Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

badreferences

If \[\left\{a_n\right\}\mid n\geq 0,a_1=5,a_{n+1}=a_n^2-2,n\in\mathbb{R}\] find \[\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_1a_2\cdots a_n}\].

  • 2 years ago
  • 2 years ago

  • This Question is Closed
  1. Mr.Math
    Best Response
    You've already chosen the best response.
    Medals 1

    This is just a guess: \(\large \frac{23}{5}\)?

    • 2 years ago
  2. badreferences
    Best Response
    You've already chosen the best response.
    Medals 0

    \(\sqrt[2]{21}\) actually.

    • 2 years ago
  3. Mr.Math
    Best Response
    You've already chosen the best response.
    Medals 1

    You didn't need to write the answer! -.-

    • 2 years ago
  4. Mr.Math
    Best Response
    You've already chosen the best response.
    Medals 1

    Okay. I will try to write a proof for that.

    • 2 years ago
  5. badreferences
    Best Response
    You've already chosen the best response.
    Medals 0

    Oh, haha. XD Apparently, the problem can be solved using repeated telescopy \(a+\frac{1}{a}=5\Rightarrow x_{n+1}={a^2}^n+\frac{1}{{a^2}^n}\), but I'm not sure I understand what they're getting at with this hint.

    • 2 years ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.