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Math Help - Summing a complex series

  1. #1
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    Summing a complex series

    Hi, here is a problem involving summing a series involving a complex equation.

    Sum this series:

    a_1 + a_2 + ... + a_{99}
    where a_n = \frac{1}{(n + 1)\sqrt{n} + n\sqrt{n + 1}} for n = 1, 2, ..., 99

    I'm not exactly sure how to start this question.

    Please help, BG
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  2. #2
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    Quote Originally Posted by BG5965 View Post
    Hi, here is a problem involving summing a series involving a complex equation.

    Sum this series:

    a_1 + a_2 + ... + a_{99}
    where a_n = \frac{1}{(n + 1)\sqrt{n} + n\sqrt{n + 1}} for n = 1, 2, ..., 99

    I'm not exactly sure how to start this question.

    Please help, BG
    Notice that if you multiply top and bottom by (n + 1)\sqrt{n} - n\sqrt{n + 1} then

    a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1) } = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}

    so the sum becomes telescopic.
    Last edited by Jester; July 28th 2009 at 03:21 PM. Reason: typo
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    Sorry, but what does telescopic mean, in terms of maths?
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  4. #4
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    Quote Originally Posted by BG5965 View Post
    Sorry, but what does telescopic mean, in terms of maths?
    If you write out the terms you'll see that most of the terms cancel. Only the first and last survive. This is what we mean as a telescopic series.
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  5. #5
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    Quote Originally Posted by Danny View Post
    Notice that if you multiply top and bottom by (n + 1)\sqrt{n} - n\sqrt{n + 1} then

    a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1)^2 } = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}

    so the sum becomes telescopic.
    Did you mean:

    a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1) }

    in the first part? And if so, the next part makes sense... but how did you get from \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } to \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}...

    EDIT: Ah, nm, got it. \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}
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  6. #6
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    Quote Originally Posted by rowe View Post
    Did you mean:

    a_n = \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1)^2 n - n^2(n + 1) }

    in the first part? And if so, the next part makes sense... but how did you get from \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } to \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}...

    EDIT: Ah, nm, got it. \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}
    I did - thanks for pointing out the typo.
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  7. #7
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    Interestingly, what happens when this becomes an infinite series? That is,

    a = \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)

    I'm guessing it gets closer and closer to 1 since \lim_{x\to\infty }{{1}\over{\sqrt{x+1}}} = 0, so a = 1.
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  8. #8
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    So does that mean the answer is:

    \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{1+1}} + \frac{1}{\sqrt{99}} - \frac{1}{\sqrt{99+1}}<br />

    I'm a little confused...
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  9. #9
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    Quote Originally Posted by rowe View Post
    Interestingly, what happens when this becomes an infinite series? That is,

    a = \sum_{n=1}^{\infty} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)

    I'm guessing it gets closer and closer to 1 since \lim_{x\to\infty }{{1}\over{\sqrt{x+1}}} = 0, so a = 1.
    Yes, since S_n = 1-\frac{1}{\sqrt{n+1}}

    @BG: \frac{1}{ \sqrt{1}} - \frac{1}{ \sqrt{1+1}} +
    \frac{1}{ \sqrt{1+1}} - \frac{1}{ \sqrt{1+2}} + \frac{1}{ \sqrt{1+2}} - ... - \frac{1}{ \sqrt{99}} + \frac{1}{ \sqrt{99}} - \frac{1}{ \sqrt{1+99}} = ?


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  10. #10
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    So its:

    \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{99+1}} ?

    Also, how did you get from:

     \frac{(n + 1)\sqrt{n} - n\sqrt{n + 1}}{ (n + 1) n } to \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} using \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{a}}?

    Thanks, BG
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