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Math Help - Inequality proof

  1. #1
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    Inequality proof

    How to proof that

    \sum_{n=1}^{\infty}\left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right)\geq2

    inequality holds for k\geq8?

    Thanks!
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  2. #2
    Member Miss's Avatar
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    Quote Originally Posted by Ben92 View Post
    How to proof that

    \sum_{n=1}^{\infty}\left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right)\geq2

    inequality holds for k\geq8?

    Thanks!

    There is something wrong here.

    \left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right) = 0 !
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  3. #3
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    Yes, you are right. The correct condition is k>8.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Ben92 View Post
    How to proof that

    \sum_{n=1}^{\infty}\left(\left\lfloor \frac{2^{k+1}}{3^{n}}\right\rfloor -2\left\lfloor \frac{2^{k}}{3^{n}}\right\rfloor \right)\geq2

    inequality holds for k\geq8?

    Thanks!
    Induction would be my first bet.
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