Proving a limit theorem of a sequence (square root)

[jbunniii]

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

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

 

[aatzn]

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 ( $\lim a_n = A$ ), if
$\lim \sqrt{a_n}$
exists finite, than it is
$\sqrt{A}$

 

[jbunniii]

So, in our case ( $\lim a_n = A$ ), if
$\lim \sqrt{a_n}$
exists finite, than it is
$\sqrt{A}$

But you didn't prove that $\lim \sqrt{a_n}$ exists.

 

[aatzn]

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?