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Math Help - Infinite Series

  1. #1
    Len
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    Infinite Series

    I wrote it in word instead :P

    Last edited by Len; April 2nd 2008 at 01:27 PM.
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  2. #2
    Len
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    Solved second part just need help with first part now!=)
    Last edited by Len; April 2nd 2008 at 04:56 PM.
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  3. #3
    Len
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    Still need some help on where to start. Thanks
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  4. #4
    Flow Master
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    Quote Originally Posted by Len View Post
    Would also be much appreciated if someone can confirm my use of the comparison test



    Forgot in the first line for all positive integers k.
    Your use of the comparison test is fine:

    \frac{\sqrt{k} 4^k}{2 + k 5^k} < \frac{k 4^k}{k 5^k} = \frac{4^k}{5^k} = \left( \frac{4}{5} \right)^k.

    Without numerical number crunching, I'm not sure how to establish how large n should be.
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  5. #5
    Len
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    Quote Originally Posted by mr fantastic View Post
    Your use of the comparison test is fine:

    \frac{\sqrt{k} 4^k}{2 + k 5^k} < \frac{k 4^k}{k 5^k} = \frac{4^k}{5^k} = \left( \frac{4}{5} \right)^k.

    Without numerical number crunching, I'm not sure how to establish how large n should be.
    Thanks

    Maybe someone else can help me with the first part.
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