aarjaneiro/planetmath huggingface dataset:
Euler product formula

Euler product formula

Theorem (Euler).

If

s

>

1

,  the infinite product

∏

p

1

1

-

1

p

s

(1)

where

p

runs the

positive

rational primes

,

converges

to the sum of the

over-harmonic series

∑

n

=

1

∞

1

n

s

=

ζ

⁢

(

s

)

.

(2)

Proof.

Denote the

sequence

of

prime numbers

by

p

1

<

p

2

<

p

3

<

…

For any

s

>

0

,  we can form

convergent

geometric series

1

1

-

1

p

1

s

=

1

+

1

p

1

s

+

1

p

1

2

⁢

s

+

…

=

∑

ν

1

=

0

∞

1

p

1

ν

1

⁢

s

,

1

1

-

1

p

2

s

=

1

+

1

p

2

s

+

1

p

2

2

⁢

s

+

…

=

∑

ν

2

=

0

∞

1

p

2

ν

2

⁢

s

.

Since these series are absolutely convergent, their product (see

multiplication of series

) may be written as

1

1

-

1

p

1

s

⋅

1

1

-

1

p

2

s

=

∑

ν

1

,

ν

2

=

0

∞

1

p

1

ν

1

⁢

s

⋅

1

p

2

ν

2

⁢

s

=

∑

ν

1

,

ν

2

=

0

∞

1

(

p

1

ν

1

⁢

p

2

ν

2

)

s

where

ν

1

and

ν

2

independently on each other run all

nonnegative integers

.  This equation can be generalised by induction to

∏

ν

=

1

k

1

1

-

1

p

ν

s

=

∑

ν

1

,

ν

2

,

…

,

ν

k

=

0

∞

1

(

p

1

ν

1

⁢

p

2

ν

2

⁢

⋯

⁢

p

k

ν

k

)

s

(3)

for

s

>

0

and for arbitrarily great

k

; the exponents

ν

1

,

ν

2

,

…

,

ν

k

run independently all nonnegative integers.

Because the

prime factorization

of

positive integers

is unique (

http://planetmath.org/FundamentalTheoremOfArithmetic

), we can rewrite (3) as

∏

ν

=

1

k

1

1

-

1

p

ν

s

=

∑

(

n

)

1

n

s

,

(4)

where

n

runs all positive integers not containing greater

prime factors

than

p

k

.  Then the inequality

∑

n

=

1

p

k

1

n

s

<

∏

ν

=

1

k

1

1

-

1

p

ν

s

,

(5)

holds for every

k

, since all the terms

1

,

1

p

1

s

,

…

,

1

p

k

s

are in the series of the right hand side of (4).  On the other hand, this series contains only a part of the terms of (2).  Thus, for

s

>

1

,  the product (3) is less than the sum

ζ

⁢

(

s

)

of the series (2), and consequently

∑

n

=

1

p

k

1

n

s

<

∏

ν

=

1

k

1

1

-

1

p

ν

s

<

ζ

⁢

(

s

)

.

(6)

Letting

k

→

∞

,  we have

p

k

→

∞

,  and the sum on the left hand side of (6) tends to the limit

ζ

⁢

(

s

)

, therefore also tends the product (3).  Hence we get the limit equation

∏

p

1

1

-

1

p

s

=

ζ

(

s

)

(

s

>

1

)

.

(7)

References

1

E. Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset III.2

.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).

Title

Euler product formula

Canonical name

EulerProductFormula

Date of creation

2013-03-22 18:39:38

Last modified on

2013-03-22 18:39:38

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40A20

Classification

msc 11M06

Classification

msc 11A51

Classification

msc 11A41

Related topic

RiemannZetaFunction

Related topic

EulerProduct

Generated on Fri Feb 9 21:23:03 2018 by

LaTeXML
infinite product of differences 1⁢ERROR(\tmspace)-.1667⁢e⁢m-ERROR(\tmspace)-.1667⁢e⁢m⁢a_i

infinite product of differences

1

⁢

\tmspace

-

.1667

⁢

e

⁢

m

-

\tmspace

-

.1667

⁢

e

⁢

m

⁢

a

i

We consider the infinite products of the form

∏

i

=

1

∞

(

1

-

a

i

)

=

(

1

-

a

1

)

⁢

(

1

-

a

2

)

⁢

(

1

-

a

3

)

⁢

⋯

(1)

and the series

a

1

+

a

2

+

a

3

+

…

where the numbers

a

i

are nonnegative reals.

