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So, lim un

n→∞

lim

l ( Sn − Sn−1 )

n→∞

= lim Sn − l

n→∞

n→∞

= S−S

= 0.

Sn−1

This gives us the important result that:

KEY POINT 29.2

∞

If the series

∑u

k =1

k

is convergent then lim

k →∞

k

0.

The converse, however, is not true: just because lim uk = 0

k →∞

∞

it is not necessarily the case that

∑u

is convergent.

k

k =1

In the next section

we will use Key point

29.2 to determine

the divergence of

some series.

∞

1

An example of the latter is ∑ . Here lim uk = 0 but, as we will

k →∞

k =1 k

∞

see in the next section

1

∑k

diverges.

k =1

Exercise 29A

1. Evaluate (correct to 3SF) the first four partial sums for the series

∞

∑u

k

where:

k =1

1

k2

1

(c) uk =

k+5

(a) uk =

(b) uk = sin

⎛ kπ ⎞

⎝ 3⎠

(d) uk = e − k

2. Write an expression (in terms of n) for the nth partial sum

of the given series, and hence decide whether the series is

convergent or divergent.

∞

(a)

∞

∑3k + 1

(b)

k =1

k =1

∞

(c)

(e)

∞

1

∑ 3k

k =1

∞

⎛ 1⎞

∑ ⎝ − 2⎠

k

∑2 − 5

(d)

∑1.2

k

k =1

4k

k =1

∞

3. Given the expression for the nth partial sum of the series ∑uk ,

k =1

find (and simplify) uk :

n2 − 1

5

(c) Sn = n − 5

4

2n − 1

(e) Sn =

3 × 2n

(a) Sn

4

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

(b) Sn = 3n − 1

(d) Sn

l n

Option 9: 29 Infinite series

29B Tests for convergence and

divergence

Divergence Test

From Key point 29.2 we note that if lim uk ≠ 0 the series cannot

k →∞

be convergent. We therefore have a possible way of checking

whether a series diverges.

KEY POINT 29.3

Divergence Test

If lim uk ≠ 0 or if the limit does not exist, the series

k →∞

∞

∑u

k

is

k =1

divergent.

Of course this will not always identify a divergent series (lim uk

k →∞

might be 0 and the series could diverge as mentioned above),

but it is often the first thing to check.

Worked example 29.2

∞

Show that the series

∑

k =1

k

3k + 1

diverges.

4k 2 + 3

To show that lim uk ≠ 0 we need

uk =

to manipulate uk into a form

that enables us to find its limit

as k → ∞

=

k →∞

k 2 3k + 1

4k 2 + 3

1 + 3k + k12

4 + k32

1

≠0

4

∞

k 2 3k + 1

Hence ∑

diverges.

2

k = 1 4k + 3

∴ lim uk =

k →∞

We will now present a number of tests for determining

convergence or divergence of series, which can be established

using the principles discussed in Section 29A.

Comparison Test

The first of these tests is similar to the Comparison Test for

improper integrals used in chapter 28 of this option.

We would like to know whether the following series converges

or not.

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

t

exam hin

it is a

In the exam

to check

good idea

nce Test

the Diverge

dering

before consi

ther tests

any of the o

apter.

from this ch

See Section 28B (Key

point 28.3) for the

Comparison Test for

improper integrals.

Option 9: 29 Infinite series

5

1 1

1

1

1 …

+ + + +

+

4 10 28 82 244

While we cannot say much about this series immediately, we do

recognise that it is similar to the geometric series:

1+

1 1 1

1

1 …

1+ + + + +

+

3 9 27 81 243

1

3

(S∞ = ). We also

3

2

note that each term of our series is less than or equal to the

corresponding term of the geometric series:

which we know converges because r =

1≤1

1 1

≤

4 3

1 1

≤

10 9

1

1

≤

28 27

Hence, each partial sum (after the first one) is less than the

corresponding partial sum of the geometric series and our series

3

must therefore converge (to a limit less than ).

2

This idea is generalised by the following result:

KEY POINT 29.4

Comparison Test

∞

Given two series of positive terms

that ak

•

∞

bk for all k ∈Z +, then if:

∑bk is convergent to a limit S,

k =1

to a limit T whereT

•

∑ak and

k =1

∞

∑b

k

such

k =1

∞

∑a

k

is also convergent

k =1

S

∞

∞

k =1

k =1

∑ak is divergent, so is ∑bk.

Just as with the Comparison Test for improper integrals in

chapter 28 of this option, this result is intuitive. Now, instead of

comparing areas under functions, we are comparing a sequence

of partial sums.

