aarjaneiro/planetmath huggingface dataset:
application of Cauchy criterion for convergence

application of Cauchy criterion for convergence

Without using the methods of the entry

determining series convergence

, we show that the real-term series

∑

n

=

0

∞

1

n

!

=

1

+

1

1

!

+

1

2

!

+

…

is

convergent

by using

Cauchy criterion for convergence

, being in in

ℝ

equipped with the usual

absolute value

|

.

|

as

http://planetmath.org/node/1604

norm.

Let

ε

be an arbitrary positive number.  For any

positive integer

n

, we have

1

n

!

≦

1

1

⋅

2

⋅

2

⁢

⋯

⁢

2

=

1

2

n

-

1

,

whence we can as follows.

|

1

(

n

+

1

)

!

+

…

+

1

(

n

+

p

)

!

|

=

1

(

n

+

1

)

!

+

…

+

1

(

n

+

p

)

!

≦

1

2

n

+

…

+

1

2

n

+

p

-

1

=

1

2

n

⁢

(

1

+

1

2

+

…

+

1

2

p

-

1

)

=

1

2

n

⋅

1

-

(

1

/

2

)

p

1

-

1

/

2

<

1

2

n

-

1

<

ε

The last

inequality

is true for all positive integers

p

, when

n

>

1

-

lb

⁢

ε

.  Thus the

Cauchy criterion

implies that the series

converges

.

Title

application of Cauchy criterion for convergence

Canonical name

ApplicationOfCauchyCriterionForConvergence

Date of creation

2013-03-22 19:03:22

Last modified on

2013-03-22 19:03:22

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Example

Classification

msc 40A05

Related topic

RealNumber

Related topic

GeometricSeries

Related topic

LogarithmusBinaris

Related topic

NapiersConstant

Generated on Fri Feb 9 21:11:57 2018 by

LaTeXML
arithmetic series

arithmetic series

An

arithmetic series

is the series,

∑

i

=

1

n

a

i

, in which each real term has the form

a

i

=

a

i

-

1

+

d

for

i

=

2

,

…

,

n

where

d

is constant. The sum of the

sequence

is given by the following

1

2

⁢

n

⁢

[

2

⁢

a

1

+

d

⁢

(

n

-

1

)

]

.

In order to find the formula above firstly we express the terms of the sequence,

a

2

,

…

,

a

n

in terms of

a

1

and the constant

d

. In this case we get

a

2

=

a

1

+

d

,

a

3

=

a

2

+

2

⁢

d

,

…

,

a

n

=

a

1

+

(

n

-

1

)

⁢

d

. Now we express the sum of the sequence by developing the series forward and we have:

S

n

=

∑

i

=

1

n

a

i

=

a

1

+

a

1

+

d

+

⋯

+

a

1

+

(

n

-

2

)

⁢

d

+

a

1

+

(

n

-

1

)

⁢

d

Reversely, we develop the series backwards and we get

S

n

=

a

n

-

d

+

a

n

-

2

⁢

d

+

⋯

+

a

n

-

(

n

-

1

)

⁢

d

It is easily seen that by adding the two

expressions

we get

2

⁢

S

n

=

n

⁢

(

a

1

+

a

n

)

(1)

S

n

=

1

2

⁢

n

⁢

(

a

1

+

a

n

)

(2)

Hence, by substituting

a

n

=

a

1

+

(

n

-

1

)

⁢

d

we get the first formula.

Title

arithmetic series

Canonical

name

ArithmeticSeries

Date of creation

2013-03-22 16:17:58

Last modified on

2013-03-22 16:17:58

Owner

georgiosl (7242)

Last modified by

georgiosl (7242)

Numerical id

10

Author

georgiosl (7242)

Entry type

Definition

Classification

msc 40A05

Generated on Fri Feb 9 21:11:59 2018 by

LaTeXML
Baroni’s theorem

Baroni’s theorem

Let

(

x

n

)

n

≥

0

be a sequence of

real numbers

such that

lim

n

→

∞

⁡

(

x

n

+

1

-

x

n

)

