n an = L = lim n cn then lim n bn = L

Example

Find

USEFUL SEQUENCES TO REMEMBER

n

nn

sinlim

.0!

lim ,constant For ).3(

.1lim ,0constant For ).2(

.1|| and 1for divergent is }{ Sequence

1 if ,1

1|| if ,0lim ,constant For ).1(

/1

1n

n

cc

cc

ccc

c

ccc

n

n

n

n

n

n

n

Example

Evaluate

2n 2n + 1 2n + 2 2. 2n, for all n 2 (2n + 1)1/n 21/n.2

By (2) above lim n 21/n.2 = 2 lim n 21/n = 2

By squeeze law lim n (2n + 1)1/n = 2

ccc

n

n

baab

n

)( :Note

)12(lim1

Series

The sum of sequence is called a series, written

Call an the nth term. Add the first n terms:

Call sn the nth partial sum.

1}{ nna

.3211

aaaak

k

.3211

n

n

kkn aaaaas

Set

We say the series converges to sum S if sn converges to a real number S.

Otherwise we say the series diverges.

Ssa nn

n

kk

lim

1

So to find

- start adding the terms one at a time, starting with the first term a1.

- if sum approach some number, S, then S is the infinite sum.

- Note:

1kka

.1

1321

nn

nnn

as

aaaaas

Examples

diverges. series the therefore1- and 0between oscillates

,0 ,1 ,0 ,1 are sums Partial

11111)1( ).1(

4321

1

ssssn

n

1. toconverges series The .11

11limlim

)1(

1

.1

11

1

11

4

1

3

1

3

1

2

1

2

11

1

11

)1(

1

1

11

)1(

1 Now

20

1

12

1

6

1

2

1

)1(

1 seriesConsider (2)

1

1 1

1

ns

kk

nnn

kkkks

kkkk

kk

nk

nn

n

k

n

kn

k

.1for to

1for

0,4

1

3

1

2

11

1

:series- )3(

1

p

p

pn

p

pppn

p

diverges

converges

The n th term test(Useful to show a series diverges)

.1 if converges and 1 if diverges

4

1

3

1

2

11

1

seriesbut ,1for 01

lim Exp.

diverges.or converges series if lcannot tel 0lim From

:Note

diverges. series then , 0lim If

1

n

1n

1n

pp

n

pn

aa

aa

pppp

np

p

nnn

nnn

Example

test. th term by thedivergent series

exist)t doesn'limit fact (in 01limlim

1 : th term the,1 )1(1

n

a

an

n

nn

n

nn

n

n

diverges. series

02

1

2

1lim

12limlim

12 :

12 )2(

1

1

nnn

nn

nn

n

na

n

na

n

n

Geometric Series: Consider series

Sequence of partial sums is given by

.0 ,1

12

aaxaxaxan

n

)1(

)1(limlim

.1for )1(

)1(

)1()1(

)(

1n

1

12

1

x

xasax

xx

xas

xasx

axaxss

axaxaxaxxs

axaxas

n

nn

n

n

n

n

nn

nnn

nnn

nn

.1|| if test th termby diverges

series e therefor,1|| if lim)2(

.1|| if converges i.e.

;1|| if 1

)1(

:Hence

1

1n

1

xn

xax

x

xx

aax

n

n

n

diverges. series theabove casesboth In

exist.not does lim so

even, for 0 and odd for then 1 If

exist.not does lim so

111 then 1 If )3(

1

1

n

n

121

n

n

n

nn

n

nn

ax

s

nsnasx

s

naaaaasx

Example

.15

),5 with series Geom. (a 1

5

3

25 )2(

.11

) with series Geom. (a 2

1

2

1

2

1 )1(

32

32

1

21

21

21

1

1

1

xa

xa

n

n

n

n

n

n

Comparison test

If 0 an bn and converges then converges.

Examples

1n nb

1n na

testcomparison by the 1

1 so

converges which 12 with series)-( 11

Now

.1

1

10 so 1 have we

1Set

1

1 where

1

1

converges. 1

1 that Show

1n2

1n1n2

2222

2

21n1n

2

1n2

n

ppnn

nnann

nb

naa

n

n

p

nn

nn