•

If the series

converges

, then also the product converges and has a value which does not depend on the order of the factors and

vanishes

only when some of the factors is 0.

•

If

lim

i

→

∞

⁡

a

i

=

0

but the series diverges, then the value of the infinite product is always zero though no of the factors were zero.

Example.

(

1

-

1

2

)

⁢

(

1

-

1

3

)

⁢

(

1

-

1

4

)

⁢

⋯

=

0

;  see the

harmonic series

.

Proof.

1

∘

.  Now we have

lim

i

→

∞

⁡

a

i

=

0

(see the necessary condition of

convergence

of series), and so

a

i

<

1

2

when

i

≧

i

0

.  We write

∏

i

=

1

∞

(

1

-

a

i

)

=

∏

i

=

1

i

0

-

1

(

1

-

a

i

)

⁢

∏

i

=

i

0

∞

(

1

-

a

i

)

(2)

and set in the last product

1

-

a

i

=

1

1

1

-

a

i

=

1

1

+

a

i

1

-

a

i

,

whence

∏

i

=

i

0

n

(

1

-

a

i

)

=

1

∏

i

=

i

0

n

(

1

+

a

i

1

-

a

i

)

.

(3)

As

a

i

<

1

2

, we have

1

1

-

a

i

<

2

and thus

0

<

a

i

1

-

a

i

<

2

⋅

a

i

,  and therefore the series

∑

i

=

i

0

∞

a

i

1

-

a

i

with nonnegative is

absolutely convergent

.  The theorem of the

http://planetmath.org/node/6204

parent entry then says that the product in the

denominator

of the right hand side of (3) tends, as

n

→

∞

,  to a finite non-zero limit, which don’t depend on the order of the factors.  Consequently, the same concerns the product of the left hand side of (3).  By (2), we now infer that the given product (1) converges, its value is on the order and it vanishes only along with some of its factors.

2

∘

.  There is an

i

0

such that

a

i

<

1

when

i

≧

i

0

,  whence

a

i

1

-

a

i

>

a

i

and the series

∑

i

=

i

0

∞

a

i

1

-

a

i

diverges.  The denominator of the right hand side of (3) tends, as

n

→

∞

,  to the infinity and thus the product of the left hand side to 0.  Hence the value of (1) is necessarily 0, also when all factors were distinct from 0.

References

1

E. Lindelöf

:

Differentiali- ja integralilasku
ja sen sovellutukset III.2

.  Mercatorin Kirjapaino Osakeyhtiö, Helsinki (1940).

Title

infinite product of

differences

1

⁢

\tmspace

-

.1667

⁢

e

⁢

m

-

\tmspace

-

.1667

⁢

e

⁢

m

⁢

a

i

Canonical name

InfiniteProductOfDifferences1ai

Date of creation

2013-03-22 18:39:45

Last modified on

2013-03-22 18:39:45

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40A20

Classification

msc 26E99

Related topic

HarmonicSeriesOfPrimes

Related topic

InfiniteProductOfSums1A_I

Related topic

InfiniteProductOfSums1a_i

Generated on Fri Feb 9 21:23:06 2018 by

LaTeXML
infinite product of sums 1⁢ERROR(\tmspace)-.1667⁢e⁢m+ERROR(\tmspace)-.1667⁢e⁢m⁢a_i

infinite product of sums

1

⁢

\tmspace

-

.1667

⁢

e

⁢

m

+

\tmspace

-

.1667

⁢

e

⁢

m

⁢

a

i

Lemma.

Let the numbers

a

i

be nonnegative reals.  The infinite product

∏

i

=

1

∞

(

1

+

a

i

)

=

(

1

+

a

1

)

⁢

(

1

+

a

2

)

⁢

(

1

+

a

3

)

⁢

⋯

(1)

converges

iff the series

a

1

+

a

2

+

a

3

+

…

is

convergent

.