=

0

. Let

A

=

{

x

n

|

n

∈

ℕ

}

and A’ the set of

limit points

of

A

.
Then A’ is a (possibly degenerate)

interval

from

ℝ

¯

, where

ℝ

¯

=

ℝ

⁢

⋃

{

-

∞

,

+

∞

}

Title

Baroni’s theorem

Canonical name

BaronisTheorem

Date of creation

2013-03-22 13:32:30

Last modified on

2013-03-22 13:32:30

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

6

Author

mathwizard (128)

Entry type

Theorem

Classification

msc 40A05

Generated on Fri Feb 9 21:12:00 2018 by

LaTeXML
Baye’s rule

Baye’s rule

Baye’s rule

Theorem

:

Let

μ

and

ν

be two

probability measures

on a

measurable space

(

Ω

,

𝒢

)

such that

d

⁢

ν

⁢

(

ω

)

=

f

⁢

(

ω

)

⁢

d

⁢

μ

⁢

(

ω

)

for some

f

∈

L

1

⁢

(

μ

)

.
Let

X

be a

random variable

on

(

Ω

,

𝒢

)

such that

E

ν

⁢

[

|

X

|

]

=

∫

Ω

|

X

⁢

(

ω

)

|

⁢

f

⁢

(

ω

)

⁢

𝑑

μ

⁢

(

ω

)

<

∞

(

ν

-integrable)

Let

ℋ

be a

σ

-

algebra

,

ℋ

⊂

𝒢

. Then,

E

ν

[

X

|

ℋ

]

.

E

μ

[

f

|

ℋ

]

=

E

μ

[

f

X

|

ℋ

]

a

.

s

.

or

E

ν

[

X

|

ℋ

]

=

E

μ

[

f

X

|

ℋ

]

E

μ

[

f

|

ℋ

]

a

.

s

.

Title

Baye’s rule

Canonical name

BayesRule

Date of creation

2013-03-11 19:52:58

Last modified on

2013-03-11 19:52:58

Owner

renato (9974)

Last modified by

(0)

Numerical id

1

Author

renato (0)

Entry type

Definition

Generated on Fri Feb 9 21:12:02 2018 by

LaTeXML
Cauchy criterion for convergence

Cauchy criterion for convergence

A series

∑

i

=

0

∞

a

i

in a

Banach space

(

V

,

∥

⋅

∥

)

is

http://planetmath.org/node/2311

convergent

iff for every

ε

>

0

there is a number

N

∈

ℕ

such that

∥

a

n

+

1

+

a

n

+

2

+

⋯

+

a

n

+

p

∥

<

ε

holds for all

n

>

N

and

p

≥

1

.

Proof:

First define

s

n

:=

∑

i

=

0

n

a

i

.

Now, since

V

is

complete

,

(

s

n

)

converges

if and only if it is a

Cauchy sequence

, so if for every

ε

>

0

there is a number

N

, such that for all

n

,

m

>

N

holds:

∥

s

m

-

s

n

∥

<

ε

.

We can assume

m

>

n

and thus set

m

=

n

+

p

. The series is iff

∥

s

n

+

p

-

s

n

∥

=

∥

a

n

+

1

+

a

n

+

2

+

⋯

+

a

n

+

p

∥

<

ε

.

Title

Cauchy criterion for convergence

Canonical name

CauchyCriterionForConvergence

Date of creation

2013-03-22 13:22:03

Last modified on

2013-03-22 13:22:03

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

14

Author

mathwizard (128)

Entry type

Theorem

Classification

msc 40A05

Generated on Fri Feb 9 21:12:04 2018 by

LaTeXML
Cauchy’s root test

Cauchy’s root test

If

∑

a

n

is a series of

positive

real terms and

a

n

n

<

k

<

1

for all

n

>

N

, then

∑

a

n

is

convergent

. If

a

n

n

≥

1

for an infinite number of values of

n

, then

∑

a

n

is divergent.

Limit form

Given a series

∑

a

n

of complex terms, set

ρ

=

lim sup

n

→

∞

⁡

|

a

n

|

n

The series

∑

a

n

is absolutely convergent if

ρ

<

1

and is divergent if

ρ

>

1

. If

ρ

=

1

, then the test is inconclusive.

