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Math Help - Series

  1. #1
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    Dec 2009
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    Series

    b_{n} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\f  rac{!}{n^2}

    a) Prove that this series have limit b ,
    b) prove that the limit b , hold \frac{49}{36} < b < 2


    I start to solve :

     \frac{1}{n^2} < \frac{1}{n(n-1)} = \frac {1}{n-1} - \frac{1}{n}

    so
     b_{n} = 1 + \frac{1}{1}-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... \frac{1}{n-1} - \frac{1}{n} = 2 - \frac{1}{n} < 2

    How I prove that the series rise ?

    and how I prove b)
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  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    715
    Quote Originally Posted by gilyos View Post
    b_{n} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\f  rac{!}{n^2}

    a) Prove that this series have limit b ,
    b) prove that the limit b , hold \frac{49}{36} < b < 2


    I start to solve :

     \frac{1}{n^2} < \frac{1}{n(n-1)} = \frac {1}{n-1} - \frac{1}{n}

    so
     b_{n} = 1 + \frac{1}{1}-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... \frac{1}{n-1} - \frac{1}{n} = 2 - \frac{1}{n} < 2

    How I prove that the series rise ?

    and how I prove b)
    O' right , you have found the upper bound ,

    the lower bound , umm ... What you have to do actually depends on what the answer is ..

    Consider the series

     S = a_1 + a_2 + a_3 + a_4 + ...

    we know any other series with a cancel of any members  a_k must be less than  S , provided  a_k > 0 .

    Therefore ,

     1 + \frac{1}{2^2 } + \frac{1}{3^2} + \frac{ 1}{4^2} + ... > 1 + \frac{1}{2^2} + \frac{1}{3^2} = 1 + \frac{1}{4} + \frac{1}{9} = \frac{ 36 + 9 + 4 }{36} = \frac{49}{36}
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