aarjaneiro/planetmath huggingface dataset:
subadditivity

subadditivity

A

sequence

{

a

n

}

n

=

1

∞

is called

subadditive

if
it

satisfies

the inequality

a

n

+

m

≤

a

n

+

a

m

for all

n

and

m

.

(1)

The major reason for use of subadditive sequences is the following
lemma due to Fekete.

Lemma

(

[

1

]

).

For every subadditive sequence

{

a

n

}

n

=

1

∞

the limit

lim

⁡

a

n

/

n

exists and is equal to

inf

⁡

a

n

/

n

.

Similarly, a

function

f

⁢

(

x

)

is

subadditive

if

f

⁢

(

x

+

y

)

≤

f

⁢

(

x

)

+

f

⁢

(

y

)

for all

x

and

y

.

The analogue of Fekete lemma holds for subadditive functions as
well.

There are

extensions

of Fekete lemma that do not require (

1

) to hold for all

m

and

n

. There are also results that allow one to deduce the rate of convergence to the limit whose

existence

is stated in Fekete lemma if some kind of both super- (

http://planetmath.org/Superadditivity

) and subadditivity is present. A good exposition of this topic may be found in

[

2

]

.

References

1

György Polya and Gábor Szegö.

Problems and

theorems

in

analysis

, volume 1.

1976.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0338.00001

Zbl
0338.00001.

2

Michael J. Steele.

Probability theory and

combinatorial optimization

, volume 69 of

CBMS-NSF Regional Conference Series in Applied Mathematics

.

SIAM, 1997.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0916.90233

Zbl
0916.90233.

Title

subadditivity

Canonical

name

Subadditivity

Date of creation

2013-03-22 13:52:22

Last modified on

2013-03-22 13:52:22

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

10

Author

bbukh (348)

Entry type

Definition

Classification

msc 39B62

Synonym

subadditive

Related topic

Superadditivity

Generated on Fri Feb 9 20:48:17 2018 by

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subdifferentiable mapping

subdifferentiable mapping

Let

X

be a

Banach space

, and let

X

*

be the

dual space

of

X

.
For a function

f

:

X

→

ℝ

, and

x

∈

X

, let
us define

∂

⁡

f

⁢

(

x

)

=

{

r

*

∈

X

*

:

f

⁢

(

x

)

-

f

⁢

(

y

)

≤

r

∗

⁢

(

x

-

y

)

⁢

for all

⁢

y

∈

X

}

.

If

∂

⁡

f

⁢

(

x

)

is non-empty, then

f

is

subdifferentiable

at

x

∈

X

, and if

∂

⁡

f

⁢

(

x

)

is non-empty for all

x

, then

f

is

subdifferentiable

[

1

,

2

]

.

References

1

C. Zalinescu,

Convex Analysis

in General Vector Spaces

,
World Scientific Publishing Company, 2002.

2

R.T. Rockafellar,

Convex Analysis

,
Princeton University Press, 1996.

Title

subdifferentiable mapping

Canonical name

SubdifferentiableMapping

Date of creation

2013-03-22 14:31:19

Last modified on

2013-03-22 14:31:19

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

13

Author

matte (1858)

Entry type

Definition

Classification

msc 39B62

Classification

msc 52-00

Generated on Fri Feb 9 20:48:19 2018 by

LaTeXML
superadditivity

superadditivity

A

sequence

{

a

n

}

n

=

1

∞

is called

superadditive

if
it

satisfies

the

inequality

a

n

+

m

≥

a

n

+

a

m

for all

n

and

m

.

(1)

The major reason for use of superadditive sequences is the following
lemma due to Fekete.

Lemma

(

[

1

]

).

For every superadditive sequence

{

a

n

}

n

=

1

∞

the limit

lim

⁡

a

n

/

n

exists and is equal to

sup

⁡

a

n

/

n

.

Similarly, a

function

f

⁢

(

x

)

is

superadditive

if

f

⁢

(

x

+

y

)

≥

f

⁢

(

x

)

+

f

⁢

(

y

)

for all

x

and

y

.

