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Thread: analysis!!

  1. #1
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    analysis!!

    $\displaystyle { For \ n \in N \ let \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }$
    $\displaystyle a- \ show \ \ that \ \ x_n \ \ is \ \ increasing . $
    $\displaystyle b- \ prove \ \ that \ \ x_n < 1 , \ \forall n \in N . $
    $\displaystyle 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
    $\displaystyle { For \ n \in N \ let \ \ x_n = \frac {1}{1+n} + \frac {1}{2+n} +...+\frac {1}{2n} : }$
    $\displaystyle a- \ show \ \ that \ \ x_n \ \ is \ \ increasing . $
    $\displaystyle b- \ prove \ \ that \ \ x_n < 1 , \ \forall n \in N . $
    $\displaystyle c- \ conclude \ \ that \ \ x_n \ \ is \ \ convergent $
    Hi

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