6Real numbers

IA Numbers and Sets

6.3 Series

In a field, the sum of two numbers is defined. By induction, the sum of finitely

many numbers is defined as well. However, infinite sums (“series”) are not. We

will define what it means to take an infinite sum. Of course, infinite sums exist

only for certain nice sums. For example, 1 + 1 + 1 + ··· does not exist.

Definition

(Series and partial sums)

.

Let (

a

n

) be a sequence. Then

s

m

=

P

m

n=1

a

n

is the mth partial sum of (a

n

). We write

∞

X

n=1

a

n

= lim

m→∞

s

m

if the limit exists.

Example. Let a

n

=

1

n(n−1)

for n ≥ 2. Then

s

m

=

m

X

n=2

1

n(n − 1)

=

m

X

n=2

1

n − 1

−

1

n

= 1 −

1

m

→ 1.

Then

∞

X

n=2

1

n(n − 1)

= 1.

Example.

Let

a

n

=

1

n

2

. Then

s

m

=

P

m

n=1

1

n

2

. We know that

s

m

is increasing.

We also know that

s

m

≤

1 +

P

1

n(n−1)

≤

2, i.e. it is bounded above. So

s

m

converges and

P

∞

n=1

1

n

2

exists (in fact it is π

2

/6).

Example.

(Geometric series) Suppose

a

n

=

r

n

, where

|r| <

1. Then

s

m

=

r ·

1−r

m

1−r

→

r

1−r

since r

n

→ 0. So

∞

X

n=1

r

n

=

r

1 − r

.

Example. (Harmonic series) Let a

n

=

1

n

. Consider

S

2

k

= 1 +

1

2

+

1

3

+

1

4

+

1

5

+

1

6

+

1

7

+

1

8

+

1

9

+ ···+

1

2

k

≥ 1 +

1

2

+

1

4

+

1

4

+

1

8

+

1

8

+

1

8

+

1

8

+

1

16

+ ···+

1

2

k

≥ 1 +

k

2

.

So

∞

X

n=1

1

n

diverges.

Decimal expansions

Definition

(Decimal expansion)

.

Let (

d

n

) be a sequence with

d

n

∈ {

0

,

1

, ···

9

}

.

Then

∞

X

n=1

d

10

n

converges to a limit

r

with 0

≤ r ≤

1 since the partial sums

s

m

are increasing and bounded by

P

9

10

n

→

1 (geometric series). We say

r = 0.d

1

d

2

d

3

···, the decimal expansion of r.

Does every

x

with 0

≤ x <

1 have a decimal expansion? Pick

d

1

maximal

such that

d

1

10

≤ x <

1. Then 0

≤ x −

d

1

10

<

1

10

since

d

1

is maximal. Then pick

d

2

maximal such that

d

2

100

≤ x −

d

1

10

. By maximality, 0

≤ x −

d

1

10

−

d

2

100

<

1

100

.

Repeat inductively, pick maximal d

n

with

d

n

10

n

≤ x −

n−1

X

j=1

d

j

10

j

so

0 ≤ x −

n

X

j=1

d

j

10

j

<

1

10

n

.

Since both LHS and RHS

→

0, by sandwich,

x−

P

∞

j=1

d

j

10

j

= 0, i.e.

x

= 0

.d

1

d

2

···

.

Since we have shown that at least one decimal expansion, can the same

number have two different decimal expansions? i.e. if 0

.a

1

a

2

···

= 0

.b

1

b

2

···

,

must a

i

= b

i

for all i?

Now suppose that the

a

j

and

b

j

are equal until

k

, i.e.

a

j

=

b

j

for

j < k

. wlog

assume a

k

< b

k

. Then

∞

X

j=k+1

a

j

10

j

≤

∞

X

j=k+1

9

10

j

=

9

10

k+1

·

1

1 − 1/10

=

1

10

k

.

So we must have

b

k

=

a

k

+ 1,

a

j

= 9 for

j > k

and

b

j

= 0 for

j > k

. For example,

0.47999 ··· = 0.48000 ···.