Proof.

Denote

∑

i

=

1

n

a

n

:=

s

n

,

∏

i

=

1

n

(

1

+

a

i

)

:=

t

n

(

n

=

1

,

2

,

…

)

.

Now

1

+

a

i

≦

1

+

a

i

1

!

+

a

i

2

2

!

+

…

=

e

a

i

,  whence

t

n

≦

∏

i

=

1

n

e

a

i

=

e

s

n

.

(2)

We can estimate also downwards:

t

n

=

(

1

+

a

1

)

⁢

(

1

+

a

2

)

⁢

⋯

⁢

(

1

+

a

n

)

=

1

+

∑

i

=

1

n

a

i

+

…

+

a

1

⁢

a

2

⁢

⋯

⁢

a

n

>

∑

i

=

1

n

a

i

=

s

n

(3)

If the series is convergent with sum

S

, then by (2),

t

n

≦

e

s

n

≦

e

S

,

and since the

monotonically nondecreasing

sequence

⟨

t

1

,

t

2

,

t

3

,

…

⟩

thus is

bounded from above

, it converges (cf.

limit of nondecreasing sequence

).  So (1) converges.

If, on the other hand, the series is divergent, then

lim

n

→

∞

⁡

s

n

=

∞

and by (3), also

lim

n

→

∞

⁡

t

n

=

∞

,  i.e. the (1) diverges.

Title

infinite product of sums

1

⁢

\tmspace

-

.1667

⁢

e

⁢

m

+

\tmspace

-

.1667

⁢

e

⁢

m

⁢

a

i

Canonical name

InfiniteProductOfSums1ai

Date of creation

2013-03-22 18:40:01

Last modified on

2013-03-22 18:40:01

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40A20

Classification

msc 26E99

Related topic

LimitOfRealNumberSequence

Related topic

DeterminingSeriesConvergence

Related topic

InfiniteProductOfDifferences1A_i

Related topic

AbsoluteConvergenceOfInfiniteProductAndSeries

Generated on Fri Feb 9 21:23:09 2018 by

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proof of infinite product of sums 1⁢ERROR(\tmspace)-.1667⁢e⁢m+ERROR(\tmspace)-.1667⁢e⁢m⁢a_i result without exponentials

proof of infinite product of sums

1

⁢

\tmspace

-

.1667

⁢

e

⁢

m

+

\tmspace

-

.1667

⁢

e

⁢

m

⁢

a

i

result without exponentials

In this entry, we show how the proof presented in the parent entry
may be modified so as to avoid use of the

exponential function

.
This modification makes it more elementary by not requiring that
one first develop the theory of the exponential function before
proving this basic result about infinite products. Note that it
is only necessary to redo the part of the result which states that,
if the series

converges

, then the

product

also converges because
the proof of the opposite implication did not involve the exponential
function.

We begin with a simple

inequality

. Suppose that

a

and

b

are

real numbers

such that

0

≤

a

and

0

≤

b

<

1

/

2

. Then we
have

2

⁢

a

⁢

b

≤

a

, hence

(

1

+

a

)

⁢

(

1

+

2

⁢

b

)

=

1

+

2

⁢

b

+

a

+

2

⁢

a

⁢

b

≤

1

+

2

⁢

b

+

a

+

a

=

1

+

2

⁢

(

a

+

b

)

.

Now suppose that the series

a

1

+

a

2

+

a

3

+

⋯

converges
to a value

S

.
Since the convergence of an infinite series or product is not
affected by removing a finite number of terms we may, without
loss of generality, assume that

S

<

1

/

2

. Then,
since the terms

a

n

are nonnegative for all

n

, for each partial
sum

s

n

we will have

0

≤

s

n

<

1

/

2

.