The analogue of Fekete lemma holds for superadditive functions as
well.

There are

extensions

of Fekete lemma that do not require (

1

) to hold for all

m

and

n

. There are also results that allow one to deduce the rate of convergence to the limit whose

existence

is stated in Fekete lemma if some kind of both super- and subadditivity is present. A good exposition of this topic may be found in

[

2

]

.

References

1

György Polya and Gábor Szegö.

Problems and

theorems

in

analysis

, volume 1.

1976.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0338.00001

Zbl 0338.00001.

2

Michael J. Steele.

Probability theory and

combinatorial optimization

, volume 69 of

CBMS-NSF Regional Conference Series in Applied Mathematics

.

SIAM, 1997.

http://www.emis.de/cgi-bin/zmen/ZMATH/en/quick.html?type=html&an=0916.90233

Zbl 0916.90233.

Title

superadditivity

Canonical name

Superadditivity

Date of creation

2013-03-22 13:52:25

Last modified on

2013-03-22 13:52:25

Owner

bbukh (348)

Last modified by

bbukh (348)

Numerical id

10

Author

bbukh (348)

Entry type

Definition

Classification

msc 39B62

Synonym

superadditive

Related topic

Subadditivity

Generated on Fri Feb 9 20:48:21 2018 by

LaTeXML
alternating series

alternating series

An

alternating series

is a series of the form

∑

i

=

0

∞

(

-

1

)

i

⁢

a

i

or

∑

i

=

0

∞

(

-

1

)

i

+

1

⁢

a

i

where

(

a

n

)

is a non-negative

sequence

. Loosely, this is just a series where the “alternate” between positive and negative.

For

convergence

issues, see the following entry:

alternating series test

(

http://planetmath.org/AlternatingSeriesTest

).

Title

alternating series

Canonical name

AlternatingSeries

Date of creation

2013-03-22 12:27:05

Last modified on

2013-03-22 12:27:05

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

6

Author

mathcam (2727)

Entry type

Definition

Classification

msc 40-00

Synonym

alternating series test

Related topic

LeibnizsTheorem

Related topic

ProofOfLeibnizsTheorem

Related topic

AlternatingSeriesTest

Generated on Fri Feb 9 21:09:28 2018 by

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Cauchy product

Cauchy product

Let

a

k

and

b

k

be two

sequences

of real or

complex numbers

for

k

∈

ℕ

0

(

ℕ

0

is the set of natural numbers containing zero).
The

Cauchy product

is defined by:

(

a

∘

b

)

⁢

(

k

)

=

∑

l

=

0

k

a

l

⁢

b

k

-

l

.

(1)

This is basically the convolution for two sequences.
Therefore the

product

of two series

∑

k

=

0

∞

a

k

,

∑

k

=

0

∞

b

k

is given by:

∑

k

=

0

∞

c

k

=

(

∑

k

=

0

∞

a

k

)

⋅

(

∑

k

=

0

∞

b

k

)

=

∑

k

=

0

∞

∑

l

=

0

k

a

l

⁢

b

k

-

l

.

(2)

A sufficient condition for the resulting series

∑

k

=

0

∞

c

k

to be

absolutely convergent

is that

∑

k

=

0

∞

a

k

and

∑

k

=

0

∞

b

k

both converge absolutely .

Title

Cauchy product

Canonical name

CauchyProduct

Date of creation

2013-03-22 13:37:14

Last modified on

2013-03-22 13:37:14

Owner

msihl (2134)

Last modified by

msihl (2134)

Numerical id

7

Author

msihl (2134)

Entry type

Definition

Classification

msc 40-00

Generated on Fri Feb 9 21:09:30 2018 by

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Cesàro mean

Cesàro mean

Definition

Let

{

a

n

}

n

=

0

∞

be a

sequence

of real (or possibly

complex numbers

).
The

Cesàro mean

of the sequence

{

a

n

}

is the sequence

{

b

n

}

n

=

0

∞

with

b

n

=

1

n

+

1

⁢

∑

i

=

0

n

a

i

.