Title

Cauchy’s root test

Canonical name

CauchysRootTest

Date of creation

2013-03-22 12:58:03

Last modified on

2013-03-22 12:58:03

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

9

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 40A05

Synonym

root test

Related topic

LambertSeries

Generated on Fri Feb 9 21:12:06 2018 by

LaTeXML
characterization of convergence of sequences in metric spaces

characterization of convergence of sequences in metric spaces

Let

(

M

,

d

)

be a

metric space

, and let

ℕ

=

P

1

⁢

⋃

⋯

⁢

⋃

P

k

be a

partition

of the set of natural numbers such that

P

i

is

infinite

for every

i

, that is, there is a

bijection

f

i

:

ℕ

→

P

i

. Then, given a

sequence

(

x

n

)

n

∈

ℕ

, it

converges

to

x

∈

M

if and only if the

subsequence

(

x

f

i

⁢

(

n

)

)

n

converges to

x

for every

i

=

1

,

⋯

,

k

.

Examples

If you have a sequence

(

x

n

)

n

and a

natural number

k

, and you know that it converges to

x

for every corresponding subsequence over the classes of

remainders

modulo

k

, then it converges to

x

.

Title

characterization of convergence of sequences in metric spaces

Canonical

name

CharacterizationOfConvergenceOfSequencesInMetricSpaces

Date of creation

2013-03-22 15:06:10

Last modified on

2013-03-22 15:06:10

Owner

gumau (3545)

Last modified by

gumau (3545)

Numerical id

4

Author

gumau (3545)

Entry type

Theorem

Classification

msc 40A05

Classification

msc 54E35

Generated on Fri Feb 9 21:12:08 2018 by

LaTeXML
comparison test

comparison test

The series

∑

i

=

0

∞

a

i

with real

a

i

is

absolutely convergent

if there is a

sequence

(

b

n

)

n

∈

ℕ

with positive real

b

n

such that

∑

i

=

0

∞

b

i

is and for all sufficiently large

k

holds

|

a

k

|

≤

b

k

.

Also, the series

∑

a

i

is divergent if there is a sequence

(

b

n

)

with positive real

b

n

, so that

∑

b

i

is divergent and

a

k

≥

b

k

for all sufficiently large

k

.

Title

comparison test

Canonical name

ComparisonTest

Date of creation

2013-03-22 13:21:48

Last modified on

2013-03-22 13:21:48

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

4

Author

mathwizard (128)

Entry type

Theorem

Classification

msc 40A05

Generated on Fri Feb 9 21:12:18 2018 by

LaTeXML
convergence of a sequence with finite upcrossings

convergence of a sequence with finite upcrossings

The following result characterizes

convergence

of a

sequence

in terms of finiteness of numbers of

upcrossings

.

Theorem

.

A sequence

x

1

,

x

2

,

…

of

real numbers

converges

to a limit in the

extended real numbers

if and only if the number of upcrossings

U

⁢

[

a

,

b

]

is finite for all

a

<

b

.

Since the number of upcrossings

U

⁢

[

a

,

b

]

differs from the number of downcrossings

D

⁢

[

a

,

b

]

by at most one, the theorem can equivalently be stated in terms of the finiteness of

D

⁢

[

a

,

b

]

.

Title

convergence of a sequence with finite upcrossings

Canonical name

ConvergenceOfASequenceWithFiniteUpcrossings

Date of creation

2013-03-22 18:49:36

Last modified on

2013-03-22 18:49:36

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Theorem

Classification

msc 40A05

Classification

msc 60G17

Related topic

UpcrossingsAndDowncrossings

Generated on Fri Feb 9 21:12:28 2018 by

LaTeXML
convergent series where not onlya_n but also n⁢a_n tends to 0

convergent series where not only

a

n

but also

n

⁢

a

n

tends to 0

Proposition.

If the terms (

http://planetmath.org/Series

)

a

n

of the

convergent series

a

1

+

a

2

+

…

are

positive

and form a

monotonically decreasing

sequence

, then

lim

n

→

∞

⁡

n

⁢

a

n

=

0

.

(1)

Proof.