Clearly,

t

1

≤

1

+

2

⁢

s

1

. Suppose that, for some

n

, we have

t

n

≤

1

+

2

⁢

s

n

. Then, using the definitions of

t

n

and

s

n

along with the inequality demonstrated above, we conclude that

t

n

+

1

=

t

n

⁢

(

1

+

a

n

+

1

)

≤

(

1

+

2

⁢

s

n

)

⁢

(

1

+

a

n

+

1

)

≤

1

+

2

⁢

(

s

n

+

a

n

+

1

)

=

1

+

2

⁢

s

n

+

1

Hence, if

t

n

≤

1

+

2

⁢

s

n

, then

t

n

+

1

≤

1

+

2

⁢

s

n

+

1

as well. By induction, we conclude that

t

n

≤

1

+

2

⁢

s

n

for all

n

.

Thus, for all

n

, we have

t

n

≤

1

+

2

⁢

s

n

≤

1

+

2

⁢

S

. Substituting
this inequality for the inequality

t

n

≤

e

s

n

≤

e

S

in the
parent entry, the rest of the proof proceeds in exactly the same manner.

Title

proof of infinite product of sums

1

⁢

\tmspace

-

.1667

⁢

e

⁢

m

+

\tmspace

-

.1667

⁢

e

⁢

m

⁢

a

i

result without

exponentials

Canonical name

ProofOfInfiniteProductOfSums1aiResultWithoutExponentials

Date of creation

2013-03-22 18:40:38

Last modified on

2013-03-22 18:40:38

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

11

Author

rspuzio (6075)

Entry type

Result

Classification

msc 40A20

Classification

msc 26E99

Generated on Fri Feb 9 21:23:12 2018 by

LaTeXML
Wallis formulae

Wallis formulae

Wallis’ formula expresses

π

as an infinite product:

π

2

=

∏

n

=

1

∞

4

⁢

n

2

4

⁢

n

2

-

1

=

2

1

⁢

2

3

⁢

4

3

⁢

4

5

⁢

⋯

It may be derived by taking the limit as

n

→

∞

of the ratio of the following
two integrals.

∫

0

π

2

sin

2

⁢

n

⁡

x

⁢

d

⁢

x

=

1

⋅

3

⁢

⋯

⁢

(

2

⁢

n

-

1

)

2

⋅

4

⁢

⋯

⁢

2

⁢

n

⁢

π

2

∫

0

π

2

sin

2

⁢

n

+

1

⁡

x

⁢

d

⁢

x

=

2

⋅

4

⁢

⋯

⁢

2

⁢

n

3

⋅

5

⁢

⋯

⁢

(

2

⁢

n

+

1

)

Title

Wallis formulae

Canonical name

WallisFormulae

Date of creation

2013-03-22 12:56:15

Last modified on

2013-03-22 12:56:15

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 40A20

Classification

msc 40A10

Related topic

Pi

Related topic

ReductionFormulasForIntegrationOfPowers

Generated on Fri Feb 9 21:23:14 2018 by

LaTeXML
Euler’s constant

Euler’s constant

Euler’s constant

γ

is defined by

γ

=

lim

n

→

∞

⁡

(

1

+

1

2

+

1

3

+

1

4

+

⋯

+

1

n

-

ln

⁡

n

)

or equivalently

γ

=

lim

n

→

∞

⁡

∑

i

=

1

n

[

1

i

-

ln

⁡

(

1

+

1

i

)

]

Euler’s constant has the value

0.57721566490153286060651209008240243104

⁢

…

It is related to the

gamma function

by

γ

=

-

Γ

′

⁢

(

1

)

It is not known whether

γ

is rational or irrational.

References

.

•

Chris Caldwell - “Euler’s Constant”,

http://primes.utm.edu/glossary/page.php/Gamma.html

http://primes.utm.edu/glossary/page.php/Gamma.html

Title

Euler’s constant

Canonical name

EulersConstant

Date of creation

2013-03-22 12:18:27

Last modified on

2013-03-22 12:18:27

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

10

Author

akrowne (2)

Entry type

Definition

Classification

msc 40A25

Synonym

Euler-Mascheroni constant

Synonym

Mascheroni constant

Generated on Fri Feb 9 21:23:16 2018 by

LaTeXML
integral representations of the Mascheroni constant

integral representations of the Mascheroni constant

Mascheroni’s constant can be expressed by the following integrals:

γ

=

-

∫

0

1

log

⁡

(

-

log

⁡

x

)

⁢

𝑑

x

γ

=

-

∫

0

∞

e

-

x

⁢

log

⁡

x

⁢

d

⁢

x

γ

=

∫

0

∞

(

1

e

t

-

1

-

1

t

⁢

e

t

)

⁢

𝑑

t

γ

=

∫

0

∞

(

1

t

-

1

1

+

t

-

1

t

⁢

e

t

)

⁢

𝑑

t

Title

integral representations of the

Mascheroni constant

Canonical name

IntegralRepresentationsOfTheMascheroniConstant

Date of creation

2013-03-22 15:53:24

Last modified on

2013-03-22 15:53:24

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 40A25

Generated on Fri Feb 9 21:23:18 2018 by

LaTeXML
proof that e is not a natural number

proof that e is not a natural number

Here, we are going to show that the natural log base

e

is not a

natural number

by showing a sharper result: that

e

is between

2

and

3

.

Proposition

.

2

<

e

<

3

.

Proof.

There are several

infinite series

representations of

e

. In this proof, we will use the most common one, the

Taylor expansion

of

e

:

∑

i

=

0

∞

1

i

!

=

1

0

!

+

1

1

!

+

1

2

!

+

⋯

+

1

n

!

+

⋯

.

(1)

We chop up the Taylor expansion of

e

into two parts: the first part

a

consists of the sum of the first two terms, and the second part

b

consists of the sum of the rest, or

e

-

a

. The proof of the proposition now lies in the estimation of

a

and

b

.

Step 1: e

>

2.

First,

a

=

1

0

!

+

1

1

!

=

1

+

1

=

2

. Next,

b

>

0

, being a sum of the terms in (1), all of which are

positive

(note also that

b

must be

bounded

because (1) is a

convergent series

). Therefore,

e

=

a

+

b

=

2

+

b

>

2

+

0

=

2

.

Step 2: e

<

3.

This step is the same as showing that

b

=

e

-

a

=

e

-

2

<

3

-

2

=

1

. With this in mind, let us compare term by term of the series (2) representing

b

and another series (3):

1

2

!

+

1

3

!

+

⋯

+

1

n

!

+

⋯

(2)

and

1

2

2

-

1

+

1

2

3

-

1

+

⋯

+

1

2

n

-

1

+

⋯

.

(3)

It is well-known that the second series (a

geometric series

) sums to 1. Because both series are convergent, the term-by-term comparisons make sense. Except for the first term, where

1

2

!

=

1

2

=

1

2

2

-

1

, we have

1

n

!

<

1

2

n

-

1

for all other terms. The

inequality

1

n

!

<

1

2

n

-

1

, for

n

a positive number can be translated into the basic inequality

n

!

>

2

n

-

1

, the proof of which, based on mathematical induction, can be found here (

http://planetmath.org/AnExampleOfMathematicalInduction

).

Because the term comparisons show

•

that the terms from (2)

≤

the corresponding terms from (3), and

•

that at least one term from (2)

<

than the corresponding term from (3),

we conclude that (2)

<

(3), or that

b

<

1

. This concludes the proof.
∎

Title

proof that e is not a natural number

Canonical

name

ProofThatEIsNotANaturalNumber

Date of creation

2013-03-22 15:39:52

Last modified on

2013-03-22 15:39:52

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

10

Author

CWoo (3771)

Entry type

Proof

Classification

msc 40A25

Classification

msc 40A05

Classification

msc 11J72

Related topic

EIsTranscendental

Generated on Fri Feb 9 21:23:20 2018 by

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proof that Euler’s constant exists

proof that Euler’s constant exists

Theorem 1

The limit

γ

=

lim

n

→

∞

⁡

(

∑

k

=

1

n

1

k

-

ln

⁡

n

)

exists.

Proof.