(1)

0.0.1

Properties

1.

A key property of the Cesàro mean is that it has the same limit as the
original sequence (when this limit exists). In other words, if

{

a

n

}

and

{

b

n

}

are as above, and

a

n

→

a

, then

b

n

→

a

.
In particular, if

{

a

n

}

converges, then

{

b

n

}

converges too.

Title

Cesàro mean

Canonical name

CesaroMean

Date of creation

2013-03-22 12:29:54

Last modified on

2013-03-22 12:29:54

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

11

Author

mathcam (2727)

Entry type

Definition

Classification

msc 40-00

Classification

msc 40G05

Synonym

Cesaro mean

Related topic

Sequence

Related topic

CesaroSummability

Related topic

StolzCesaroTheorem

Generated on Fri Feb 9 21:09:32 2018 by

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converging alternating series not satisfying all Leibniz’ conditions

converging alternating series not satisfying all Leibniz’ conditions

The

alternating series

∑

n

=

1

∞

(

-

1

)

n

-

1

n

+

(

-

1

)

n

-

1

=

1

2

-

1

1

+

1

4

-

1

3

+

1

6

-

1

5

+

-

…

(1)

satisfies the other requirements of

Leibniz test

except the monotonicity of the

absolute values

of the terms.  The

convergence

may however be shown by manipulating the terms as follows.

We first multiply the

numerator

and the denominator of the general term by the

difference

n

-

(

-

1

)

n

-

1

, getting from (1)

∑

n

=

1

∞

(

-

1

)

n

-

1

n

+

(

-

1

)

n

-

1

=

1

2

+

∑

n

=

2

∞

n

-

(

-

1

)

n

-

1

n

2

-

1

⁢

(

-

1

)

n

-

1

=

1

2

+

∑

n

=

2

∞

(

(

-

1

)

n

-

1

⁢

n

n

2

-

1

-

1

n

2

-

1

)

.

(2)

One can that the series

∑

n

=

2

∞

(

-

1

)

n

-

1

⁢

n

n

2

-

1

(3)

satisfies all requirements of Leibniz test and thus is

convergent

.  Since

0

<

1

n

2

-

1

<

1

n

2

-

1

2

⁢

n

2

=

2

⋅

1

n

2

for

n

≧

2

,

and the

over-harmonic series

∑

n

=

2

∞

1

n

2

converges

, the

comparison test

guarantees the convergence of the series

∑

n

=

2

∞

1

n

2

-

1

.

(4)

Therefore the difference series of (3) and (4) and consequently, by (2), the given series (1) is convergent.

Title

converging alternating series not satisfying all Leibniz’ conditions

Canonical name

ConvergingAlternatingSeriesNotSatisfyingAllLeibnizConditions

Date of creation

2013-03-22 19:00:45

Last modified on

2013-03-22 19:00:45

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 40-00

Classification

msc 40A05

Related topic

SumOfSeriesDependsOnOrder

Related topic

LeibnizEstimateForAlternatingSeries

Generated on Fri Feb 9 21:09:35 2018 by

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example of converging increasing sequence

example of converging increasing sequence

Let

a

be a positive real number and

q

an

integer

greater

than 1.  Set

x

1

:=

a

q

,

x

2

:=

a

+

x

1

q

=

a

+

a

q

q

,

x

3

:=

a

+

x

2

q

=

a

+

a

+

a

q

q

q

,

and generally

x

n

:=

a

+

x

n

-

1

q

.

(1)

Since

x

1

>

0

,  the two first above equations imply that

x

1

<

x

2

.  By induction on

n

one can show that

x

1

<

x

2

<

x

3

<

…

<

x

n

<

…

The numbers

x

n

are all below a finite bound

M

.  For demonstrating this, we write the

inequality

x

n

<

x

n

+

1

in the form

x

n

<

a

+

x

n

q

, which implies

x

n

q

<

a

+

x

n

,  i.e.

x

n

q

-

x

n

-

a

<

0

(2)

for all

n

.  We study the

polynomial

f

⁢

(

x

)

:=

x

q

-

x

-

a

=

x

⁢

(

x

q

-

1

-

1

)

-

1

.