Let

ε

be any positive number.  By the

Cauchy criterion for convergence

and the positivity of the terms, there is a

positive integer

m

such that

0

<

a

m

+

1

+

…

+

a

m

+

p

<

ε

2

(

p

=

1

,

2

,

…

)

.

Since the sequence

a

1

,

a

2

,

…

is

decreasing

, this implies

0

<

p

a

m

+

p

<

ε

2

(

p

=

1

,

2

,

…

)

.

(2)

Choosing here especially

p

:=

m

,  we get

0

<

m

⁢

a

m

+

m

<

ε

2

,

whence again due to the decrease,

0

<

m

a

m

+

p

<

ε

2

(

p

=

m

,

m

+

1

,

…

)

.

(3)

Adding the

inequalities

(2) and (3) with the common values

p

=

m

,

m

+

1

,

…

then yields

0

<

(

m

+

p

)

⁢

a

m

+

p

<

ε

for

p

≧

m

.

This may be written also in the form

0

<

n

⁢

a

n

<

ε

for

n

≧

2

⁢

m

which means that

lim

n

→

∞

⁡

n

⁢

a

n

=

0

.

Remark.

The assumption of monotonicity in the Proposition is essential.  I.e., without it, one cannot gererally get the limit result (1).  A

counterexample

would be the series

a

1

+

a

2

+

…

where

a

n

:=

1

n

for any

perfect square

n

but 0 for other values of

n

.  Then this series is

convergent

(cf. the

over-harmonic series

), but

n

⁢

a

n

=

1

for each perfect square

n

; so

n

⁢

a

n

↛

0

as

n

→

∞

.

Title

convergent series where not only

a

n

but also

n

⁢

a

n

tends to 0

Canonical name

ConvergentSeriesWhereNotOnlyanButAlsoNanTendsTo0

Date of creation

2013-03-22 19:03:29

Last modified on

2013-03-22 19:03:29

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Theorem

Classification

msc 40A05

Synonym

Olivier’s theorem

Related topic

NecessaryConditionOfConvergence

Related topic

AGeneralisationOfOlivierCriterion

Generated on Fri Feb 9 21:12:31 2018 by

LaTeXML
Cooperative and noncooperative elliptic systems

Cooperative and noncooperative elliptic systems

Cooperative and noncooperative elliptic systems

Cooperative (+) and noncooperative (-) elliptic systems are of the
form

(

P

±

)

⁢

{

-

Δ

⁢

u

=

λ

⁢

u

±

δ

⁢

v

+

G

u

⁢

(

x

;

u

,

v

)

,

x

∈

Ω

-

Δ

⁢

v

=

δ

⁢

u

+

γ

⁢

v

±

G

v

⁢

(

x

;

u

,

v

)

,

x

∈

Ω

u

=

v

=

0

⁢

or

⁢

∂

⁡

u

∂

⁡

n

=

∂

⁡

v

∂

⁡

n

=

0

x

∈

∂

⁡

Ω

where

Ω

⊂

ℝ

N

(

N

≥

1

)

is an open bounded domain,

λ

,

γ

,

δ

are real

parameters

,

G

⁢

(

x

;

u

,

v

)

∈

𝒞

1

⁢

(

Ω

¯

×

ℝ

2

;

ℝ

)

in the

variables

(

u

,

v

)

∈

ℝ

2

with

∇

⁡

G

=

(

G

u

,

G

v

)

.

P

+

and

P

-

are called cooperative and noncooperative, respectively. In

addition

,

δ

is assumed to
be

positive

for the noncooperative case. Note that systems

(

P

±

)

are closely related to reaction-diffusion systems arising in various chemical/physical and biological phenomena.

Title

Cooperative and noncooperative elliptic systems

Canonical

name

CooperativeAndNoncooperativeEllipticSystems1

Date of creation

2013-03-11 19:28:25

Last modified on

2013-03-11 19:28:25

Owner

linor (11198)

Last modified by

(0)

Numerical id

1

Author

linor (0)

Entry type

Definition

Generated on Fri Feb 9 21:12:34 2018 by

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determining series convergence

determining series convergence

Consider a series

Σ

⁢

a

n

. To determine whether

Σ

⁢

a

n

converges

or diverges, several tests are available. There is no precise rule indicating which of test to use with a given series. The more

obvious

approaches are collected below.

