ch12 (PDF)

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