6

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

Option 9: 29 Infinite series

∞

If

∑b

k

converges to S and ak

bk for all k ∈Z + we have the

k =1

following for the partial sums of each series:

n

If Sn = k=1 bk converges to S then

n

Tn = k=1 ak converges to T < S

S

T

S3

S2

S1

T3

T2

T1

n

∞

and if

∑a

k

diverges, we have:

convergent

k =1

If Tn =

n

then Sk =

∞

∑b

k

k =1

a bk for

such that k

is often

all k, first it

dering

worth consi

ries.

geometric se

k=1 ak diverges

n

k=1 bk diverges

S3

S2

S1

t

exam hin

ra

In looking fo

We will see shortly

that the p-series is

also commonly used

in the Comparison

Test.

T3

T2

T1

n

Worked example 29.3

∞

Establish whether or not the series

∑2

k =1

k

1

converges.

+3

We will try first of all to show that

the series converges by finding

a series we know converges and

which satisfies the conditions of the

Comparison Test

If we do not have any success with

this approach we will then consider

trying to establish divergence by

finding a known divergent series

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

Option 9: 29 Infinite series

7

continued . . .

2k

∞

1

∑ 2k which

k =1

we know converges (it is a geometric

The series is similar to

series with r =

0 for all k

and

2k 3 > 2k for all k

1

1

∴ k

< k for all k ∈Z +

2 +3 2

1

), so let’s start by

2

considering this

∞

and since

∞

1

∑2

k =1

converges, so does

k

∑2

k =1

k

1

by

+3

the Comparison test.

Limit Comparison Test

Sometimes we meet a series which is very similar to a known

convergent (or divergent) series, but which does not fulfil the

conditions of the Comparison Test.

∞

For example the series

∑2

k =1

k

1

1 1 1 1

1

= 1+ + + + + +…

−1

3 7 15 31 63

∞

has terms similar to our known convergent series

1

∑2

k =1

k

but

each of its terms are larger than those of the known convergent

series and not smaller. Although the standard Comparison Test

is therefore no use, the following extension of the test is.

KEY POINT 29.5

Limit Comparison Test

Given two series of positive terms

∞

∞

∑ak and ∑bk , where

=1

k =1

a

lim k = l > 0, then if one series converges so does the

k →∞ b

k

other and if one series diverges so does the other.

exam hint

Choose as bk the general term of the series to which you

had hoped to apply the Comparison Test.

8

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

Option 9: 29 Infinite series

Worked example 29.4

∞

∑2

Show that the series

k =1

k

1

is convergent.

−1

We know that this series isn’t suitable

∞

1

for the Comparison Test with ∑ k

k =1 2

but we hope it satisfies the conditions

Let

ak =

Then

ak

1

2k

= k

×

bk 2 − 1 1

2k

= k

2 −1

1

=

k

⎛ 1⎞

1−

⎝ 2⎠

a

lim k = 1

k →∞ b

k

of the Limit Comparison Test

and so

∞

Hence

1

1

and bk = k

2 −1

2

k

∑2

k =1

k

1

converges by the Limit Comparison

−1

Test.

Integral Test

In the previous chapter we found upper and lower bounds

(sums) for the improper integral

∫

∞

a

f x ) dx in the separate

cases where f(x) was increasing and decreasing. Here we will

use this result for the case where f(x) is a decreasing positive

function and a = 1, but now put bounds on the infinite sum and

not the improper integral.

So, from chapter 28 we have:

∞

∑

∞

∞

k a +1

f ( k ) < ∫ f ( x ) dxx

∑ f (k)

a

Look back at the

graphical method

used in Section 28C

of this option to

establish the upper

and lower sums, and

at Worked example

28.10 where we

placed bounds on

∞

1

the sum ∑ 2 .

k =3 k

k a

For the lower sum, letting a = 1 and noting that f ( ) is positive:

∞

∑ f (k ) < ∫

1

k =1+1

∞

⇒∑

k =1

( ( )) − f

∞

⇒ ∑f (

k =1

∞

f ( x ) dx

∞

<∫ f(

1

∞

) < f ( ) + ∫1 f ( ) dx

∞

Combining this with the upper sum ∫1

we now have:

∫

∞

1

f ( x ) dx

) dx

∞

∑

k =1

∞

f ( x ) dx < ∑ f k )

k =1

∞

f ( ) + ∫ f ( x ) dx

1

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

Option 9: 29 Infinite series

9

Then, if the improper integral

t

exam hin

, you

In the exam

uired to

may be req

od to

use this meth n

ds o

place boun

m that

an infinite su

valuated

cannot be e

directly.

