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MarcLeclair Group Title

Can anybody help me with this sequence question?

  • one year ago
  • one year ago

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  1. MarcLeclair Group Title
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    I tried to use proof by induction by I don't know how to show its bounded above x.x I used the rule \[a _{1}= \sqrt{2} , a _{n+1}= \sqrt{2a _{n}}\]

    • one year ago
  2. electrokid Group Title
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    did you try the ratio test?

    • one year ago
  3. electrokid Group Title
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    \[L=\lim_{n\to\infty}\left|a_{n+1}\over a_n\right|=\lim_{n\to\infty}\sqrt{2}>1\]

    • one year ago
  4. MarcLeclair Group Title
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    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)

    • one year ago
  5. Zarkon Group Title
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    use induction to show that \[a_n\le 2\] for all \(n\)

    • one year ago
  6. MarcLeclair Group Title
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    So I just calculate its lowest upper bound limit by plugging in n?

    • one year ago
  7. MarcLeclair Group Title
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    oh nvm it cant be greater than 2 because 2= root of 4, right?

    • one year ago
  8. Zarkon Group Title
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    \[\sqrt{2}\le 2\] assume \(a_n\le 2\) \(a_{n+1}=\sqrt{2\cdot a_n}\le \sqrt{2\cdot 2}=2\)

    • one year ago
  9. Zarkon Group Title
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    then use monotone convergence theorem

    • one year ago
  10. MarcLeclair Group Title
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    makes very much sense. Thanks a lot, I forgot to think of that. ^^ and yeah, a function is bounded above if its increasing.

    • one year ago
  11. Zarkon Group Title
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    "...a function is bounded above if its increasing" that is not true

    • one year ago
  12. MarcLeclair Group Title
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    I mean sequence* I thought if it was a sequence that was only increasing it was bounded above

    • one year ago
  13. Zarkon Group Title
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    \[a_n=n\] is an increasing sequence

    • one year ago
  14. Zarkon Group Title
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    it is not bounded above

    • one year ago
  15. MarcLeclair Group Title
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    So, is it IF a sequence is bounded, it is converging? Im sorry Im pretty new to this stuff

    • one year ago
  16. Zarkon Group Title
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    if the sequence is bounded and it is either increasing or decreasing then the sequence converges.

    • one year ago
  17. MarcLeclair Group Title
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    thank you for your patience!

    • one year ago
  18. Zarkon Group Title
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    Monotone convergence theorem

    • one year ago
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