From its latter form we see that the function

f

attains

negative

values when

0

≦

x

≦

1

and that

f

increases

monotonically

and boundlessly when

x

increases from 1 to

∞

.  Because

f

as a polynomial function is also

continuous

, we infer that the equation

x

q

-

x

-

a

=

0

(3)

has exactly one root (

http://planetmath.org/Equation

)

x

=

M

>

1

,  and that

f

is negative for

0

<

x

<

1

and positive for

x

>

M

.  Thus we can conclude by (2) that

x

n

<

M

for all values of

n

.

The proven facts

x

1

<

x

2

<

x

3

<

…

<

x

n

<

…

<

M

settle, by the theorem of the parent entry (

http://planetmath.org/NondecreasingSequenceWithUpperBound

), that the

sequence

x

1

,

x

2

,

x

3

,

…

,

x

n

,

…

converges

to a limit

x

′

≦

M

.

Taking limits of both sides of (1) we see that

x

′

=

a

+

x

′

q

,  i.e.

x

′

⁣

q

-

x

′

-

a

=

0

,  which means that

x

′

=

M

,  in other words: the limit of the sequence is the only

M

of the equation (3).

References

1

E. Lindelöf:

Johdatus korkeampaan analyysiin

. Neljäs painos.  Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).

Title

example of converging

increasing sequence

Canonical name

ExampleOfConvergingIncreasingSequence

Date of creation

2013-03-22 17:40:44

Last modified on

2013-03-22 17:40:44

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Example

Classification

msc 40-00

Related topic

NthRoot

Related topic

BolzanosTheorem

Generated on Fri Feb 9 21:09:38 2018 by

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examples for limit comparison test

examples for limit comparison test

Example 1

.

Does the following series

converge

?

∑

n

=

1

∞

1

n

2

+

n

+

1

The series is similar to

∑

n

=

1

∞

1

/

n

2

which converges (use

p

-test, for example). Next we compute the limit:

lim

n

→

∞

⁡

1

n

2

+

n

+

1

1

n

2

=

lim

n

→

∞

⁡

n

2

n

2

+

n

+

1

=

1

Therefore, since

1

≠

0

, by the

Limit Comparison Test

(with

a

n

=

1

/

(

n

2

+

n

+

1

)

and

b

n

=

1

/

n

2

), the series converges.

Example 2

.

Does the following series converge?

∑

n

=

1

∞

n

3

+

n

+

1

n

4

+

n

+

1

If we “forget” about the lower order terms of

n

:

n

3

+

n

+

1

n

4

+

n

+

1

∼

n

3

n

4

=

1

n

and

∑

n

=

1

∞

1

/

n

is the

harmonic series

which diverges (by the

p

-test). Thus, we take

b

n

=

1

/

n

and compute:

lim

n

→

∞

⁡

n

3

+

n

+

1

n

4

+

n

+

1

1

n

=

lim

n

→

∞

⁡

n

⁢

(

n

3

+

n

+

1

)

n

4

+

n

+

1

=

lim

n

→

∞

⁡

n

4

+

n

2

+

n

n

4

+

n

+

1

=

lim

n

→

∞

⁡

1

+

1

/

n

2

+

1

/

n

3

1

+

1

/

n

3

+

1

/

n

4

=

1

Therefore the series diverges like the harmonic does.

Title

examples for limit comparison test

Canonical name

ExamplesForLimitComparisonTest

Date of creation

2013-03-22 15:08:48

Last modified on

2013-03-22 15:08:48

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Example

Classification

msc 40-00

Generated on Fri Feb 9 21:09:40 2018 by

LaTeXML
examples using comparison test without limit

examples using comparison test without limit

Do the following series

converge

?