∫

∞

1

f ( x ) dx converges to a limit

L, the sum is bounded between the two constants L and f(1) + L:

∞

∑ f (kk

L

k =1

f (1) + L

Also, as f(x) is positive, the sum increases as more and more

terms are added. It seems clear that an increasing series that is

bounded above (here by the constant f(1) + L) must converge to

a limit that is less than or equal to this bound:

y

∞

Area under blue curve: f(1) + 1 f(x) dx

∞

Area under red curve: 1 f(x) dx

∞

Area in grey boxes:

k=1 f (k)

y = f(x)

y = f(x + 1)

x

1

Do we need to have a

formal proof of a result

like the Monotone

Convergence Theorem

when it seems so clear that it is

true? Does a proof add to our

understanding?

2

3

4

5

6

7

8

9

10

The general result that any increasing series bounded above or

decreasing series bounded below has a limit is known as the

Monotone Convergence Theorem. Although this result is not

part of the syllabus it helps us understand the following test for

convergence of series:

KEY POINT 29.6

Integral Test

Given a positive decreasing function f ( x ) , x ≥ 1 ,

∞

if ∫ f x ) dx is:

1

• convergent then

∞

∑f

k) is convergent

k =1

• divergent then

∞

∑f

k) is divergent.

k =1

The following example of the use of the Integral Test resembles

closely the method of Worked example 28.5 when we examined

the convergence of the improper integral

10

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

∫

∞

1

x p dx .

Option 9: 29 Infinite series

Worked example 29.5

∞

Determine for which values of p the series

∑k

p

converges.

k =1

We note that when p ≥ 0 it looks

like the series must be divergent so

use the Divergence Test to confirm

this

For p > 0

lim uk

k →∞

lim k p = ∞ ≠ 0

k →∞

For p = 0

lim uk = lim 1 = 1 ≠ 0

k →∞

k →∞

In both cases the series diverges by the

Divergence Test.

Now we turn our attention to

p < 0. We see that k p is positive

and decreasing for k >1 so we can

use the Integral Test.

We realise that there will be two

different cases: if p = −1 we need to

integrate with ln x

∫

∞

1

x p dx

lim

b→

b→∞

∫

b

1

x p dx

⎧ ⎡ x p +1 ⎤b

⎪⎪ lim ⎢

⎥

= ⎨b →∞ ⎣ p + 1 ⎦ 1

b

⎪

[

]

1

⎪⎩ blim

→∞

if p ≠ −1

if p = −1

⎧

⎛ b p + 1 − 1⎞

⎪blim

⎜

⎟ if p ≠ 1

= ⎨ →∞ ⎝ p + 1 ⎠

⎪ lim lnb

if p ≠ −1

⎩b →∞

⎧

∞

⎪

1

⎪

= ⎨−

p

+

1

⎪

⎪⎩ ∞

At this stage we can see that the

case p ≠ −1 needs to be split into

two separate cases. When p +1> 0,

b p+1 → ∞ but when p +1< 0 , b p+1 → 0

if −1 < p < 0

if p < −1

if p = −1

∞

∴ by the Integral test

∑k

p

k =1

is:

• divergent for −1 ≤ p < 0

• convergent for p < −1

∞

So,

∑k

p

k =1

is convergent only when p < −1 or

equivalently,

∞

1

∑k

k =1

p

is convergent only when p > 1 .

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

Option 9: 29 Infinite series

11

∞

t

exam hin

for

The result

ce of the

convergen

l

very usefu

p-series is

ften be

and will o

njunction

used in co ts such

tes

with other

parison

m

o

as the C

blish

Test to esta

rgence

the conve

ise) of

(or otherw

of other

a number

series.

The series

1

∑k

k =1

will be referred to as the p-series. When p = 1

p

this is the harmonic series.

Alternating series

∞

We have just seen that the harmonic series

1

∑k

diverges,

k =1

but what about the same series with alternating positive and

negative terms?

∞

∑ ( −1)

k +1

k =1

1

1 1 1

=1− + − +…

k

2 3 4

It certainly has a greater chance of being convergent due to

the negative terms but we cannot use any of the tests we have

seen thus far as these all require positive terms. Any series with

alternately positive and negative signs is known as an alternating

series. The following result allows us to analyse the convergence

of this kind of series.

KEY POINT 29.7

Alternating Series Test

∞

If for an alternating series

∑u :

k

k =1

• uk

< uk for sufficiently large k

• lim uk = 0

k →∞

then the series is convergent.

The fact that the magnitude of each uk is decreasing to 0 (while

the signs are alternately positive and negative) means that the

sequence of partial sums Sk alternates either side of its eventual

limit:

Sk

S1

S3

S5

S

S4

S6

S2

k

12

Cambridge Mathematics for the IB Diploma Higher Level © Cambridge University Press, 2012

Option 9: 29 Infinite series

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