•

When the terms in

Σ

⁢

a

n

are

positive

, there are several possibilities:

–

comparison test

,

–

root test

(Cauchy’s root test),

–

ratio test

of d’Alembert (

http://planetmath.org/RatioTestOfDAlembert

),

–

ratio test,

–

p

-test (

http://planetmath.org/PTest

),

–

integral test

,

–

Raabe’s criteria.

•

limit comparison test

.

•

the

divergence test

(

http://planetmath.org/ThenA_kto0IfSum_k1inftyA_kConverges

).

•

If the series is an

alternating series

, then the

alternating series test

(

http://planetmath.org/AlternatingSeriesTest

) may be used.

•

Abel’s test for

convergence

can be used when terms in

Σ

⁢

a

n

can be obained as the product of terms of a

convergent series

with terms of a

monotonic

convergent sequence

.

The root test and the ratio test are direct applications of the comparison test to the

geometric series

with terms

(

|

a

n

|

)

1

/

n

and

a

n

+

1

a

n

, respectively.

For a paper about tests for convergence, please see

http://planetmath.org/?op=getobj&from=lec&id=37

this article.

Title

determining series convergence

Canonical

name

DeterminingSeriesConvergence

Date of creation

2013-03-22 13:24:45

Last modified on

2013-03-22 13:24:45

Owner

CWoo (3771)

Last modified by

CWoo (3771)

Numerical id

16

Author

CWoo (3771)

Entry type

Topic

Classification

msc 40A05

Related topic

ThenA_kto0IfSum_k1inftyA_kConverges

Related topic

LimitComparisonTest

Related topic

AbsoluteConvergence

Related topic

InfiniteProductOfSums1a_i

Generated on Fri Feb 9 21:12:52 2018 by

LaTeXML
difference equations

difference equations

difference equations

.

Title

difference equations

Canonical

name

DifferenceEquations1

Date of creation

2013-03-11 19:32:04

Last modified on

2013-03-11 19:32:04

Owner

srajakumar (16963)

Last modified by

(0)

Numerical id

1

Author

srajakumar (0)

Entry type

Definition

Generated on Fri Feb 9 21:12:54 2018 by

LaTeXML
Dirichlet’s convergence test

Dirichlet’s convergence test

Theorem. Let

{

a

n

}

and

{

b

n

}

be

sequences

of

real numbers

such that

{

∑

i

=

0

n

a

i

}

is

bounded

and

{

b

n

}

decreases with

0

as limit.
Then

∑

n

=

0

∞

a

n

⁢

b

n

converges

.

Proof. Let

A

n

:=

∑

i

=

0

n

a

n

and let

M

be an

upper bound

for

{

|

A

n

|

}

. By Abel’s lemma,

∑

i

=

m

n

a

i

⁢

b

i

=

∑

i

=

0

n

a

i

⁢

b

i

-

∑

i

=

0

m

-

1

a

i

⁢

b

i

=

∑

i

=

0

n

-

1

A

i

⁢

(

b

i

-

b

i

+

1

)

-

∑

i

=

0

m

-

2

A

i

⁢

(

b

i

-

b

i

+

1

)

+

A

n

⁢

b

n

-

A

m

-

1

⁢

b

m

-

1

=

∑

i

=

m

-

1

n

-

1

A

i

⁢

(

b

i

-

b

i

+

1

)

+

A

n

⁢

b

n

-

A

m

-

1

⁢

b

m

-

1

|

∑

i

=

m

n

a

i

⁢

b

i

|

≤

∑

i

=

m

-

1

n

-

1

|

A

i

⁢

(

b

i

-

b

i

+

1

)

|

+

|

A

n

⁢

b

n

|

+

|

A

m

-

1

⁢

b

m

-

1

|

≤

M

⁢

∑

i

=

m

-

1

n

-

1

(

b

i

-

b

i

+

1

)

+

|

A

n

⁢

b

n

|

+

|

A

m

-

1

⁢

b

m

-

1

|

Since

{

b

n

}

converges to

0

, there is an

N

⁢

(

ϵ

)

such that both

∑

i

=

m

-

1

n

-

1

(

b

i

-

b

i

+

1

)

<

ϵ

3

⁢

M

and

b

i

<

ϵ

3

⁢

M

for

m

,

n

>

N

⁢

(

ϵ

)

. Then, for

m

,

n

>

N

⁢

(

ϵ

)

,

|

∑

i

=

m

n

a

i

⁢

b

i

|

<

ϵ

and

∑

a

n

⁢

b

n

converges.

