Proving a limit theorem of a sequence (square root)

  • Thread starter shoescreen
  • Start date
  • #26
1
0
was answered
 
  • #27
6
0
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!
 
Last edited:
  • #28
6
0
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 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
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
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
6
0
I used the theorem about the limit of products of sequence, as I wrote.
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
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,473
255
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
6
0
The theorem says that the limit of a sequence product of two other sequences
(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
jbunniii
Science Advisor
Homework Helper
Insights Author
Gold Member
3,473
255
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
6
0
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?
 
Last edited:

Related Threads on Proving a limit theorem of a sequence (square root)

  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
9
Views
3K
Replies
4
Views
9K
Replies
4
Views
1K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
3
Views
13K
Replies
3
Views
2K
Top