Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Can anybody help me with this sequence question?

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I tried to use proof by induction by I don't know how to show its bounded above x.x I used the rule \[a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}\]
did you try the ratio test?
\[L=\lim_{n\to\infty}\left|a_{n+1}\over a_n\right|=\lim_{n\to\infty}\sqrt{2}>1\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Oops my post erased. My book / cal class does not and will not go over the ratio test I believe. I found the equation myself ( i mean the formula), when I use \[\lim_{n \rightarrow \infty} f(a _{n} ) = f(l)\], the answer it gives me is right but induction doesn't seem to work because I don't know if its bounded above ( well I'm assuming its above because it's increasing)
use induction to show that \[a_n\le 2\] for all \(n\)
So I just calculate its lowest upper bound limit by plugging in n?
oh nvm it cant be greater than 2 because 2= root of 4, right?
\[\sqrt{2}\le 2\] assume \(a_n\le 2\) \(a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2\)
then use monotone convergence theorem
makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.
"...a function is bounded above if its increasing" that is not true
I mean sequence* I thought if it was a sequence that was only increasing it was bounded above
\[a_n=n\] is an increasing sequence
it is not bounded above
So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff
if the sequence is bounded and it is either increasing or decreasing then the sequence converges.
thank you for your patience!
Monotone convergence theorem

Not the answer you are looking for?

Search for more explanations.

Ask your own question