Title

Dirichlet’s

convergence test

Canonical name

DirichletsConvergenceTest

Date of creation

2013-03-22 13:19:53

Last modified on

2013-03-22 13:19:53

Owner

lieven (1075)

Last modified by

lieven (1075)

Numerical id

5

Author

lieven (1075)

Entry type

Theorem

Classification

msc 40A05

Generated on Fri Feb 9 21:12:58 2018 by

LaTeXML
double series

double series

Theorem

.

If the

double series

∑

m

=

1

∞

∑

n

=

1

∞

a

m

⁢

n

=

∑

n

=

1

∞

a

1

⁢

n

+

∑

n

=

1

∞

a

2

⁢

n

+

∑

n

=

1

∞

a

3

⁢

n

+

…

(1)

converges

and if it remains

convergent

when the of the partial series are replaced with their

absolute values

, i.e. if the series

∑

n

=

1

∞

|

a

1

⁢

n

|

+

∑

n

=

1

∞

|

a

2

⁢

n

|

+

∑

n

=

1

∞

|

a

3

⁢

n

|

+

…

(2)

has a finite sum

M

, then the

addition

in (1) can be performed in reverse , i.e.

∑

m

=

1

∞

∑

n

=

1

∞

a

m

⁢

n

=

∑

n

=

1

∞

∑

m

=

1

∞

a

m

⁢

n

=

∑

m

=

1

∞

a

m

⁢

1

+

∑

m

=

1

∞

a

m

⁢

2

+

∑

m

=

1

∞

a

m

⁢

3

+

…

Proof.

The

assumption

on (2) implies that the sum of an arbitrary finite amount of the numbers

|

a

m

⁢

n

|

is always

≦

M

.  This means that (1) is absolutely convergent, and thus the order of

summing

is insignificant.

Note.

The series satisfying the assumptions of the theorem is often denoted by

∑

m

,

n

=

1

∞

a

m

⁢

n

and this may by interpreted to an arbitrary summing .  One can use e.g. the

diagonal summing

:

a

11

+

a

12

+

a

21

+

a

13

+

a

22

+

a

31

+

…

Title

double series

Canonical name

DoubleSeries

Date of creation

2013-03-22 16:32:54

Last modified on

2013-03-22 16:32:54

Owner

PrimeFan (13766)

Last modified by

PrimeFan (13766)

Numerical id

6

Author

PrimeFan (13766)

Entry type

Theorem

Classification

msc 40A05

Classification

msc 26A06

Synonym

double series theorem

Related topic

FourierSineAndCosineSeries

Related topic

AbsoluteConvergenceOfDoubleSeries

Related topic

PerfectPower

Defines

diagonal summing

Generated on Fri Feb 9 21:13:00 2018 by

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every bounded sequence has limit along an ultrafilter

every bounded sequence has limit along an ultrafilter

Theorem 1

.

Let

F

be an

ultrafilter

on

N

and

(

x

n

)

be a real

bounded

sequence

. Then

F

⁢

-

⁢

lim

⁡

x

n

exists.

Proof.

Let

(

x

n

)

be a bounded sequence. Choose

a

0

and

b

0

such
that

a

0

≤

x

n

≤

b

0

. Put

c

0

:=

a

0

+

b

0

2

. Then
precisely one of the sets

{

n

∈

ℕ

;

x

n

∈

⟨

a

0

,

c

0

⟩

}

,

{

n

∈

ℕ

;

x

n

∈

⟨

c

0

,

b

0

⟩

}

belongs to the filter

ℱ

.
(Their union is

ℕ

and the filter

ℱ

is an ultrafilter.) We
choose

⟨

a

1

,

b

1

⟩

as that

subinterval

from

⟨

a

0

,

c

0

⟩

and

⟨

c

0

,

b

0

⟩

for which

C

:=

{

n

∈

ℕ

;

x

n

∈

⟨

a

1

,

b

1

⟩

}

belongs to

ℱ

.

