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Quick sequence question

  1. Nov 23, 2004 #1
    [tex]\frac{n}{2^{n+2}}[/tex]

    I know it's monotonic decreasing, a sub n < a sub n+1 and so has an upper bound of 1/8.

    Can you then use L'Hopital's rule to determine that the sequence converges to 0, it's lower bound?
     
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  3. Nov 23, 2004 #2

    Galileo

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    No. I don't think you'll ever have to use L'hospital to solve a sequence question. In this case you're not allowed to differentiate, since the function is only defined on the integers.

    If a sequence is monotonic decreasing, then it is sufficient to show it has a LOWER bound for it to converge.
     
  4. Nov 23, 2004 #3

    HallsofIvy

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    But [itex]\frac{n}{2^{n+2}}[/itex] has the same limit (as n-> infinity) as the continuous function [itex]\frac{x}{2^{x+2}}[/itex] (as x-> infinity). Certainly you can use L'Hopital to find the limit of the continuous function and then assert that as the limit of the sequence.

    It is sufficient to show it has a lower bound (and 0 is an obvious lower bound) if you only want to prove that it does converge. That doesn't determine what it converges to.
     
  5. Nov 23, 2004 #4

    Galileo

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    Yes ofcourse, but you'll have to state that explicitly and not differentiate numerator and denominator w.r.t. n immediately.

    In general, L'hospitals rule is a source for obtaining many fraudulent proofs. I'm just not a fan of using L'Hospital :yuck:
     
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