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Math Help - please help me solve this equation

  1. #1
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    please help me solve this equation

    show that

    9/8<summation(1/(n^3))<5/4
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  2. #2
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    Quote Originally Posted by nzm88 View Post
    show that

    9/8<summation(1/(n^3))<5/4

    Do you mean  \sum_{n=1}^{\infty} \frac{1}{n^3} is bounded between  \frac{9}{8} and  \frac{5}{4} ?


    Consider  \sum_{n=1}^{\infty} \frac{1}{n^3}= 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... > 1 + \frac{1}{2^3 } = 1 + \frac{1}{8} = \frac{9}{8}  , the lower bound .


    Now , we are going to find the upper bound :

     \sum_{n=1}^{\infty} \frac{1}{n^3}

     = 1 + \sum_{n=2}^{\infty} \frac{1}{n^3}

    Consider  0 < n^3 - n < n^3  , n \geq 2

     \frac{1}{ n^3 - n} > \frac{1}{n^3}

     \sum_{n=2}^{\infty}\frac{1}{ n^3 - n}  >   \sum_{n=2}^{\infty}\frac{1}{n^3}

       \sum_{n=2}^{\infty} \left [ \frac{1}{n(n^2 - 1)} \right ] >   \sum_{n=2}^{\infty}\frac{1}{n^3}

      \sum_{n=2}^{\infty} \left [ \frac{1}{2} \left ( \frac{1}{ n -1} + \frac{1}{n + 1} \right ) - \frac{1}{n}  \right ] >   \sum_{n=2}^{\infty}\frac{1}{n^3}

    The left hand side is a telescoping series , the value is  \frac{1}{4}

    Therefore ,

     \frac{1}{4} >  \sum_{n=2}^{\infty}\frac{1}{n^3}

     1 + \frac{1}{4 } = \frac{5}{4} >  1 + \sum_{n=2}^{\infty}\frac{1}{n^3} = \sum_{n=1}^{\infty}\frac{1}{n^3}
    Last edited by simplependulum; January 31st 2010 at 10:57 PM.
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  3. #3
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    Quote Originally Posted by simplependulum View Post
    Do you mean  \sum_{n=1}^{\infty} \frac{1}{n^3} is bounded between  \frac{9}{8} and  \frac{5}{4} ?


    Consider  \sum_{n=1}^{\infty} \frac{1}{n^3}= 1 + \frac{1}{2^3} + \frac{1}{3^3} + ... > 1 + \frac{1}{2^3 } = 1 + \frac{1}{8} = \frac{1}{9}  , the lower bound .
    Typo: 1+ \frac{1}{8}= \frac{8}{9}, of course.


    Now , we are going to find the upper bound :

     \sum_{n=1}^{\infty} \frac{1}{n^3}

     = 1 + \sum_{n=2}^{\infty} \frac{1}{n^3}

    Consider  0 < n^3 - n < n^3  , n \geq 2

     \frac{1}{ n^3 - n} > \frac{1}{n^3}

     \sum_{n=2}^{\infty}\frac{1}{ n^3 - n}  >   \sum_{n=2}^{\infty}\frac{1}{n^3}

       \sum_{n=2}^{\infty} \left [ \frac{1}{n(n^2 - 1)} \right ] >   \sum_{n=2}^{\infty}\frac{1}{n^3}

      \sum_{n=2}^{\infty} \left [ \frac{1}{2} \left ( \frac{1}{ n -1} + \frac{1}{n + 1} \right ) - \frac{1}{n}  \right ] >   \sum_{n=2}^{\infty}\frac{1}{n^3}

    The left hand side is a telescoping series , the value is  \frac{1}{4}

    Therefore ,

     \frac{1}{4} >  \sum_{n=2}^{\infty}\frac{1}{n^3}

     1 + \frac{1}{4 } = \frac{5}{4} >  1 + \sum_{n=2}^{\infty}\frac{1}{n^3} = \sum_{n=1}^{\infty}\frac{1}{n^3}
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  4. #4
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    thanx for your answer. i really appreciate it
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Typo: 1+ \frac{1}{8}= \frac{8}{9}, of course.
    Is this another typo?
    its \frac{9}{8}.
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  6. #6
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    Quote Originally Posted by TWiX View Post
    Is this another typo?
    its \frac{9}{8}.
    fractional dyslexia ... many are afflicted.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    fractional dyslexia ... many are afflicted.
    Do not use hard words
    Use simple words
    My English is horrible
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  8. #8
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    HaHa , I am just wondering what if my post is published .
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