Now we again bisect the

interval

⟨

a

1

,

b

1

⟩

by putting

c

1

=

a

1

+

b

1

2

. Denote

A

:=

{

n

∈

ℕ

;

x

n

∈

⟨

a

1

,

c

1

⟩

}

,

B

:=

{

n

∈

ℕ

;

x

n

∈

⟨

c

1

,

b

1

⟩

}

. It
holds

B

∪

A

∪

(

ℕ

∖

C

)

=

ℕ

. By the

alternative
characterization of ultrafilters

we get that one of these sets is
in

ℱ

. The set

ℕ

∖

C

doesn’t belong to

ℱ

, therefore
it must be one of the sets

A

and

B

. We choose the
corresponding interval for

⟨

a

2

,

b

2

⟩

.

By

induction

we obtain the monotonous sequences

(

a

n

)

,

(

b

n

)

with the same limit

lim

n

→

∞

⁡

a

n

=

lim

n

→

∞

⁡

b

n

:=

L

such that
for any

n

∈

ℕ

it holds

{

n

∈

ℕ

;

x

n

∈

⟨

a

1

,

b

1

⟩

}

∈

ℱ

.

We claim that

ℱ

⁢

-

⁢

lim

⁡

x

n

=

L

. Indeed, for any

ε

>

0

there is

n

∈

ℕ

such that

⟨

a

n

,

b

n

⟩

⊆

(

L

-

ε

,

L

+

ε

)

, thus

{

n

∈

ℕ

;

x

n

∈

⟨

a

n

,

b

n

⟩

}

⊆

A

(

ε

)

. The set

{

n

∈

ℕ

;

x

n

∈

⟨

a

1

,

b

1

⟩

}

belongs to

ℱ

, hence

A

⁢

(

ε

)

∈

ℱ

as well.
∎

Note that, if we modify the definition of

ℱ

-limit in a such way
that we admit the values

±

∞

, then every sequence has

ℱ

-limit along an ultrafilter

ℱ

. (The limit is

+

∞

if
for each

neighborhood

V

of

infinity

, the set

{

n

∈

ℕ

;

x

n

∈

V

}

belongs to

ℱ

. Similarly for

-

∞

.)

References

1

M. A. Alekseev, L. Yu. Glebsky, and E. I. Gordon,

On
approximations of
groups, group actions and Hopf algebras

, Journal of Mathematical Sciences

107

(2001), no. 5, 4305–4332.

2

B. Balcar and P. Štěpánek,

Teorie množin

,
Academia,
Praha, 1986 (Czech).

3

K. Hrbacek and T. Jech,

Introduction

to

set theory

,
Marcel Dekker,
New York, 1999.

Title

every bounded sequence has limit along an ultrafilter

Canonical name

EveryBoundedSequenceHasLimitAlongAnUltrafilter

Date of creation

2013-03-22 15:32:26

Last modified on

2013-03-22 15:32:26

Owner

kompik (10588)

Last modified by

kompik (10588)

Numerical id

4

Author

kompik (10588)

Entry type

Theorem

Classification

msc 40A05

Classification

msc 03E99

Related topic

Ultrafilter

Generated on Fri Feb 9 21:13:03 2018 by

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example of integral test

example of integral test

Consider the series

∑

k

=

1

∞

1

k

⁢

log

⁡

k

.

Since the integral

∫

1

∞

1

x

⁢

log

⁡

x

⁢

𝑑

x

=

lim

M

→

∞

⁡

[

log

⁡

(

log

⁡

(

x

)

)

]

1

M

is

divergent

also the series considered is divergent.

Title

example of

integral test

Canonical name

ExampleOfIntegralTest

Date of creation

2013-03-22 13:37:11

Last modified on

2013-03-22 13:37:11

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

5

Author

paolini (1187)

Entry type

Example

Classification

msc 40A05

Generated on Fri Feb 9 21:13:05 2018 by

LaTeXML