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Math Help - convergents of sequence

  1. #1
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    convergents of sequence

    question: what number does \frac{2^{n}}{3^{n+1}} converge to?
    answer: 0
    what I've done: sort of use L'Hopital's rule by taking derivative of numerator and denominator => \frac{n2^{n-1}}{(n+1)3^n} so wouldn't this diverge?
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  2. #2
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    I think it's easier to just write it as ~ \frac{1}{3} \cdot \left( \frac{2}{3} \right)^n

    Now, 0< \frac{2}{3} < 1 , can you see where to go from here?
    Last edited by pomp; July 15th 2009 at 07:01 PM. Reason: typo
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  3. #3
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    Quote Originally Posted by superdude View Post
    question: what number does \frac{2^{n}}{3^{n+1}} converge to?
    answer: 0
    what I've done: sort of use L'Hopital's rule by taking derivative of numerator and denominator => \frac{n2^{n-1}}{(n+1)3^n} so wouldn't this diverge?
    If you wanted to use L'Hospital's rule, you can, because the numerator and denominator both tend to \infty.

    Your derivatives are also correct, BUT, the numerator and denominator still both tend to \infty.

    In fact, you would need to work out the numerator's and denominator's derivatives exactly n times to make something not tend to \infty.


    After n derivatives of the numerator and denominator have been taken, you should have

    \lim_{n \to \infty}{\frac{n\cdot (n - 1)\cdot (n - 2)\cdot\dots \cdot 3\cdot 2\cdot 1 \cdot 2^{0}}{(n + 1)\cdot n \cdot (n - 1) \cdot (n - 2)\cdot \dots \cdot 3 \cdot 2 \cdot 3^1}}

     = \lim_{n \to \infty}\frac{1}{3(n + 1)}

     = 0 since the denominator tends to \infty.



    Of course, like pomp said, it's easier to just write the fraction as \frac{1}{3}\cdot \left(\frac{2}{3}\right)^n.
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    Quote Originally Posted by Prove It View Post
    If you wanted to use L'Hospital's rule, you can, because the numerator and denominator both tend to \infty.

    Your derivatives are also correct, BUT, the numerator and denominator still both tend to \infty.
    Sorry but when you're taking these derivatives, what are you taking them with respect to?

    Surely, \frac{d}{dn} \left( 2^n \right) = 2^n \log{2}
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  5. #5
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    you're right. thanks
    I guess I got Prove It on the wrong track
    Last edited by superdude; July 16th 2009 at 12:48 PM. Reason: added 2nd sentence
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  6. #6
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    the denominator is tending to infinity faster than the numerator since 3>2, so this sequence converges to zero.
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  7. #7
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    Quote Originally Posted by dr.tea View Post
    the denominator is tending to infinity faster than the numerator since 3>2, so this sequence converges to zero.
    Simpler is the fact that 2/3< 1, as pomp pointed out initially, so the sequence converges to 0.
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