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Math Help - analysis!!

  1. #1
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    analysis!!

     { For \  n \in N  \ let  \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }
     a- \ show \  \ that   \ \ x_n  \ \ is \  \ increasing .
     b- \ prove \ \  that \  \ x_n < 1 , \  \forall n \in N .
     c- \ conclude  \ \  that \   \ x_n  \  \ is  \  \ convergent
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  2. #2
    MHF Contributor
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    Quote Originally Posted by flower3 View Post
     { For \  n \in N  \ let  \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }
     a- \ show \  \ that   \ \ x_n  \ \ is \  \ increasing .
     b- \ prove \ \  that \  \ x_n < 1 , \  \forall n \in N .
     c- \ conclude  \ \  that \   \ x_n  \  \ is  \  \ convergent
    Hi

    a- Compute x_{n+1}-x_{n}
    b- x_n is the sum of n term, each of them being lower than ...
    c- Use the appropriate theorem
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  3. #3
    Junior Member
    Joined
    May 2008
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    a) just compute x_{n+1}-x_n and conclude that \ge0
    b) \frac{1}{n+1}\le\frac{1}{n+1}, \frac{1}{n+2}<\frac{1}{n+1}, ...,\frac{1}{2n}<\frac{1}{n+1} after we sum them we get x_n<\frac{n}{n+1}<1
    c) we have an increasing sequence (x_n)_n and bounded by 0 and 1, so (x_n)_n is convergent.
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