Sequences

Sequences A sequence is a list of numbers in a definite order:

a1, a2,,an, or {an} or

Suppose n1<n2<<nk<, then we call

a subsequence of {an}, denoted by Some sequences can be defined by giving a formula for the

general n-th term. A sequence can be thought as a function f(n)=an, with

domain N={1,2,}

1}{ nna

,,,,21 knnn aaa

}.{kna

Example Ex. Find a formula for the general term of the sequence

Sol.

Ex. The Fibonacci sequence is defined recursively by

The first few terms are

3 4 5 6 7, , , , ,

5 25 125 625 3125

na

2( 1)

5n

n n

na

{ }nf

1 2 2 11, 1, ( 0)n n nf f f f f n

1,1,2,3,5,8,13,21,

Example Ex. Find a formula for the general term of the Fibonacc

i sequence. Sol. Assume that On comparing

coefficients, we have

Solving the quadratic equation, we get

Since is a geometric sequence, we find

1, 1.

nf

1n nf f

1 5,

2

2 1 1( ).n n n nf f f f

1 11 2 1( ) (1 ) , 1, 2,k k

k kf f f f k

2 11 1 3 2 2 1

1 2 2 1

[ ] [ ] [ ] [ ]

(1 )( )(1 )[ ]

n nn n n n

n nn n n n

f f f f f f f f

Example This gives

or,

Therefore,

Question: find a formula for the general term of

1 1

(1 )( )n nn

nf f

1 1

1

(1 )( ) ( ) ( )n n n n n nn

nf

1 1

1

( ) ( )n n n n

nf

1 2 2 1, , ( 0)n n nf a f b f f f n

Limit of a sequence Definition A sequence {an} has the limit L and we write

if 8>0, 9 N >0 such that |an-L|< whenever n >N. If the limit of a sequence exists, we say the sequence converges, otherwise we say it diverges.

A sequence is a special function, so all the properties for function limits are also true for sequence limits.

lim or asn nn

a L a L n

Function limit and sequence limit The following theorem is obvious:

Theorem If and an=f(n), thenLxfx

)(lim .lim Lann

Properties The Squeeze Theorem holds also true:

If an· bn· cn for n¸ n0 and then

Ex. Find the limit of the sequence an=n!/nn.

Sol. Since 0<an·1/n, by the Squeeze Theorem,

,limlim Lca nn

nn

.lim Lbnn

.0lim n

na

Example Ex. Discuss the convergence an=rn, r2(-1,+1).

Sol. (i)When |r|>1, |r|n increasingly goes to infinity.

(ii)When |r|<1, |r|n decreasingly goes to zero.

(iii)When r=1, an1, so the limit is 1.

(iv)When r=-1, an oscillates between 1 and –1 infinitely

often, so it diverges.

To summarize, {rn} is convergent if –1<r·1 and divergent

for all other values of r. 1

11

1

0lim

r

r

if

ifr n

n

Example Ex. Find the limit

Sol.

.1

2

1

1

1lim

222

nnnnn

nn

n

nnnnnnnn

222222

111

2

1

1

1

11

1

1

11

2

1

1

1222222

n

n

nnnnnn

11

1

1lim

1lim,1

11

1limlim

2

22

nn

n

nnn

nnnnn

Monotonicity and boundedness Definition A sequence is called increasing if

for all It is called decreasing if for all

It is called monotonic if it is either increasing or decreasing.

Definition A sequence is bounded above if there is a

number M such that for all It is bounded

below if there is a number m such that for all

{ }na 1n na a 1.n 1n na a 1.n

{ }na1.nna M

na m 1.n

Monotonic sequence theoremTheorem Every bounded, monotonic sequence is convergent.

In particular, an increasing sequence that is bounded above

is convergent; a decreasing sequence that is bounded below

is convergent.

Ex prove that {an} is convergent.

Sol. It is easy to see and

That is, {an} is increasing and

bounded above, and thus converges.

1 1 1,

1 2 2 4 2n na

n

nnnn an

aa

11 2)1(

1

.12

11

2

1

4

1

2

1

nnna

Example

Ex. Suppose {an} is defined by the recursive relationship

Find the limit of {an}.