∑

n

=

1

∞

1

n

2

+

n

+

1

(1)

∑

n

=

1

∞

n

3

+

n

+

1

n

4

+

n

+

1

(2)

The general of (1) may be estimated upwards:

0

<

1

n

2

+

n

+

1

<

1

n

2

+

0

+

0

=

1

n

2

Because

∑

n

=

1

∞

1

n

2

(an

over-harmonic series

) converges, then also (1) converges.

The general of (2) may be estimated downwards:

n

3

+

n

+

1

n

4

+

n

+

1

>

n

3

+

0

+

0

n

4

+

n

4

+

n

4

=

1

3

⋅

1

n

>

0

Because

∑

n

=

1

∞

1

3

⁢

1

n

(the

harmonic series

divided by 3) diverges, then also (2) diverges.

Title

examples using comparison test without limit

Canonical name

ExamplesUsingComparisonTestWithoutLimit

Date of creation

2013-03-22 15:08:55

Last modified on

2013-03-22 15:08:55

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

9

Author

pahio (2872)

Entry type

Example

Classification

msc 40-00

Related topic

PTest

Defines

over-harmonic series

Generated on Fri Feb 9 21:09:42 2018 by

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Fibonacci polynomials

Fibonacci polynomials

Fibonacci polynomials

are defined recursively as follow

F

n

⁢

(

x

)

:=

{

1

if

⁢

n

=

1

;

x

if

⁢

n

=

2

;

x

⁢

F

n

-

1

⁢

(

x

)

+

F

n

-

2

⁢

(

x

)

,

if

⁢

n

>

2

.

The main (and

obvious

)

property

of this

polynomials

is that

F

n

⁢

(

1

)

is

n

th

number in

Fibonacci sequence

.

Title

Fibonacci polynomials

Canonical

name

FibonacciPolynomials

Date of creation

2013-03-22 18:03:39

Last modified on

2013-03-22 18:03:39

Owner

veselin (20506)

Last modified by

veselin (20506)

Numerical id

4

Author

veselin (20506)

Entry type

Definition

Classification

msc 40-00

Generated on Fri Feb 9 21:09:43 2018 by

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Furstenberg-Kesten theorem

Furstenberg-Kesten theorem

Consider

μ

a

probability measure

, and

f

:

M

→

M

a

measure preserving

dynamical system

. Consider

A

:

M

→

G

⁢

L

⁢

(

d

,

𝐑

)

, a

measurable

transformation, where GL(d,

R

) is the space of

invertible

square matrices

of size

d

.
Consider the multiplicative cocycle

(

ϕ

n

⁢

(

x

)

)

n

defined by the transformation

A

.

If

log

+

⁡

|

|

A

|

|

is

integrable

, where

log

+

⁡

|

|

A

|

|

=

max

⁡

{

log

⁡

|

|

A

|

|

,

0

}

, then:

λ

max

⁢

(

x

)

=

lim

n

⁡

1

n

⁢

log

⁡

|

|

ϕ

n

⁢

(

x

)

|

|

exists almost everywhere, and

λ

max

+

is integrable and

∫

λ

max

⁢

𝑑

μ

=

lim

n

⁡

1

n

⁢

∫

log

⁡

|

|

ϕ

n

|

|

⁢

d

⁢

μ

=

inf

n

⁡

1

n

⁢

∫

log

⁡

|

|

ϕ

n

|

|

⁢

d

⁢

μ

If

log

+

⁡

|

|

A

-

1

|

|

is integrable, then:

λ

min

⁢

(

x

)

=

lim

n

-

1

n

⁢

log

⁡

|

|

ϕ

-

n

⁢

(

x

)

|

|

exists almost everywhere, and

λ

min

+

is integrable and

∫

λ

min

⁢

𝑑

μ

=

lim

n

-

1

n

⁢

∫

log

⁡

|

|

ϕ

-

n

|

|

⁢

d

⁢

μ

=

sup

n

-

1

n

⁢

∫

log

⁡

|

|

ϕ

-

n

|

|

⁢

d

⁢

μ

Furthermore, both

λ

min

and

λ

max

are

invariant

for the tranformation

f

, that is,

λ

min

∘

f

⁢

(

x

)

=

λ

min

⁢

(

x

)

and

λ

max

∘

f

⁢

(

x

)

=

λ

max

⁢

(

x

)

, for

μ

almost everywhere.

