real part series and imaginary part series

real part series and imaginary part series

Theorem 1.

Given the series

z

1

+

z

2

+

z

3

+

…

(1)

with the

real parts

of its terms

ℜ

⁡

z

n

=

a

n

and the

imaginary parts

of its terms

ℑ

⁡

z

n

=

b

n

(

n

=

1

,

2

,

3

,

…

). If the series (1)

converges

and its sum is

A

+

i

⁢

B

, where

A

and

B

are real, then also the series

a

1

+

a

2

+

a

3

+

…

⁢

and

⁢

b

1

+

b

2

+

b

3

+

…

converge and their sums are

A

and

B

, respectively. The converse is valid as well.

Proof.

Let

ε

be an arbitrary positive number. Denote the

partial sum

of (1) by

S

n

=

z

1

+

…

+

z

n

=

(

a

1

+

i

⁢

b

1

)

+

…

+

(

a

n

+

i

⁢

b

n

)

=

(

a

1

+

…

+

a

n

)

+

i

⁢

(

b

1

+

…

+

b

n

)

:=

A

n

+

i

⁢

B

n

(

n

=

1

,

2

,

3

,

…

). When (1) converges to the sum

A

+

i

⁢

B

, then there is a number

n

ε

such that  for any integer

n

>

n

ε

we have

|

(

A

n

-

A

)

+

i

⁢

(

B

n

-

B

)

|

=

|

(

A

n

+

i

⁢

B

n

)

-

(

A

+

i

⁢

B

)

|

<

ε

.

But a

complex number

is always absolutely at least equal to the real part (see the

inequalities

in

modulus of complex number

), and therefore

|

A

n

-

A

|

≦

|

(

A

n

-

A

)

+

i

⁢

(

B

n

-

B

)

|

<

ε

, similarly

|

B

n

-

B

|

≦

|

(

A

n

-

A

)

+

i

⁢

(

B

n

-

B

)

|

<

ε

as soon as

n

>

n

ε

.  Hence,

A

n

→

A

and

B

n

→

B

as

n

→

∞

.  This means the

convergences

a

1

+

a

2

+

a

3

+

…

=

A

⁢

and

⁢

b

1

+

b

2

+

b

3

+

…

=

B

,

Q.E.D. The converse part is straightforward.

Theorem 2.

Notations same as in the preceding theorem. The series

|

z

1

|

+

|

z

2

|

+

|

z

3

|

+

…

converges if and only if the series

a

1

+

a

2

+

a

3

+

…

⁢

and

⁢

b

1

+

b

2

+

b

3

+

…

converge absolutely

(

http://planetmath.org/AbsoluteConvergence

).

Proof.

Use the inequalities

0

≦

|

a

n

|

≦

|

z

n

|

,

0

≦

|

b

n

|

≦

|

z

n

|

and

0

≦

|

z

n

|

≦

|

a

n

|

+

|

b

n

|

for using the

comparison test

.

Theorem 3.

If the series

∑

n

=

1

∞

|

z

n

|

converges, then also the series

∑

n

=

1

∞

z

n

converges and we have

|

∑

n

=

1

∞

z

n

|

≦

∑

n

=

1

∞

|

z

n

|

.

Proof.

By theorem 2, the convergence of

∑

|

z

n

|

implies the convergence of

∑

a

n

and

∑

b

n

, which, by theorem 1, in turn imply the convergence of

∑

z

n

. Since for every

n

the

triangle inequality

guarantees the inequality

|

∑

j

=

1

n

z

j

|

≦

∑

j

=

1

n

|

z

j

|

,

then we must have the asserted limit inequality, too.

Title

real part series and imaginary part series

Canonical name

RealPartSeriesAndImaginaryPartSeries

Date of creation

2013-03-22 17:28:08

Last modified on

2013-03-22 17:28:08

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40-00

Related topic

SumOfSeries

Related topic

ModulusOfComplexNumber

Related topic

AbsoluteConvergenceTheorem

Related topic

RealAndImaginaryPartsOfContourIntegral

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