aarjaneiro/planetmath huggingface dataset:
Proof of Baroni’s theorem

Proof of Baroni’s theorem

Let

m

=

inf

⁡

A

′

and

M

=

sup

⁡

A

′

. If

m

=

M

we are done since the

sequence

is

convergent

and

A

′

is the degenerate interval composed of the point

l

∈

ℝ

¯

, where

l

=

lim

n

→

∞

⁡

x

n

.

Now , assume that

m

<

M

. For every

λ

∈

(

m

,

M

)

, we will construct
inductively two

subsequences

x

k

n

and

x

l

n

such that

lim

n

→

∞

⁡

x

k

n

=

lim

n

→

∞

⁡

x

l

n

=

λ

and

x

k

n

<

λ

<

x

l

n

From the

definition

of

M

there is an

N

1

∈

ℕ

such that :

λ

<

x

N

1

<

M

Consider the set of all such values

N

1

. It is

bounded from below

(because it
consists only of

natural numbers

and has at least one

element

) and thus it has a
smallest element . Let

n

1

be the smallest such element and from its definition we
have

x

n

1

-

1

≤

λ

<

x

n

1

. So , choose

k

1

=

n

1

-

1

,

l

1

=

n

1

.
Now, there is an

N

2

>

k

1

such that :

λ

<

x

N

2

<

M

Consider the set of all such values

N

2

. It is bounded from below and it has a
smallest element

n

2

. Choose

k

2

=

n

2

-

1

and

l

2

=

n

2

. Now , proceed by

induction

to construct the sequences

k

n

and

l

n

in the same fashion . Since

l

n

-

k

n

=

1

we have :

lim

n

→

∞

⁡

x

k

n

=

lim

n

→

∞

⁡

x

l

n

and
thus they are both equal to

λ

.

Title

Proof of Baroni’s

theorem

Canonical

name

ProofOfBaronisTheorem

Date of creation

2013-03-22 13:32:33

Last modified on

2013-03-22 13:32:33

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:16:48 2018 by

LaTeXML
proof of Bolzano-Weierstrass Theorem

proof of Bolzano-Weierstrass Theorem

To prove the

Bolzano-Weierstrass theorem

, we will first need two lemmas.

Lemma 1.

All bounded monotone sequences

converge

.

proof.

Let

(

s

n

)

be a bounded, nondecreasing

sequence

. Let

S

denote the set

{

s

n

:

n

∈

ℕ

}

. Then let

b

=

sup

⁡

S

(the supremum of

S

.)

Choose some

ϵ

>

0

. Then there is a corresponding

N

such that

s

N

>

b

-

ϵ

. Since

(

s

n

)

is nondecreasing, for all

n

>

N

,

s

n

>

b

-

ϵ

. But

(

s

n

)

is bounded, so we have

b

-

ϵ

<

s

n

≤

b

. But this implies

|

s

n

-

b

|

<

ϵ

, so

lim

⁡

s

n

=

b

.

□

(The proof for

nonincreasing

sequences is analogous.)

Lemma 2.

Every sequence has a

monotonic

subsequence.

proof.

First a definition: call the

n

th term of a sequence

dominant

if it is

greater

than every term following it.

For the proof, note that a sequence

(

s

n

)

may have finitely many or infinitely many dominant terms.

First we suppose that

(

s

n

)

has infinitely many dominant terms. Form a subsequence

(

s

n

k

)

solely of dominant terms of

(

s

n

)

. Then

s

n

k

+

1

<

s

n

k

k

by definition of “dominant”, hence

(

s

n

k

)

is a decreasing (

monotone

) subsequence of (

s

n

).

For the second case, assume that our sequence

(

s

n

)

has only finitely many dominant terms. Select

n

1

such that

n

1

is beyond the last dominant term. But since

n

1

is not dominant, there must be some

m

>

n

1

such that

s

m

>

s

n

1

. Select this

m

and call it

n

2

. However,

n

2

is still not dominant, so there must be an

n

3

>

n

2

with

s

n

3

>

s

n

2

, and so on, inductively. The resulting sequence

s

1

,

s

2

,

s

3

,

…

is monotonic (nondecreasing).

