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Math Help - Series

  1. #1
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    Series

    What can be said about this sequence:

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

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

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

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

    <br />
S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n <br />

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

    <br />
\int_{x=1}^{n+1} \sqrt{x-1} dx<S_n<\int_{x=1}^{n+1} \sqrt{x} dx<br />

    So doing the integrals gives:

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

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

    <br />
S_n= 1 + \surd2 + \surd3 + 2 + \surd5 + \surd6 + ..... + \surd n <br />

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

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

    <br />
\sqrt{x-1}<\sqrt{n}<\sqrt{x}<br />

    for 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:

    <br />
\sqrt{x-1}<\sqrt{n}<\sqrt{x}<br />

    for 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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