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Math Help - Solving infinite series inequality

  1. #1
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    Solving infinite series inequality

    How do you solve

    x+2<1+1/x^2 +1/x^3+..., where x>0?

    Thanks
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  2. #2
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    Are you sure it's not \displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots?
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  3. #3
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    Yes, sorry it should be that--I tried to copy and paste from another document where it was laid out correctly, but the formatting got mangled in the process...
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  4. #4
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    \displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots

    \displaystyle x + 2 < \sum_{n = 0}^{\infty}\frac{1}{x^n}

    \displaystyle x + 2 < \sum_{n = 0}^{\infty}\left(\frac{1}{x}\right)^n.


    The right hand side is an infinite geometric series with common ratio \displaystyle \frac{1}{x}. Recall that the sum of an infinite geometric series is \displaystyle \frac{a}{1 - r} if |r| < 1, i.e. \displaystyle \left|\frac{1}{x}\right| < 1, i.e. \displaystyle |x| > 1.

    So assuming that \displaystyle x > 1, the right hand side is \displaystyle \frac{1}{1 - \frac{1}{x}} = \frac{1}{\frac{x - 1}{x}} = \frac{x}{x - 1}.


    Thus

    \displaystyle x + 2 < \frac{x}{x - 1}

    \displaystyle (x + 2)(x - 1) < x (we can do this since we know \displaystyle x > 1)

    \displaystyle x^2 + x - 2 < x

    \displaystyle x^2 - 2 < 0

    \displaystyle x^2 < 2

    \displaystyle |x| < \sqrt{2}

    \displaystyle -\sqrt{2} < x < \sqrt{2}.
    Last edited by Prove It; October 31st 2010 at 05:48 AM.
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  5. #5
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    Thank you!
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Prove It View Post
    \displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots

    \displaystyle x + 2 < \sum_{n = 0}^{\infty}\frac{1}{x^n}

    \displaystyle x + 2 < \sum_{n = 0}^{\infty}\left(\frac{1}{x}\right)^n.


    The right hand side is an infinite geometric series with common ratio \displaystyle \frac{1}{x}. Recall that the sum of an infinite geometric series is \displaystyle \frac{a}{1 - r} if |r| < 1, i.e. \displaystyle \left|\frac{1}{x}\right| < 1, i.e. \displaystyle |x| > 1.

    So assuming that \displaystyle x > 1, the right hand side is \displaystyle \frac{1}{1 - \frac{1}{x}} = \frac{1}{\frac{x - 1}{x}} = \frac{x}{x - 1}.


    Thus

    \displaystyle x + 2 < \frac{x}{x - 1}

    \displaystyle (x + 2)(x - 1) < x (we can do this since we know \displaystyle x > 1)

    \displaystyle x^2 + x - 2 < x

    \displaystyle x^2 - 2 < 0

    \displaystyle x^2 < 2

    \displaystyle |x| < \sqrt{2}

    \displaystyle -\sqrt{2} < x < \sqrt{2}.
    The 'identity' \displaystyle \sum_{n=0}^{\infty}\frac{1}{x^{n}} = \frac{x}{x-1} is valid only for |x|>1 , so that the inequality \displaystyle x+2<  \sum_{n=0}^{\infty}\frac{1}{x^{n}} is valid, because is also x>0, 'only' for 1<x<\sqrt{2}...

    Kind regards

    \chi \sigma
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