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Math Help - Limit

  1. #1
    Member Em Yeu Anh's Avatar
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    Red face Limit

    \lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}
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  2. #2
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    Quote Originally Posted by Em Yeu Anh View Post
    \lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}
    You should show some effort. We aren't suppose to just solve problems for you.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    \lim_{n\to\infty}\sum_{i=1}^{n}\frac{3}{n} \displaystyle{[(}1+\frac{3i}{n}\displaystyle{)}^3-2(1+\frac{3i}{n}\displaystyle{)]}
    I would assume you haven't done the fundamental theorem of calculus yet, because this Riemann sum appears to be \int_1^4x^3-2x\,dx=\frac{195}{4}

    Pull the \frac{3}{n} out and expand the summand. You get

    \lim_{n\to\infty}\frac{3}{n}\sum_{i=1}^n\left[(3i/n)^3+3(3i/n)^2+(3i/n)-1\right] =\lim_{n\to\infty}\frac{3}{n}\left[\frac{27}{n^3}\sum_{i=1}^n i^3+\frac{27}{n^2}\sum_{i=1}^n i^2+\frac{3}{n}\sum_{i=1}^n i-\sum_{i=1}^n 1\right]

    Evaluate that using summation formulas and then take the limit.
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  4. #4
    Member Em Yeu Anh's Avatar
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    Smile

    I have studied the fundamental thereom although it was quite some time ago, would you mind explaining how you obtained 4 on the upper limit of integration for \int_1^4x^3-2xdx?

    Sorry, think I may have just found it:
    \frac{3}{n} = \frac{b-1}{n} is that the formula?
    Last edited by Em Yeu Anh; November 10th 2009 at 06:45 PM.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Yup. The reason I thought you hadn't done the FTC yet was that I assumed you wanted to evaluate this limit in order to evaluate the integral. I guess not.
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  6. #6
    Member Em Yeu Anh's Avatar
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    Alright I got 195/4! Thank you very much redsoxfan you have been most helpful
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  7. #7
    Super Member redsoxfan325's Avatar
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    You're welcome.
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