□

proof of Bolzano-Weierstrass.

The proof of the Bolzano-Weierstrass theorem is now simple: let

(

s

n

)

be a bounded sequence. By Lemma 2 it has a monotonic subsequence. By Lemma 1, the subsequence converges.

□

Title

proof of Bolzano-Weierstrass Theorem

Canonical name

ProofOfBolzanoWeierstrassTheorem

Date of creation

2013-03-22 12:22:26

Last modified on

2013-03-22 12:22:26

Owner

akrowne (2)

Last modified by

akrowne (2)

Numerical id

5

Author

akrowne (2)

Entry type

Proof

Classification

msc 40A05

Classification

msc 26A06

Generated on Fri Feb 9 21:16:50 2018 by

LaTeXML
proof of Cauchy’s root test

proof of Cauchy’s root test

If for all

n

≥

N

a

n

n

<

k

<

1

then

a

n

<

k

n

<

1

.

Since

∑

i

=

N

∞

k

i

converges

so does

∑

i

=

N

∞

a

n

by the

comparison test

. If

a

n

n

>

1

then by comparison with

∑

i

=

N

∞

1

the series is

divergent

.

Absolute convergence

in case of nonpositive

a

n

can be proven in exactly the same way using

|

a

n

|

n

.

Title

proof of Cauchy’s

root test

Canonical name

ProofOfCauchysRootTest

Date of creation

2013-03-22 13:23:43

Last modified on

2013-03-22 13:23:43

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

5

Author

mathwizard (128)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:16:52 2018 by

LaTeXML
proof of comparison test

proof of comparison test

Assume

|

a

k

|

≤

b

k

for all

k

>

n

. Then we define

s

k

:=

∑

i

=

k

∞

|

a

i

|

and

t

k

:=

∑

i

=

k

∞

b

i

.

Obviously

s

k

≤

t

k

for all

k

>

n

. Since by assumption

(

t

k

)

is

http://planetmath.org/node/601

convergent

(

t

k

)

is

bounded

and so is

(

s

k

)

. Also

(

s

k

)

is

monotonic

and therefore . Therefore

∑

i

=

0

∞

a

i

is absolutely convergent.

Now assume

b

k

≤

a

k

for all

k

>

n

. If

∑

i

=

k

∞

b

i

is divergent then so is

∑

i

=

k

∞

a

i

because otherwise we could apply the test we just proved and show that

∑

i

=

0

∞

b

i

is convergent, which is is not by assumption.

Title

proof of comparison test

Canonical name

ProofOfComparisonTest

Date of creation

2013-03-22 13:22:06

Last modified on

2013-03-22 13:22:06

Owner

mathwizard (128)

Last modified by

mathwizard (128)

Numerical id

4

Author

mathwizard (128)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:16:53 2018 by

LaTeXML
proof of convergence of a sequence with finite upcrossings

proof of convergence of a sequence with finite upcrossings

We show that 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

.

Denoting the

infimum limit

and

supremum limit

by

l

=

lim inf

n

→

∞

⁡

x

n

,

u

=

lim sup

n

→

∞

⁡

x

n

,

then

l

≤

u

and the sequence converges to a limit if and only if

l

=

u

.

We first show that if the sequence converges then

U

⁢

[

a

,

b

]

is finite for

a

<

b

. If

l

>

a

then there is an

N

such that

x

n

>

a

for all

n

≥

N

. So, all upcrossings of

[

a

,

b

]

must start before time

N

, and we may conclude that

U

⁢

[

a

,

b

]

≤

N

is finite. On the other hand, if

l

≤

a

then

u

=

l

<

b

and we can infer that

x

n

<

b

for all

n

≥

N

and some

N

. Again, this gives

U

⁢

[

a

,

b

]

≤

N

.

