11
INFINITE SEQUENCES AND SERIES
INFINITE SEQUENCES AND SERIES
11.2
Series
In this section, we will learn about:
Various types of series.
Series 1
SERIES
If we try to add the terms of an infinite

n n 1
sequence {a }
of the form
we get an expression
a1 + a2 + a3 + ··· + an + ∙·∙
INFINITE SERIES
This is called an infinite series
(or just a series).
 It is denoted, for short, by the symbol

a
n 1
n
or
a
n
INFINITE SERIES
However, does it make sense
to talk about the sum of infinitely
many terms?
INFINITE SERIES
It would be impossible to find a finite sum
for the series
1 + 2 + 3 + 4 + 5 + ∙∙∙ + n + ···
 If we start adding the terms, we get
the cumulative sums 1, 3, 6, 10, 15, 21, . . .
 After the nth term, we get n(n + 1)/2,
which becomes very large as n increases.
INFINITE SERIES
However, if we start to add the terms
of the series
1 1 1 1 1
1
1
        n  
2 4 8 16 32 64
2
we get:
1
2
, , , , , ,   ,1 1/ 2 , 
3
4
7 15
8 16
31
32
63
64
n
INFINITE SERIES
The table shows that, as we add more
and more terms, these partial sums become
closer and closer to 1.
 In fact, by adding sufficiently
many terms of the series,
we can make the partial sums
as close as we like to 1.
INFINITE SERIES
So, it seems reasonable to say that
the sum of this infinite series is 1 and
to write:

1 1 1 1 1
1
       n    1

n
2 4 8 16
2
n 1 2
INFINITE SERIES
We use a similar idea to
determine whether or not a general
series (Series 1) has a sum.
INFINITE SERIES
We consider the partial sums
s1 = a1
s2 = a1 + a2
s3 = a1 + a2 + a3
s3 = a1 + a2 + a3 + a4
n
 In general,
sn  a1  a2  a3     an   ai
i 1
INFINITE SERIES
These partial sums form
a new sequence {sn}, which
may or may not have a limit.
SUM OF INFINITE SERIES
If lim sn  s exists (as a finite number),
n 
then, as in the preceding example,
we call it the sum of the infinite series Σ an.
SUM OF INFINITE SERIES
Definition 2
Given a series

a
n 1
n
 a1  a2  a3  
let sn denote its nth partial sum:
n
sn   ai  a1  a2    an
i 1
SUM OF INFINITE SERIES
Definition 2
If the sequence {sn} is convergent and
lim sn  s exists as a real number, then the
n 
series Σ an is called convergent and we write:
a1  a2    an    s or

a
n 1
 The number s is called the sum of the series.
 Otherwise, the series is called divergent.
n
s
SUM OF INFINITE SERIES
Thus, the sum of a series is the limit
of the sequence of partial sums.

 So, when we write
a
n 1
n
 s,
we mean that, by adding sufficiently many terms
of the series, we can get as close as we like to
the number s.
SUM OF INFINITE SERIES
Notice that:

a
n 1
n
n
 lim ai
n 
i 1
SUM OF INFINITE SERIES VS. IMPROPER INTEGRALS
Compare with the improper integral


1
t
f ( x) dx  lim f ( x) dx
t  1
 To find this integral, we integrate from 1 to t
and then let t → ∞.
 For a series, we sum from 1 to n and then let n → ∞.
Example 1
GEOMETRIC SERIES
An important example of an infinite series
is the geometric series
a  ar  ar  ar    ar
2
3
n 1

 
  ar
n 1
n 1
a0
GEOMETRIC SERIES
Example 1
Each term is obtained from the preceding
one by multiplying it by the common ratio r.
 We have already considered the special case
where a = ½ and r = ½ earlier in the section.
GEOMETRIC SERIES
Example 1
If r = 1, then
sn = a + a + ∙∙∙ + a = na → ±∞
sn doesn’t exist, the geometric series
 Since lim
n 
diverges in this case.
GEOMETRIC SERIES
Example 1
If r ≠ 1, we have
sn = a + ar + ar2 + ∙∙∙ + ar n–1
and
rsn = ar + ar2 + ∙∙∙ +ar n–1 + ar n
E. g. 1—Equation 3
GEOMETRIC SERIES
Subtracting these equations,
we get:
sn – rsn = a – ar n
a(1  r )
sn 
1 r
n
Example 1
GEOMETRIC SERIES
If –1 < r < 1, we know from Result 9 in
Section 11.1 that r n → 0 as n → ∞.
So,
a(1  r )
a
a
a
n
lim sn  lim


lim r 
n 
n 
1 r
1  r 1  r n
1 r
n
 Thus, when |r | < 1, the series is convergent
and its sum is a/(1 – r).
Example 1
GEOMETRIC SERIES
If r ≤ –1 or r > 1, the sequence {r n}
is divergent by Result 9 in Section 11.1
So, by Equation 3, lim sn does not exist.
n 
 Hence, the series diverges in those cases.
GEOMETRIC SERIES
The figure provides
a geometric
demonstration
of the result in
Example 1.
GEOMETRIC SERIES
If s is the sum of the
series, then, by similar
triangles,
s
a

a a  ar
So,
a
s
1 r
GEOMETRIC SERIES
We summarize the results
of Example 1 as follows.
Result 4
GEOMETRIC SERIES
The geometric series

 ar
n 1
 a  ar  ar  
n 1
is convergent if |r | < 1.
2
Result 4
GEOMETRIC SERIES
The sum of the series is:

 ar
n 1
n 1
a

1 r
r 1
If |r | ≥ 1, the series is divergent.
Example 2
GEOMETRIC SERIES
Find the sum of the geometric series
5 
10
3
20
9