Sol. so an>1 for all n.

so {an} is decreasing. Let

Using the recursive relationship, we have

Solving the equation for L, we get L=1 or L=0. Since an>1,

we eliminate L=0. Therefore

).1(1

2,1 11

n

a

aaa

n

nn

,12

1

21,1 11

kk

k

k

kkk aa

a

a

aaaa

,11

2

1

21 n

n

n

nn a

a

a

aa

.1lim n

na

.1

2

1

2limlim 1

L

L

a

aaL

n

n

nn

n

.lim Lann

Example {an} is defined by Show that {an} is

convergent and find its limit.

Sol. We can easily prove an>1 for all n by induction. Then we

have an+1<an. So {an} is decreasing and bounded below.

Let Taking limits on both sides of the recurrence

equality, we obtain

Solving the equation for L, we get Since an>1, we

discard L=-1 to get L=1.

).1(3

13,1 11

na

aaa

n

nn

.lim Lann

.3

13

3

13limlim 1

L

L

a

aaL

n

n

nn

n

.1L

Series If is a sequence, then is called an

infinite series (or just a series) and is denoted by

We call the n-th partial sum of the series.

The partial sums form a new sequence If it converges,

then the series is called convergent and the number

is called the sum of the series. We denote

{ }na

1

orn nn

a a

1 2 na a a

1 21

n

n n ii

s a a a a

{ }.ns

nalim nn

s s

1

nn

s a

Example Ex. Discuss the convergence of the geometric series

Sol. If r=1, then so the series

diverges.

If then so the series converges

when and diverges otherwise. To summarize, the

geometric series is convergent if and only if and the

sum is

2 1 1

1

0n n

n

a ar ar ar ar a

1

1

(1 ),

1

nni

ni

a rs ar

r

1

,n

ni

s a na

1,r 1 1r

| | 1r .

1

a

r

Example Ex. Is the series convergent or divergent?

Sol. Since

it is a series with thus is divergent. Ex. Find the sum of the series Sol.

2 1

1

2 3n n

n

4 / 3 1r

2 1 1 1

1 1 1

42 3 4 3 4( )

3n n n n n

n n n

1 1 1.

1 3 2 4 ( 2)n n

1

1 1 1 1 1 1 1( ) (1 )

2 2 2 2 1 2

n

nk

sk k n n

3lim

4nn

s s

Example Ex. Show that the harmonic series

is divergent? Sol.

1

1 1 1 11

2 3 4n n

1 2

1 1 1 2 11, 1 ,

2 3 4 4 2s s

1 1 1 1 4 1 1 1 8 1,

5 6 7 8 8 2 9 16 16 2

21

2n

ns

Necessary condition for convergence Theorem. If the series is convergent, then

Proof.

The test for divergence If does not exist or if

then the series is divergent.

Remark. is only a necessary condition, but not sufficient.

1n

n

a

1 lim 0n n n nna s s a s s

lim 0.nn

a

lim 0,nn

a

lim nn

a

nalim 0nn

a

Properties Theorem If and are convergent, then

Question: converges, diverges, then

both and diverges, then

Remark A finite number of terms doesn’t affect the convergence or divergence of a series.

na nb

1 1 1 1 1

(i) (ii) ( )n n n n n nn n n n n

ca c a a b a b

na nb ( )?n na bna nb ( )?n na b

Convergence test for positive series Theorem Suppose is a series with positive terms,

then it converges if and only if the sequence of the partial

sum is bounded, i.e., there exists a constant M such that

for all n. Proof. Note that for a positive series, the sequence of the p

artial sum is always increasing. So the theorem follows immediately from the monotonic sequence theorem.

na

ns

ns M

The integral test Theorem Suppose f is a continuous, positive, decreasing

function on and let Then the series

is convergent if and only if the improper integral

is convergent.

Ex. For what values of p is the series convergent?

Sol. The p-series is convergent if p>1 and divergent

if

na1

( )f x dx

( ).na f n[1, )

1

1p

n n

1

1p

n n

1.p

Example Ex. Determine whether the series converges

or diverges. Sol. The improper integral

So the series diverges.

2

11

ln ln

2

x xdx

x

1

ln

n

n

n

1

ln

n

n

n

Question Ex. Determine whether the series converges or

diverges. Sol.

2

1

lnqn n n

diverge for 1q

converge for 1q

Homework 23 Section 11.1: 22, 35, 36, 51, 61, 62

Section 11.2: 23, 44

Section 11.3: 7, 20, 24