Conversely, suppose that the sequence does not converge, so that

u

>

l

. Then choose

a

<

b

in the

interval

(

l

,

u

)

. For any

integer

n

, there is then an

m

>

n

such that

x

m

>

b

and an

m

>

n

with

x

m

<

a

. This allows us to define infinite sequences

s

k

,

t

k

by

t

0

=

0

and

s

k

=

inf

⁡

{

m

≥

t

k

-

1

:

X

m

<

a

}

,

t

k

=

inf

⁡

{

m

≥

s

k

:

X

m

>

b

}

,

for

k

≥

1

. Clearly,

s

1

<

t

1

<

s

2

<

⋯

and

x

s

k

<

a

<

b

<

x

t

k

for all

k

≥

1

, so

U

⁢

[

a

,

b

]

=

∞

.

Title

proof of

convergence

of a sequence with finite upcrossings

Canonical name

ProofOfConvergenceOfASequenceWithFiniteUpcrossings

Date of creation

2013-03-22 18:49:39

Last modified on

2013-03-22 18:49:39

Owner

gel (22282)

Last modified by

gel (22282)

Numerical id

4

Author

gel (22282)

Entry type

Proof

Classification

msc 40A05

Classification

msc 60G17

Generated on Fri Feb 9 21:16:56 2018 by

LaTeXML
proof of divergence of harmonic series (by grouping terms)

proof of divergence of harmonic series (by grouping terms)

The

harmonic series

can be shown to diverge by a simple argument involving grouping terms. Write

∑

n

=

1

2

M

1

n

=

∑

m

=

1

M

∑

n

=

2

m

-

1

+

1

2

m

1

n

.

Since

1

/

n

≥

1

/

N

when

n

≤

N

, we have

∑

n

=

2

m

-

1

+

1

2

m

1

n

≥

∑

n

=

2

m

-

1

+

1

2

m

2

-

m

=

(

2

m

-

2

m

-

1

)

⁢

2

-

m

=

1

2

Hence,

∑

n

=

1

2

M

1

n

≥

M

2

so the series diverges in the limit

M

→

∞

.

Title

proof of divergence of harmonic series (by grouping terms)

Canonical name

ProofOfDivergenceOfHarmonicSeriesbyGroupingTerms

Date of creation

2013-03-22 15:08:39

Last modified on

2013-03-22 15:08:39

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:16:58 2018 by

LaTeXML
proof of divergence of harmonic series (by splitting odd and even terms)

proof of divergence of harmonic series (by splitting odd and even terms)

Suppose that the series

∑

n

=

1

∞

1

/

n

converged. Since all the terms are

positive

,
we could regroup them as we please, in particular, split the series into two series, that of
even terms and that of odd terms:

∑

n

=

1

∞

1

n

=

∑

n

=

1

∞

1

2

⁢

n

+

∑

n

=

1

∞

1

2

⁢

n

-

1

Since

∑

n

=

1

∞

1

/

n

=

2

⁢

∑

n

=

1

∞

1

/

(

2

⁢

n

)

, we would conclude that

∑

n

=

1

∞

1

2

⁢

n

=

∑

n

=

1

∞

1

2

⁢

n

-

1

.

But

2

⁢

n

-

1

<

2

⁢

n

, hence

1

/

(

2

⁢

n

)

<

1

/

(

2

⁢

n

-

1

)

, so we would also have

∑

n

=

1

∞

1

2

⁢

n

<

∑

n

=

1

∞

1

2

⁢

n

-

1

,

which contradicts the previous

conclusion

. Thus, the

assumption

that the
series converged is untenable.