40
27
 
 The first term is a = 5 and the common ratio
is r = –2/3
Example 2
GEOMETRIC SERIES
 Since |r | = 2/3 < 1, the series is convergent
by Result 4 and its sum is:
10 20 40
5
5       
3 9 27
1  ( 23 )

5
5
3
3
GEOMETRIC SERIES
What do we really mean when we say
that the sum of the series in Example 2
is 3?
 Of course, we can’t literally add an infinite number
of terms, one by one.
GEOMETRIC SERIES
However, according to Definition 2,
the total sum is the limit of the sequence
of partial sums.
 So, by taking the sum of sufficiently many terms,
we can get as close as we like to the number 3.
GEOMETRIC SERIES
The table shows the first ten partial sums sn.
The graph shows how the sequence of
partial sums approaches 3.
Example 3
GEOMETRIC SERIES
Is the series

2
2 n 1 n
3
n 1
convergent or divergent?
Example 3
GEOMETRIC SERIES
Let’s rewrite the nth term of the series
in the form ar n-1:

2
n 1
2 n 1 n
3

  (2 ) 3
n 1
2 n  ( n 1)


4
4
  n1   4  
n 1 3
n 1  3 
n
 We recognize this series as a geometric series
with a = 4 and r = 4/3.
 Since r > 1, the series diverges by Result 4.
n 1
GEOMETRIC SERIES
Example 4
Write the number 2.317  2.3171717...
as a ratio of integers.
17 17 17
 2.3171717…  2.3  3  5  7  
10 10 10
 After the first term, we have a geometric series
with a = 17/103 and r = 1/102.
GEOMETRIC SERIES
Example 4
 Therefore,
17
17
3
2.317  2.3  10  2.3  1000
1
99
1 2
10
100
23 17


10 990
1147

495
GEOMETRIC SERIES
Example 5
Find the sum of the series
where |x| < 1.

x
n
n0
 Notice that this series starts with n = 0.
 So, the first term is x0 = 1.
 With series, we adopt the convention that x0 = 1
even when x = 0.
Example 5
GEOMETRIC SERIES
Thus,

x
n
 1  x  x  x  x  
2
n 0
 This is a geometric series
with a = 1 and r = x.
3
4
E. g. 5—Equation 5
GEOMETRIC SERIES
Since |r | = |x| < 1, it converges, and
Result 4 gives:

1
x 

1 x
n 0
n
Example 6
SERIES
Show that the series

1

n 1 n(n  1)
is convergent, and find its sum.
Example 6
SERIES
This is not a geometric series.
 So, we go back to the definition of a convergent
series and compute the partial sums:
n
1
sn  
i 1 i (i  1)
1
1
1
1



   
1 2 2  3 3  4
n(n  1)
Example 6
SERIES
We can simplify this expression if we use
the partial fraction decomposition.
1
1 1
 
i (i  1) i i  1
 See Section 7.4
SERIES
Example 6
Thus, we have:
n
1
sn  
i 1 i (i  1)
1 1 
  

i 1 
i 1  i
1 
 1 1 1 1 1
1
 1               

 2  2 3  3 4
 n n 1 
1
 1
n 1
n
Example 6
SERIES
Thus,
1 

lim sn  lim 1 
 1 0  1

n 
n 
 n 1 
 Hence, the given series is convergent
and

1
1

n 1 n( n  1)
SERIES
The figure illustrates Example 6 by
showing the graphs of the sequence of terms
an =1/[n(n + 1)] and the sequence {sn}
of partial sums.
 Notice that an → 0
and sn → 1.
HARMONIC SERIES
Example 7
Show that the harmonic series

1
1 1 1
 1     

2 3 4
n 1 n
is divergent.
Example 7
HARMONIC SERIES
For this particular series it’s convenient to
consider the partial sums s2, s4, s8, s16, s32, …
and show that they become large.
s1  1
s2  1 
1
2
s4  1  
1
2
 1
2
2

