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  1. #1
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    Series

    What can be said about this sequence:

    $\displaystyle 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n $

    Hint: it might be helpful to consider $\displaystyle \int_{a}^{b} \surd x dx$ for suitable a and b
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  2. #2
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    Quote Originally Posted by Natasha1
    What can be said about this sequence:

    $\displaystyle 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n $

    Hint: it might be helpful to consider $\displaystyle \int_{a}^{b} \surd x dx$ for suitable a and b
    Let:

    $\displaystyle
    S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n
    $

    Then as $\displaystyle \{\sqrt{r}, r=1,2, \dots,\ n\}$ is an increasing sequence:

    $\displaystyle
    \int_{x=1}^{n+1} \sqrt{x-1} dx<S_n<\int_{x=1}^{n+1} \sqrt{x} dx
    $

    So doing the integrals gives:

    $\displaystyle
    \frac{2}{3}(n)^{3/2}<S_n<\frac{2}{3} ((n+1)^{3/2}-1)
    $

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack
    Let:

    $\displaystyle
    S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n
    $

    Then as $\displaystyle \{\sqrt{r}, r=1,2, \dots,\ n\}$ is an increasing sequence:

    $\displaystyle
    \int_{x=1}^{n+1} \sqrt{x-1} dx<S_n<\int_{x=1}^{n+1} \sqrt{x} dx
    $
    In explanation, this is because:

    $\displaystyle
    \sqrt{x-1}<\sqrt{n}<\sqrt{x}
    $

    for $\displaystyle x \in (n,n+1)$

    RonL
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  4. #4
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    Quote Originally Posted by CaptainBlack
    In explanation, this is because:

    $\displaystyle
    \sqrt{x-1}<\sqrt{n}<\sqrt{x}
    $

    for $\displaystyle x \in (n,n+1)$

    RonL

    RonL,

    At the bottom of your integrals you put x=1, should it not just be 1? Could I just put 1 or not?
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  5. #5
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    Quote Originally Posted by Natasha1
    RonL,

    At the bottom of your integrals you put x=1, should it not just be 1? Could I just put 1 or not?
    Yes you can put 1

    RonL
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