Let

C

n

=

1

1

+

1

2

+

⋯

+

1

n

-

ln

⁡

n

and

D

n

=

C

n

-

1

n

Then

C

n

+

1

-

C

n

=

1

n

+

1

-

ln

⁡

(

1

+

1

n

)

and

D

n

+

1

-

D

n

=

1

n

-

ln

⁡

(

1

+

1

n

)

Now, by considering the

Taylor series

for

ln

⁡

(

1

+

x

)

, we see that

1

n

+

1

<

ln

⁡

(

1

+

1

n

)

<

1

n

and so

C

n

+

1

-

C

n

<

0

<

D

n

+

1

-

D

n

Thus, the

C

n

decrease

monotonically

, while the

D

n

increase monotonically, since the

differences

are

negative

(positive for

D

n

). Further,

D

n

<

C

n

and thus

D

1

=

0

is a

lower bound

for

C

n

. Thus the

C

n

are

monotonically decreasing

and bounded below, so they must

converge

.

References

1

E. Artin,

The

Gamma Function

, Holt, Rinehart, Winston 1964.

Title

proof that Euler’s constant exists

Canonical name

ProofThatEulersConstantExists

Date of creation

2013-03-22 16:34:48

Last modified on

2013-03-22 16:34:48

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Proof

Classification

msc 40A25

Generated on Fri Feb 9 21:23:23 2018 by

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Abel’s limit theorem

Abel’s limit theorem

Suppose that

∑

a

n

⁢

x

n

has a

radius of convergence

r

and that

∑

a

n

⁢

r

n

is

convergent

. Then

lim

x

→

r

-

⁡

∑

a

n

⁢

x

n

=

∑

a

n

⁢

r

n

=

∑

(

lim

x

→

r

-

⁡

a

n

⁢

x

n

)

Title

Abel’s limit theorem

Canonical name

AbelsLimitTheorem

Date of creation

2013-03-22 12:57:49

Last modified on

2013-03-22 12:57:49

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

5

Author

CWoo (3771)

Entry type

Theorem

Classification

msc 40A30

Related topic

PowerSeries

Related topic

AbelsMultiplicationRuleForSeries

Related topic

AbelSummability

Related topic

NielsHenrikAbel

Generated on Fri Feb 9 21:23:25 2018 by

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absolute convergence implies uniform convergence

absolute convergence implies uniform convergence

Theorem 1

.

Let

T

be a

topological space

,

f

be a

continuous function

from

T

to

[

0

,

∞

)

, and
let

{

f

k

}

k

=

0

∞

be a

sequence

of continuous functions from

T

to

[

0

,

∞

)

such that, for all

x

∈

T

, the sum

∑

k

=

0

∞

f

k

⁢

(

x

)

converges

to

f

⁢

(

x

)

. Then
the

convergence

of this sum is uniform on

compact subsets

of

T

.

Proof.

Let

X

be a compact subset of

T

and let

ϵ

be a positive real number. We will
construct an

open cover

of

X

. Because the series is assumed to converge

pointwise

, for
every

x

∈

X

, there exists an integer

n

x

such that

∑

k

=

n

x

∞

f

k

⁢

(

x

)

<

ϵ

/

3

. By

continuity

, there exists an open neighborhood

N

1

of

x

such that

|

f

⁢

(

x

)

-

f

⁢

(

y

)

|

<

ϵ

/

3

when

y

∈

N

1

and an open neighborhood

N

2

of

x

such that

|

∑

k

=

0

n

x

f

k

⁢

(

x

)

-

∑

k

=

0

n

f

k

⁢

(

y

)

|

<

ϵ

/

3

when

y

∈

N

2

.
Let

N

x

be the

intersection

of

N

1

and

N

2

. Then, for every

y

∈

N

, we have

f

⁢

(

y

)

-

∑

k

=

0

n

x

f

k

⁢

(

y

)

<

|

f

⁢

(

y

)

-

f

⁢

(

x

)

|

+

|

f

⁢

(

x

)

-

∑

k

=

0

n

x

f

k

⁢

(

x

)

|

+

|

∑

k

=

0

n

x

f

k

⁢

(

x

)

-

∑

k

=

0

n

x

f

k

⁢

(

y

)

|

<

ϵ

.