Title

proof of divergence of harmonic series (by splitting odd and even terms)

Canonical

name

ProofOfDivergenceOfHarmonicSeriesbySplittingOddAndEvenTerms

Date of creation

2013-03-22 17:38:26

Last modified on

2013-03-22 17:38:26

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

4

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 40A05

Generated on Fri Feb 9 21:17:00 2018 by

LaTeXML
proof of integral test

proof of integral test

Consider the function (see the definition of floor)

g

⁢

(

x

)

=

a

⌊

x

⌋

.

Clearly for

x

∈

[

n

,

n

+

1

)

, being

f

non

increasing

we have

g

⁢

(

x

+

1

)

=

a

n

+

1

=

f

⁢

(

n

+

1

)

≤

f

⁢

(

x

)

≤

f

⁢

(

n

)

=

a

n

=

g

⁢

(

x

)

hence

∫

M

+

∞

g

⁢

(

x

+

1

)

⁢

𝑑

x

=

∫

M

+

1

+

∞

g

⁢

(

x

)

⁢

𝑑

x

≤

∫

M

+

∞

f

⁢

(

x

)

≤

∫

M

+

∞

g

⁢

(

x

)

⁢

𝑑

x

.

Since the

integral

of

f

and

g

on

[

M

,

M

+

1

]

is finite we notice that

f

is

integrable

on

[

M

,

+

∞

)

if and only if

g

is integrable on

[

M

,

+

∞

)

.

On the other hand

g

is locally constant so

∫

n

n

+

1

g

⁢

(

x

)

⁢

𝑑

x

=

∫

n

n

+

1

a

n

⁢

𝑑

x

=

a

n

and hence for all

N

∈

ℤ

∫

N

+

∞

g

⁢

(

x

)

=

∑

n

=

N

∞

a

n

that is

g

is integrable on

[

N

,

+

∞

)

if and only if

∑

n

=

N

∞

a

n

is

convergent

.

But, again,

∫

M

N

g

⁢

(

x

)

⁢

𝑑

x

is finite hence

g

is integrable on

[

M

,

+

∞

)

if and only if

g

is integrable on

[

N

,

+

∞

)

and also

∑

n

=

0

N

a

n

is finite so

∑

n

=

0

∞

a

n

is convergent if and only if

∑

n

=

N

∞

a

n

is convergent.

Title

proof of integral test

Canonical name

ProofOfIntegralTest

Date of creation

2013-03-22 13:32:07

Last modified on

2013-03-22 13:32:07

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

4

Author

paolini (1187)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:17:02 2018 by

LaTeXML
proof of Leibniz’s theorem (using Dirichlet’s convergence test)

proof of Leibniz’s theorem (using Dirichlet’s convergence test)

Proof.

Let us define the

sequence

α

n

=

(

-

1

)

n

for

n

∈

ℕ

=

{

0

,

1

,

2

,

…

}

.

Then

∑

i

=

0

n

α

i

=

{

1

for even

⁢

n

,

0

for odd

⁢

n

,

so the sequence

∑

i

=

0

n

α

i

is

bounded

.
By

assumption

{

a

n

}

n

=

1

∞

is a bounded

decreasing
sequence

with limit

0

.
For

n

∈

ℕ

we set

b

n

:=

a

n

+

1

.
Using Dirichlet’s

convergence test

, it follows that the series

∑

i

=

0

∞

α

i

⁢

b

i

converges

. Since

∑

i

=

0

∞

α

i

⁢

b

i

=

∑

n

=

1

∞

(

-

1

)

n

+

1

⁢

a

n

,

the claim follows.

