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Thread: 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 $\displaystyle \displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$?
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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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    $\displaystyle \displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$

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

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


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

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


    Thus

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

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

    $\displaystyle \displaystyle x^2 + x - 2 < x$

    $\displaystyle \displaystyle x^2 - 2 < 0$

    $\displaystyle \displaystyle x^2 < 2$

    $\displaystyle \displaystyle |x| < \sqrt{2}$

    $\displaystyle \displaystyle -\sqrt{2} < x < \sqrt{2}$.
    Last edited by Prove It; Oct 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
    $\displaystyle \displaystyle x + 2 < 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} + \dots$

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

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


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

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


    Thus

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

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

    $\displaystyle \displaystyle x^2 + x - 2 < x$

    $\displaystyle \displaystyle x^2 - 2 < 0$

    $\displaystyle \displaystyle x^2 < 2$

    $\displaystyle \displaystyle |x| < \sqrt{2}$

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

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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