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Math Help - tricky prob;em of summation

  1. #1
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    tricky prob;em of summation

    X=1/1001 + 1/1002 + ........ + 1/3001 then..

    a) X<1
    b) X>3/2
    c) 1<X<3/2
    d) None of the above
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  2. #2
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    Quote Originally Posted by biplab View Post
    X=1/1001 + 1/1002 + ........ + 1/3001 then..

    a) X<1
    b) X>3/2
    c) 1<X<3/2
    d) None of the above
    c)
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  3. #3
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    Thanks..but can anybody please hep me in detail solution...
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    Quote Originally Posted by biplab View Post
    X=1/1001 + 1/1002 + ........ + 1/3001 then..

    a) X<1
    b) X>3/2
    c) 1<X<3/2
    d) None of the above
    1+\frac{1}{2}+...+\frac{1}{n} \approx \ln n. I can give you an error estimate but you probably are not interested in that. Thus, 1+\frac{1}{2}+...+\frac{1}{3001} \approx \ln 3001. But 1+\frac{1}{2}+...+\frac{1}{1000}\approx \ln 1000. This means, \frac{1}{1001}+...+\frac{1}{3001}\approx \ln 3001 - \ln 1001 = \ln (3001/1001) = 1.09....
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  5. #5
    Moo
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    Hello,

    Quote Originally Posted by ThePerfectHacker View Post
    1+\frac{1}{2}+...+\frac{1}{n} \approx \ln n
    I'm only curious : how do you get this approximation ? ?
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    Quote Originally Posted by Moo View Post
    I'm only curious : how do you get this approximation
    Since you ask I would have to give you the error term as well.

    Consider the function f(x) = \frac{1}{x} on the interval [1,n] where n\geq 2 is an integer.
    Then, approximating by Riemann sums,
    \sum_{k=1}^{n-1} \frac{1}{k} \geq \int_1^n \frac{dx}{x} \geq \sum_{k=2}^n\frac{1}{k}
    Thus, H_n - \frac{1}{n} \geq \log n \geq H_n - 1 \implies |H_n  - \log n| \leq \frac{1}{n}.

    This is Mine 94th Post!!!
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  7. #7
    Moo
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    Oh I see, I was looking towards power series and Taylor series ^^'

    Thanks
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