□

Title

proof of Leibniz’s

theorem

(using Dirichlet’s convergence test)

Canonical name

ProofOfLeibnizsTheoremusingDirichletsConvergenceTest

Date of creation

2013-03-22 13:22:17

Last modified on

2013-03-22 13:22:17

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

7

Author

mathcam (2727)

Entry type

Proof

Classification

msc 40A05

Related topic

AlternatingSeries

Generated on Fri Feb 9 21:17:03 2018 by

LaTeXML
proof of ratio test

proof of ratio test

Assume

k

<

1

. By

definition

∃

N

such that

n

>

N

⟹

|

a

n

+

1

a

n

-

k

|

<

1

-

k

2

⟹

|

a

n

+

1

a

n

|

<

1

+

k

2

<

1

i.e.

eventually

the series

|

a

n

|

becomes less than a

convergent

geometric series

, therefore a shifted

subsequence

of

|

a

n

|

converges

by the

comparison test

. Note that a general

sequence

b

n

converges iff a shifted subsequence of

b

n

converges. Therefore, by the

absolute convergence theorem

, the series

a

n

converges.

Similarly for

k

>

1

a shifted subsequence of

|

a

n

|

becomes

greater

than a geometric series tending to

∞

, and so also tends to

∞

. Therefore

a

n

diverges.

Title

proof of ratio test

Canonical

name

ProofOfRatioTest

Date of creation

2013-03-22 12:24:46

Last modified on

2013-03-22 12:24:46

Owner

vitriol (148)

Last modified by

vitriol (148)

Numerical id

6

Author

vitriol (148)

Entry type

Proof

Classification

msc 40A05

Classification

msc 26A06

Generated on Fri Feb 9 21:17:05 2018 by

LaTeXML
proof of slower divergent series

proof of slower divergent series

Let us show that, if the series

∑

i

=

1

∞

a

i

of

positive

terms is

divergent

, then Abel’s series

∑

i

=

1

∞

a

i

∑

j

=

1

i

a

j

also

diverges

.

Since the series

∑

i

=

1

∞

a

i

diverges, we can find an

increasing sequence

(

n

i

)

i

=

0

∞

of

integers

such that

∑

j

=

1

n

i

+

1

a

j

>

2

⁢

∑

j

=

1

n

i

a

j

for all

i

. By convention, set

n

0

=

0

. Then we can group the sum like so:

∑

i

=

1

n

m

a

i

∑

j

=

1

i

a

j

=

∑

i

=

0

m

-

1

∑

k

=

n

i

+

1

n

i

-

1

a

k

∑

j

=

1

k

a

j

Because

∑

j

=

1

k

a

j

≤

∑

j

=

1

n

i

-

1

a

j

, we have

∑

k

=

n

i

+

1

n

i

-

1

a

k

∑

j

=

1

k

a

j

≥

∑

k

=

n

i

+

1

n

i

-

1

a

k

∑

j

=

1

n

i

-

1

a

j

=

∑

k

=

n

i

+

1

n

i

-

1

a

k

∑

j

=

1

n

i

-

1

a

j

By the way we chose the

sequence

(

n

i

)

i

=

0

∞

, we have

2

⁢

∑

k

=

n

i

+

1

n

i

-

1

a

k

>

∑

j

=

1

n

i

-

1

a

j

and, hence,

∑

k

=

n

i

+

1

n

i

-

1

a

k

∑

j

=

1

k

a

j

≥

∑

k

=

n

i

+

1

n

i

-

1

a

k

∑

j

=

1

n

i

-

1

a

j

>

1

2

.

Therefore,

∑

i

=

1

n

m

a

i

∑

j

=

1

i

a

j

>

∑

i

=

1

n

m

1

2

=

m

2

,

so the sum diverges in the limit

m

→

∞

.

Title

proof of slower divergent series

Canonical name

ProofOfSlowerDivergentSeries

Date of creation

2013-03-22 15:08:42

Last modified on

2013-03-22 15:08:42

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

9

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:17:08 2018 by

LaTeXML
Proof of Stolz-Cesaro theorem

Proof of Stolz-Cesaro theorem

From the definition of

convergence

, for every

ϵ

>

0

there is

N

⁢

(

ϵ

)

∈

ℕ

such that

(

∀

)

⁡

n

≥

N

⁢

(

ϵ

)