This theorem is a direct consequence of Kingman’s subadditive

ergodic theorem

, by observing that both

log

⁡

|

|

ϕ

n

⁢

(

x

)

|

|

and

log

⁡

|

|

ϕ

-

n

⁢

(

x

)

|

|

are subadditive sequences.

The results in this theorem are strongly improved by Oseledet’s multiplicative ergodic theorem, or Oseledet’s

decomposition

.

Title

Furstenberg-Kesten theorem

Canonical name

FurstenbergKestenTheorem

Date of creation

2014-03-19 22:14:18

Last modified on

2014-03-19 22:14:18

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

3

Author

Filipe (28191)

Entry type

Theorem

Related topic

Oseledet’s decomposition

Related topic

multiplicative cocycle

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geometric sequence

geometric sequence

A

sequence

of the form

a

,

a

⁢

r

,

a

⁢

r

2

,

a

⁢

r

3

,

…

of real or

complex numbers

is called

geometric sequence

.  of the geometric sequence is thus that every two consecutive members of the sequence have the constant ratio

r

, called usually the

common ratio

of the sequence (if

a

⁢

r

=

0

, speaking the ratio of members does not exist).

The

n

th

member of the geometric sequence has the

a

n

=

a

⁢

r

n

-

1

.

Let

a

≠

0

.  The sequence is

convergent

for

|

r

|

<

1

having the limit (

http://planetmath.org/LimitOfRealNumberSequence

) 0, and for

r

=

1

having as constant sequence the limit

a

.

When the members of the sequence are positive numbers, each member is the

geometric mean

of the preceding and the following member; the name “geometric sequence”(or “

geometric series

”) is due to this fact (a fact is true for the

harmonic series

and

harmonic mean

).

Title

geometric sequence

Canonical name

GeometricSequence

Date of creation

2013-03-22 14:38:52

Last modified on

2013-03-22 14:38:52

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

14

Author

pahio (2872)

Entry type

Definition

Classification

msc 40-00

Related topic

GeometricSeries

Related topic

LimitOfRealNumberSequence

Defines

common ratio

Generated on Fri Feb 9 21:09:47 2018 by

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if ∑@⁢ERROR(\slimits)⁢@⁢@⁢@_{k=1}^∞⁢a_k converges then a_k→0

if

∑

@

⁢

\slimits

⁢

@

⁢

@

⁢

@

k

=

1

∞

⁢

a

k

converges then

a

k

→

0

Theorem 1

.

Suppose

a

1

,

a

2

,

…

is a

sequence

of real or

complex numbers

.
If the series

∑

k

=

1

∞

a

k

converges

, then

lim

k

→

∞

⁡

a

k

=

0

.

Remarks

1.

The harmonic series

∑

k

=

1

∞

1

/

k

shows that the

implication

can not be reversed.

2.

This result can be used as a first test for convergence of a series

∑

k

=

1

∞

a

k

. If

a

k

does not converge to

0

, then

∑

k

=

1

∞

a

k

does not converge either.

Proof.

Let

S

∈

ℂ

be the value of the sum, and let

ε

>

0

be arbitrary. Then there exists an

N

≥

1

such that

|

∑

k

=

1

M

a

k

-

S

|

<

ε

2

for all

M

≥

N

. For

j

≥

N

we then have

|

a

j

+

1

|

=

|

∑

k

=

1

j

+

1

a

k

-

∑

k

=

1

j

a

k

|

≤

|

∑

k

=

1

j

+

1

a

k

-

S

|

+

|

S

-

∑

k

=

1

j

a

k

|

<

ε

,

and the claim follows.
∎

Title

if

∑

@

⁢

\slimits

⁢

@

⁢

@

⁢

@

k

=

1

∞

⁢

a

k

converges then

a

k

→

0

Canonical name

Ifsumk1inftyAkConvergesThenAkto0

Date of creation

2013-03-22 15:00:38

Last modified on

2013-03-22 15:00:38

Owner

matte (1858)

