- #26

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was answered

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- Thread starter shoescreen
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- #26

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was answered

- #27

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We know that

[itex]\lim a_n = A[/itex]

but this is the same as

[itex]\lim (\sqrt{a_n})^2 = (\sqrt{A})^2[/itex]

or

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})[/itex]

and, for the theorem about the limit of products of sequenze:

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})[/itex]

than

[itex]\lim \sqrt{a_n} = \sqrt{A}[/itex]

QED

NB You can generalize this to other roots!

[itex]\lim a_n = A[/itex]

but this is the same as

[itex]\lim (\sqrt{a_n})^2 = (\sqrt{A})^2[/itex]

or

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})[/itex]

and, for the theorem about the limit of products of sequenze:

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})[/itex]

than

[itex]\lim \sqrt{a_n} = \sqrt{A}[/itex]

QED

NB You can generalize this to other roots!

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- #28

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[itex]\lim a_n = A[/itex]

but this is the same as

[itex]\lim (\sqrt{a_n})^2 = (\sqrt{A})^2[/itex]

or

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})[/itex]

and, for the theorem about the limit of products of sequence:

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})[/itex]

then

[itex]\lim \sqrt{a_n} = \sqrt{A}[/itex]

QED

NB You can generalize to other roots

- #29

- 3,473

- 255

and, for the theorem about the limit of products of sequence:

[itex]\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})[/itex] then

[itex]\lim \sqrt{a_n} = \sqrt{A}[/itex]

What theorem are you using here? It is certainly not the case in general that

[tex]\lim [(x_n)(x_n)] = (x)(x)\implies \lim x_n = x[/tex]

For example, take [itex]x_n = (-1)^n[/itex] and [itex]x = 1[/itex].

Also, this is a very old thread.

- #30

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You can use it for sequences with finite limits.

Anyway 1 is not the limit of your sequence, there is no limit for your example.

Do you mean it is a product of two other regular sequences?

- #31

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Anyway 1 is not the limit of your sequence, there is no limit for your example.

You do not know that your sequence [itex]\sqrt{a_n}[/itex] has a limit, either. That is what you are trying to prove! Please state precisely what your theorem says.

- #32

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(with finite limits!) is the product of their limits.

So, in our case ( [itex]\lim a_n = A[/itex] ), if

[itex]\lim \sqrt{a_n}[/itex]

exists finite, than it is

[itex]\sqrt{A}[/itex]

- #33

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So, in our case ( [itex]\lim a_n = A[/itex] ), if

[itex]\lim \sqrt{a_n}[/itex]

exists finite, than it is

[itex]\sqrt{A}[/itex]

But you didn't prove that [itex]\lim \sqrt{a_n}[/itex] exists.

- #34

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You are right.

My weaker demonstration shows how that limit should be, if it exists.

On the other hand, with the previous demonstration not only we know the form of the hypothetical limit, but we are sure that it always exists.

But your sample? what did you mean with post n. #30?

Anyway don’t you think my simple method could be a useful aid for the study of more complicate sequences with nth roots?

My weaker demonstration shows how that limit should be, if it exists.

On the other hand, with the previous demonstration not only we know the form of the hypothetical limit, but we are sure that it always exists.

But your sample? what did you mean with post n. #30?

Anyway don’t you think my simple method could be a useful aid for the study of more complicate sequences with nth roots?

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