1
3

1
4


1 
1
2

1
4

1
4

Example 7
HARMONIC SERIES
Similarly,


s8  1  
1
3

1
4
 1
1
4

1
4
1
2
1
2
 1  
1
2
 1
3
2
1
2
1
2


1

5
  
1

8
1
8
  
1
8
1
6
1
8
1
7
1
8

Example 7
HARMONIC SERIES
Similarly,


s16  1  
1
2
 1
1
2
1
3

1
4
1
4

1
4


 1   
1
2
 1
4
2
1
2
1
2
1
2


1

5
  
1

8
1
8
  
1
8

1
9
  
1
16
1
16
  

1
16

HARMONIC SERIES
Example 7
Similarly, s32 > 1 + 5/2, s64 > 1 + 6/2,
and, in general,
n
s2 n  1 
2
 This shows that s2n → ∞ as n → ∞,
and so {sn} is divergent.
 Therefore, the harmonic series diverges.
HARMONIC SERIES
The method used in Example 7 for
showing that the harmonic series diverges
is due to the French scholar Nicole Oresme
(1323–1382).
SERIES
If the series

is convergent,
a
 n
n 1
then lim an  0
n 
Theorem 6
SERIES
Let
Then,
Theorem 6—Proof
sn = a1 + a2 + ∙∙∙ + an
an = sn – sn–1
 Since Σ an is convergent, the sequence {sn}
is convergent.
Theorem 6—Proof
SERIES
Let lim sn  s
n 
Since n – 1 → ∞ as n → ∞,
we also have:
lim sn 1  s
n 
Theorem 6—Proof
SERIES
Therefore,
lim an  lim  sn  sn 1 
n 
n 
 lim sn  lim sn1
n 
 ss
0
n 
SERIES
Note 1
With any series Σ an we associate
two sequences:
 The sequence {sn} of its partial sums
 The sequence {an} of its terms
SERIES
Note 1
If Σ an is convergent, then
 The limit of the sequence {sn} is s (the sum
of the series).
 The limit of the sequence {an}, as Theorem 6
asserts, is 0.
SERIES
Note 2
The converse of Theorem 6 is not true
in general.
 If lim an  0, we cannot conclude
n 
that Σ an is convergent.
SERIES
Note 2
Observe that, for the harmonic series Σ 1/n,
we have an = 1/n → 0 as n → ∞.
 However, we showed in Example 7 that Σ 1/n
is divergent.
THE TEST FOR DIVERGENCE
Test 7
If lim an does not exist or if lim an  0 ,
n 
n 

then the series
a
n 1
is divergent.
n
TEST FOR DIVERGENCE
The Test for Divergence follows from
Theorem 6.
 If the series is not divergent, then it is convergent.
 Thus,
lim an  0
n 
TEST FOR DIVERGENCE
Show that the series

Example 8

2
n
diverges.

2
n 1 5n  4
n2
1
1
lim an = lim 2
 lim
 0
2
n 
n  5n  4
n  5  4 / n
5
 So, the series diverges by the Test for Divergence.
SERIES
Note 3
If we find that lim an  0 , we know that Σ an
n 
is divergent.
If we find that lim an  0 , we know nothing
n 
about the convergence or divergence of Σ an.
SERIES
Note 3
Remember the warning in Note 2:
 If lim an  0 , the series Σ an might
n 
converge or diverge.
Theorem 8
SERIES
If Σ an and Σ bn are convergent series, then
so are the series Σ can (where c is a constant),
Σ (an + bn), and Σ (an – bn), and


n 1
n 1
i. can  c an



n 1
n 1
n 1



n 1
n 1
n 1
ii.  an  bn    an  bn
iii.  an  bn    an  bn
SERIES
These properties of convergent series
follow from the corresponding Limit Laws
for Sequences in Section 11.1
 For instance, we prove part ii of Theorem 8
as follows.
THEOREM 8 ii—PROOF
Let
n
sn   ai
i 1
n
tn   bi
i 1

s  an
n 1

t  bn
n 1
THEOREM 8 ii—PROOF
The nth partial sum for the series
Σ (an + bn) is:
n
un    a i  bi 
i 1
THEOREM 8 ii—PROOF
Using Equation 10 in Section 5.2,
we have:
n
lim un  lim   ai  bi 
n 
n 
i 1


 lim   ai   bi 
n 
i 1
 i 1

n
n
n
n
 lim  ai  lim  bi
n 
i 1
n 
i 1
 lim sn  lim tn  s  t
n 
n 
THEOREM 8 ii—PROOF
Hence, Σ (an + bn) is convergent, and
its sum is:

a
n 1
n
 bn   s  t


n 1
n 1
  an   bn
Example 9
SERIES
Find the sum of the series
 3
1
 n


2 
n 1  n(n  1)

 The series Σ 1/2n is a geometric series
with a = ½ and r = ½.
 Hence, 
1
1
2


1

n
1
2
1

n 1
2
Example 9
SERIES
 In Example 6, we found that:

1
1

n 1 n( n  1)
 So, by Theorem 8, the given series is convergent
and
 3
1
 n


2
n 1  n ( n  1)




1
1
 n
  3
n 1 n ( n  1)
n 1 2

 3 1  1
4
SERIES
Note 4
A finite number of terms doesn’t
affect the convergence or divergence
of a series.
Note 4
SERIES
For instance, suppose that we were able to

n
show that the series 
is convergent.
3
n4 n  1

 Since

n
1 2 3
n
    3

3
2 9 28 n4 n  1
n 1 n  1

it follows that the entire series
is convergent.
n

3
n
1
n 1
Note 4
SERIES
Similarly, if it is known that the series
N
 a  a
n 1
n
is also convergent.
n 1
n


a
n  N 1

n  N 1
converges, then the full series


n
an