, we have :

l

-

ϵ

<

a

n

+

1

-

a

n

b

n

+

1

-

b

n

<

l

+

ϵ

Because

b

n

is

strictly increasing

we can multiply the last equation with

b

n

+

1

-

b

n

to get :

(

l

-

ϵ

)

⁢

(

b

n

+

1

-

b

n

)

<

a

n

+

1

-

a

n

<

(

l

+

ϵ

)

⁢

(

b

n

+

1

-

b

n

)

Let

k

>

N

⁢

(

ϵ

)

be a

natural number

. Summing the last

relation

we get :

(

l

-

ϵ

)

⁢

∑

i

=

N

⁢

(

ϵ

)

k

(

b

i

+

1

-

b

i

)

<

∑

i

=

N

⁢

(

ϵ

)

k

(

a

n

+

1

-

a

n

)

<

(

l

+

ϵ

)

⁢

∑

i

=

N

⁢

(

ϵ

)

k

(

b

i

+

1

-

b

i

)

⇒

(

l

-

ϵ

)

⁢

(

b

k

+

1

-

b

N

⁢

(

ϵ

)

)

<

a

k

+

1

-

a

N

⁢

(

ϵ

)

<

(

l

+

ϵ

)

⁢

(

b

k

+

1

-

b

N

⁢

(

ϵ

)

)

Divide the last relation by

b

k

+

1

>

0

to get :

(

l

-

ϵ

)

⁢

(

1

-

b

N

⁢

(

ϵ

)

b

k

+

1

)

<

a

k

+

1

b

k

+

1

-

a

N

⁢

(

ϵ

)

b

k

+

1

<

(

l

+

ϵ

)

⁢

(

1

-

b

N

⁢

(

ϵ

)

b

k

+

1

)

⇔

(

l

-

ϵ

)

⁢

(

1

-

b

N

⁢

(

ϵ

)

b

k

+

1

)

+

a

N

⁢

(

ϵ

)

b

k

+

1

<

a

k

+

1

b

k

+

1

<

(

l

+

ϵ

)

⁢

(

1

-

b

N

⁢

(

ϵ

)

b

k

+

1

)

+

a

N

⁢

(

ϵ

)

b

k

+

1

This means that there is some

K

such that for

k

≥

K

we have :

(

l

-

ϵ

)

<

a

k

+

1

b

k

+

1

<

(

l

+

ϵ

)

(since the other terms who were left out

converge

to 0)

This obviously means that :

lim

n

→

∞

⁡

a

n

b

n

=

l

and we are done .

Title

Proof of Stolz-Cesaro theorem

Canonical name

ProofOfStolzCesaroTheorem

Date of creation

2013-03-22 13:17:45

Last modified on

2013-03-22 13:17:45

Owner

slash (33)

Last modified by

slash (33)

Numerical id

4

Author

slash (33)

Entry type

Proof

Classification

msc 40A05

Generated on Fri Feb 9 21:17:11 2018 by

LaTeXML
proof that a metric space is compact if and only if it is complete and totally bounded

proof that a metric space is compact if and only if it is complete and totally bounded

Theorem

:

A

metric space

is

compact

if and only if it is

complete

and

totally bounded

.

Proof.

Let

X

be a metric space with metric

d

. If

X

is compact, then it is

sequentially compact

and thus complete. Since

X

is compact, the covering of

X

by all

ϵ

-balls must have a finite

subcover

, so that

X

is totally bounded.

Now assume that

X

is complete and totally bounded. For metric spaces, compact and sequentially compact are

equivalent

; we prove that

X

is sequentially compact. Choose a

sequence

p

n

∈

X

; we will find a Cauchy

subsequence

(and hence a

convergent

subsequence, since

X

is complete).

Cover

X

by finitely many balls of radius

1

(since

X

is totally bounded). At least one of those balls must contain an

infinite

number of the

p

i

. Call that ball

B

1

, and let

S

1

be the set of integers

i

for which

p

i

∈

B

1

.