In this way, we associate to every point

x

an

neighborhood

N

x

and an integer

n

x

.
Since

X

is

compact

, there will exist a finite number of points

x

1

,

…

⁢

x

m

such
that

X

⊆

N

x

1

∪

⋯

∪

N

x

m

. Let

n

be the greatest of

n

x

1

,

…

,

n

x

m

. Then we have

f

⁢

(

y

)

-

∑

k

=

0

n

f

k

⁢

(

y

)

<

ϵ

for all

y

∈

X

,
so, the

functions

f

k

being

positive

,

f

⁢

(

y

)

-

∑

k

=

0

h

f

k

⁢

(

y

)

<

ϵ

for all

h

≥

n

, which means that the sum

converges uniformly

.
∎

Note: This result can also be deduced from Dini’s

theorem

,
since the

partial sums

of positive functions are

monotonically increasing

.

Title

absolute convergence implies uniform convergence

Canonical name

AbsoluteConvergenceImpliesUniformConvergence

Date of creation

2013-03-22 18:07:27

Last modified on

2013-03-22 18:07:27

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Theorem

Classification

msc 40A30

Generated on Fri Feb 9 21:23:28 2018 by

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Bohr’s theorem

Bohr’s theorem

(Bohr 1914).

If the

power series

∑

n

=

0

∞

a

n

⁢

z

n

satisfies

|

∑

n

=

0

∞

a

n

⁢

z

n

|

<

1

(1)

in the

unit disk

|

z

|

<

1

,  then (1) and the inequality

∑

n

=

0

∞

|

a

n

⁢

z

n

|

<

1

(2)

is true in the disk

|

z

|

<

1

3

.  Here, the radius

1

3

is the best possible.

Proof.

One needs

Carathéodory’s inequality

which says that if the

real part

of a

holomorphic function

g

⁢

(

z

)

:=

∑

n

=

0

∞

b

n

⁢

z

n

is positive in the unit disk, then

|

b

n

|

≦

2

⁢

Re

⁢

b

0

for

⁢

n

=

1

,

2

,

…

Choosing now

g

⁢

(

z

)

:=

1

-

e

i

⁢

φ

⁢

f

⁢

(

z

)

where

φ

is any real number and

f

⁢

(

z

)

the sum

function

of the series in the theorem, we get

|

a

n

|

≦

2

⁢

Re

⁢

(

1

-

e

i

⁢

φ

⁢

a

0

)

=

2

⁢

(

1

-

a

0

⁢

cos

⁡

φ

)

,

and especially

|

a

n

|

≦

2

⁢

(

1

-

|

a

0

|

)

,

for

⁢

n

=

1

,

2

,

…

If

f

⁢

(

z

)

≢

a

0

,  in the disk

|

z

|

<

1

3

we thus have

∑

n

=

0

∞

|

a

n

⁢

z

n

|

<

|

a

0

|

+

2

⁢

(

1

-

|

a

0

|

)

⁢

∑

n

=

1

∞

(

1

3

)

n

=

1

.

Take then in particular the function defined by

f

⁢

(

z

)

:=

z

-

c

1

-

c

⁢

z

with

0

<

c

<

1

.  Its

series expansion

f

⁢

(

z

)

=

∑

n

=

0

∞

a

n

⁢

z

n

=

-

c

+

(

1

-

c

2

)

⁢

z

+

(

1

-

c

2

)

⁢

c

⁢

z

2

+

(

1

-

c

2

)

⁢

c

2

⁢

z

3

+

…

shows that

∑

n

=

0

∞

|

a

n

⁢

z

n

|

=

f

⁢

(

|

z

|

)

+

2

⁢

c

,

which last form can be seen to become greater than 1 for

|

z

|

>

1

1

+

2

⁢

c

.  Because

c

may come from below arbitrarily to 1, one sees that the value

1

3

in the theorem cannot be increased.

References

1

Harald Bohr

: “A theorem concerning power series”. –

Proc. London Math. Soc.

13

(1914).

2

Harold P. Boas

: “Majorant series”. –

J. Korean Math. Soc.

37

(2000).

Title

Bohr’s theorem

Canonical name

BohrsTheorem

Date of creation

2015-04-13 12:52:55

Last modified on

2015-04-13 12:52:55

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40A30

Classification

msc 30B10

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