Last modified by

matte (1858)

Numerical id

13

Author

matte (1858)

Entry type

Theorem

Classification

msc 40-00

Synonym

necessary condition of convergence

Related topic

DeterminingSeriesConvergence

Related topic

CompleteUltrametricField

Related topic

ConvergenceConditionOfInfiniteProduct

Related topic

LambertSeries

Related topic

AbsoluteConvergenceOfIntegralAndBoundednessOfDerivative

Related topic

ConvergentSeriesWhereNotOnlyA_nButAlsoNa_nTendsTo0

Generated on Fri Feb 9 21:09:50 2018 by

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limit comparison test

limit comparison test

The following theorem is a powerful test for

convergence

of series.

Theorem 1

(Limit ).

Let

∑

n

=

0

∞

a

n

and

∑

n

=

0

∞

b

n

be two series of positive numbers.

1.

If the limit

lim

n

→

∞

⁡

a

n

b

n

=

L

exists and

L

≠

0

is a
non-zero finite number, then both series

∑

n

=

0

∞

a

n

and

∑

n

=

0

∞

b

n

converge

or both diverge.

2.

If

L

=

0

and

∑

n

=

0

∞

b

n

converges then

∑

n

=

0

∞

a

n

converges as well. If

L

=

0

and

∑

n

=

0

∞

a

n

diverges then

∑

n

=

0

∞

b

n

diverges as well.

3.

Similarly, if the limit is infinite (“

L

=

∞

”) and

∑

n

=

0

∞

a

n

converges then

∑

n

=

0

∞

b

n

converges as well. If

L

=

∞

and

∑

n

=

0

∞

b

n

diverges then

∑

n

=

0

∞

a

n

diverges as well.

Title

limit comparison test

Canonical name

LimitComparisonTest

Date of creation

2013-03-22 15:01:31

Last modified on

2013-03-22 15:01:31

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

4

Author

alozano (2414)

Entry type

Theorem

Classification

msc 40-00

Related topic

DeterminingSeriesConvergence

Related topic

SequenceDeterminingConvergenceOfSeries

Generated on Fri Feb 9 21:09:51 2018 by

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limit of geometric sequence

limit of geometric sequence

As mentionned in the

geometric sequence

entry,

lim

n

→

∞

⁡

a

⁢

r

n

=

0

(1)

for

|

r

|

<

1

.  We will prove this for real or

complex

values of

r

.

We first remark, that for the values

s

>

1

we have

lim

n

→

∞

⁡

s

n

=

∞

(cf.

limit of real number sequence

).  In fact, if

M

is an arbitrary positive number, the

binomial theorem

(or Bernoulli’s

inequality

) implies that

s

n

=

(

1

+

s

-

1

)

n

>

1

n

+

(

n

1

)

⁢

(

s

-

1

)

=

1

+

n

⁢

(

s

-

1

)

>

n

⁢

(

s

-

1

)

>

M

as soon as

n

>

M

s

-

1

.

Let now

|

r

|

<

1

and

ε

be an arbitrarily small positive number.  Then

|

r

|

=

1

s

with

s

>

1

.  By the above remark,

|

r

n

|

=

|

r

|

n

=

1

s

n

<

1

n

⁢

(

s

-

1

)

<

ε

when

n

>

1

(

s

-

1

)

⁢

ε

.  Hence,

lim

n

→

∞

⁡

r

n

=

0

,

which easily implies (1) for any

real number

a

.

Title

limit of geometric sequence

Canonical name

LimitOfGeometricSequence

Date of creation

2013-03-22 18:32:43

Last modified on

2013-03-22 18:32:43

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Proof

Classification

msc 40-00

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