Proceeding inductively, it is clear that we can define, for each

positive integer

k

>

1

, a ball

B

k

of radius

1

/

k

containing an infinite number of the

p

i

for which

i

∈

S

k

-

1

; define

S

k

to be the set of such

i

.

Each of the

S

k

is infinite, so we can choose a sequence

n

k

∈

S

k

with

n

k

<

n

k

+

1

for all

k

. Since the

S

k

are nested, we have that whenever

i

,

j

≥

k

, then

n

i

,

n

j

∈

S

k

. Thus for all

i

,

j

≥

k

,

p

n

i

and

p

n

j

are both

contained

in a ball of radius

1

/

k

. Hence the sequence

p

n

k

is Cauchy.

References

1

J. Munkres,

Topology

, Prentice Hall, 1975.

Title

proof that a metric space is compact if and only if it is complete and totally bounded

Canonical

name

ProofThatAMetricSpaceIsCompactIfAndOnlyIfItIsCompleteAndTotallyBounded

Date of creation

2013-03-22 18:01:06

Last modified on

2013-03-22 18:01:06

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

5

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 40A05

Classification

msc 54D30

Generated on Fri Feb 9 21:17:13 2018 by

LaTeXML
Proof that every absolutely convergent series is unconditionally convergent

Proof that every absolutely convergent series is unconditionally convergent

Suppose the series

∑

k

=

1

∞

a

k

converges

to a

A

and that it is also

absolutely convergent

. So, as

∑

k

=

1

∞

|

a

k

|

converges, we have

∀

ϵ

>

0

:

∃

N

>

0

:

∑

k

=

n

+

1

∞

|

a

k

|

<

ϵ

2

,

∀

n

>

N

As the series converges to

A

, we may choose

n

such that:

|

A

-

∑

k

=

1

n

a

k

|

<

ϵ

2

Now suppose that

∑

k

=

1

∞

b

k

is a rearrangement of

∑

k

=

1

∞

a

k

, that is, it exists a bijection

σ

:

\mathbb

⁢

N

→

\mathbb

⁢

N

, such that

b

k

=

a

σ

⁢

(

k

)

Let

J

=

m

⁢

a

⁢

x

⁢

{

j

:

σ

⁢

(

j

)

≤

n

}

. Now take

m

≥

J

. Then:

∑

k

=

1

m

b

k

=

b

1

+

…

+

b

m

=

a

σ

⁢

(

1

)

+

…

+

a

σ

⁢

(

m

)

This sum must include the terms

a

1

,

…

,

a

n

, otherwise

J

wouldn’t be

maximum

. Thus, for

m

≥

J

, we have that

∑

j

=

1

m

b

j

-

∑

k

=

1

n

a

k

is a finite sum of terms

a

k

, with

k

≥

n

+

1

. So:

|

∑

j

=

1

m

b

j

-

∑

k

=

1

n

a

k

|

≤

∑

k

=

n

+

1

∞

|

a

k

|

<

ϵ

2

Finally, taking

m

≥

J

, we have:

|

A

-

∑

k

=

1

m

b

k

|

≤

|

A

-

∑

k

=

1

n

a

k

|

+

|

∑

j

=

1

m

b

j

-

∑

k

=

1

n

a

k

|

<

ϵ

2

+

ϵ

2

=

ϵ

So

∑

k

=

1

∞

b

k

=

A

Title

Proof that every absolutely convergent series is unconditionally convergent

Canonical name

ProofThatEveryAbsolutelyConvergentSeriesIsUnconditionallyConvergent

Date of creation

2013-03-11 19:17:17

Last modified on

2013-03-11 19:17:17

Owner

Filipe (28191)

Last modified by

(0)

Numerical id

7

Author

Filipe (0)

Entry type

Proof

Classification

msc 40A05

Synonym

Related topic

Defines

Generated on Fri Feb 9 21:17:15 2018 by

LaTeXML