multiplication of series

multiplication of series

Theorem

(Franz Mertens).
If the series

∑

k

=

1

∞

a

k

and

∑

k

=

1

∞

b

k

with real or

complex

converge

and have the sums (

http://planetmath.org/SumOfSeries

)

A

and

B

, respectively, and at least one of them

converges absolutely

, then also the series

a

1

⁢

b

1

+

(

a

1

⁢

b

2

+

a

2

⁢

b

1

)

+

(

a

1

⁢

b

3

+

a

2

⁢

b

2

+

a

3

⁢

b

1

)

+

⋯

(1)

is

convergent

and its is equal to

A

⁢

B

.

Proof.

Denote the

partial sums

of the series

A

n

:=

a

1

+

a

2

+

⋯

+

a

n

,

B

n

:=

b

1

+

b

2

+

⋯

+

b

n

and

s

n

:=

a

1

⁢

b

1

+

(

a

1

⁢

b

2

+

a

2

⁢

b

1

)

+

(

a

1

⁢

b

3

+

a

2

⁢

b

2

+

a

3

⁢

b

1

)

+

⋯

+

(

a

1

⁢

b

n

+

a

2

⁢

b

n

-

1

+

⋯

+

a

n

⁢

b

1

)

for each

n

.  Then we have

lim

n

→

∞

⁡

A

n

=

A

and

lim

n

→

∞

⁡

B

n

=

B

.  Suppose that e.g. the series

∑

a

n

converges absolutely and that at least one

a

n

is distinct from zero; so the

∑

n

=

1

∞

|

a

n

|

is a real positive number

M

.  Let

ε

be an arbitrary positive number.

Now we can write the

identities

A

⁢

B

=

(

A

-

A

n

)

⁢

B

+

a

1

⁢

B

+

a

2

⁢

B

+

⋯

+

a

n

⁢

B

,

s

n

=

a

1

⁢

(

b

1

+

b

2

+

⋯

+

b

n

)

+

a

2

⁢

(

b

1

+

b

2

+

⋯

+

b

n

-

1

)

+

⋯

+

a

n

⁢

b

1

=

a

1

⁢

B

n

+

a

2

⁢

B

n

-

1

+

⋯

+

a

n

⁢

B

1

,

A

⁢

B

-

s

n

=

(

A

-

A

n

)

⁢

B

+

a

1

⁢

(

B

-

B

n

)

+

a

2

⁢

(

B

-

B

n

-

1

)

+

⋯

+

a

n

⁢

(

B

-

B

1

)

=

(

A

-

A

n

)

⁢

B

+

[

a

1

⁢

(

B

-

B

n

)

+

a

2

⁢

(

B

-

B

n

-

1

)

+

⋯

+

a

k

⁢

(

B

-

B

n

-

k

+

1

)

]

+

a

k

+

1

⁢

(

B

-

B

n

-

k

)

+

⋯

+

a

n

⁢

(

B

-

B

1

)

.

There is a positive number

n

1

⁢

(

ε

)

such that

|

A

-

A

n

|

<

ε

3

⁢

(

|

B

|

+

1

)

when

n

>

n

1

⁢

(

ε

)

.  Then

|

(

A

-

A

n

)

⁢

B

|

=

|

A

-

A

n

|

⋅

|

B

|

<

ε

3

⁢

(

|

B

|

+

1

)

⁢

(

|

B

|

+

1

)

=

ε

3

.

(2)

The

convergence

of

∑

b

n

implies that there is a number

n

2

⁢

(

ε

)

such that

|

B

-

B

n

|

<

ε

3

⁢

M

when

n

>

n

2

⁢

(

ε

)

.  Thus we have

|

[

…

]

|

≤

|

a

1

|

⋅

|

B

-

B

n

|

+

⋯

+

|

a

k

|

⋅

|

B

-

B

n

-

k

+

1

|

<

(

|

a

1

|

+

⋯

+

|

a

k

|

)

⁢

ε

3

⁢

M

≤

M

⋅

ε

3

⁢

M

=

ε

3

(3)

if

n

-

k

+

1

>

n

2

⁢

(

ε

)

.  Because

lim

n

→

∞

⁡

B

n

=

B

,  the numbers

|

B

n

|

are

bounded

, i.e. there is a positive number

K

such that for each

j

we have

|

B

j

|

<

K

and consequently

|

B

|

≤

K

.  It follows that

|

B

-

B

j

|

≤

|

B

|

+

|

B

j

|

<

K

+

K

=

2

⁢

K

for every

j

.  We apply

Cauchy criterion for convergence

to the series

∑

n

=

1

∞

|

a

n

|

getting a number

n

3

⁢

(

ε

)

such that for each

m

, one has the

inequality

|

a

k

+

1

|

+

⋯

+

|

a

m

|

<

ε

6

⁢

K

if

k

>

n

3

⁢

(

ε

)

.  Accordingly we obtain the estimation

|

a

k

+

1

⁢

(

B

-

B

n

-

k

)

+

⋯

+

a

n

⁢

(

B

-

B

1

)

|

≤

|

a

k

+

1

|

⁢

|

B

-

B

n

-

k

|

+

⋯

+

|

a

n

|

⁢

|

B

-

B

1

|

<

2

⁢

K

⋅

ε

6

⁢

K

=

ε

3

(4)

which is valid when

k

>

n

3

⁢

(

ε

)

.

If we choose

n

>

max

{

n

1

(

ε

}

,

n

2

(

ε

)

+

n

3

(

ε

)

}

and

k

such that

n

≥

k

>

n

3

⁢

(

ε

)

+

1

,  then the inequalities (2), (3) and (4) are satisfied, ensuring that

|

A

⁢

B

-

s

n

|

<

ε

3

+

ε

3

+

ε

3

=

ε

.

This means that the assertion of the theorem has been proved.

Remark.

The mere convergence of both series does not
suffice for convergence of (1).  This is seen in the following
example by Cauchy where both series are

1

-

1

2

+

1

3

-

+

⋯

They converge by virtue of

Leibniz test

, but not absolutely (see the

p

-test (

http://planetmath.org/PTest

)).  In their product series

1

-

(

1

2

+

1

2

)

+

(

1

3

+

1

2

1

2

+

1

3

)

-

+

⋯

the

absolute value

of the

n

th

is

1

⋅

1

n

+

1

2

⁢

1

n

-

1

+

⋯

+

1

n

⋅

1

,  having

n

summands which all are greater than

2

n

+

1

(this is seen when one looks at the half circle

y

=

x

⁢

n

+

1

-

x

or

(

x

-

n

+

1

2

)

2

+

y

2

=

(

n

+

1

2

)

2

,  which shows that

y

≤

n

+

1

2

and thus

1

y

≥

2

n

+

1

).  Because

lim

n

→

∞

⁡

n

⋅

2

n

+

1

=

2

≠

0

, the product series does not satisfy the necessary condition of convergence (

http://planetmath.org/ThenA_kto0IfSum_k1inftyA_kConverges

) and therefore the series diverges.

References

1

Ernst Lindelöf:

Johdatus funktioteoriaan

.  Mercatorin Kirjapaino Osakeyhtiö. Helsinki (1936).

Title

multiplication of series

Canonical name

MultiplicationOfSeries

Date of creation

2014-10-31 20:23:29

Last modified on

2014-10-31 20:23:29

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

21

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40A05

Synonym

Cauchy multiplication rule

Related topic

ManipulatingConvergentSeries

Related topic

AlternatingHarmonicSeries

Related topic

ErnstLindelof

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