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Math Help - finding the limit for convergence

  1. #1
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    finding the limit for convergence

    I can't seem to figure out how this sequence converges to a value.

    \lim_{n\to\infty}\frac{4^{n+2}}{5^n}

    I tried simplifying it to this:

    \lim_{n\to\infty}\frac{(16)4^{n}}{5^n}=
    16\lim_{n\to\infty}(\frac{4}{5})^n

    But i dont know if that helps it still seems like it diverges, but it doesn't.
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  2. #2
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    If |r|<1 then \left(r^n\right)\to 0.
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  3. #3
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    Quote Originally Posted by mmattson07 View Post
    I can't seem to figure out how this sequence converges to a value.

    \lim_{n\to\infty}\frac{4^{n+2}}{5^n}

    I tried simplifying it to this:

    \lim_{n\to\infty}\frac{(16)4^{n}}{5^n}=
    16\lim_{n\to\infty}(\frac{4}{5})^n

    But i dont know if that helps it still seems like it diverges, but it doesn't.

    Very good! Now you only need to remember from H.S. that the limit of an infinite geometric sequence whose quotient is between -1 and 1 is zero and you're done.

    Tonio
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  4. #4
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    ah yes thanks guys!
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