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Math Help - Limits question

  1. #1
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    Limits question

    Limit[(Sum[i^0.5, {i, n}] - n^0.5)/n^1.5, n -> \[Infinity]]

    F[x] = [ 1^0.5 + 2^0.5 ....... (n-1)^0.5 ] / n*n^0.5


    Find Limit f(x) where n tends to infinity.

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  2. #2
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    \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-1}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \cdots +  (n-2)^{\frac{1}{2}} + (n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}

    = \lim_{n \to \infty} \left( \frac{\sqrt{1}}{n^{\frac{3}{2}}} + \frac{\sqrt{2}}{n^{\frac{3}{2}}} + \cdots + \frac{(n-2)^{\frac{1}{2}}}{n^{\frac{3}{2}}} + \frac{(n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right)

    You can see that since the degrees of the denominators is greater than that of the numerators, they dominate (for each term).
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    Quote Originally Posted by o_O View Post
    \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-1}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \cdots +  (n-2)^{\frac{1}{2}} + (n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}

    = \lim_{n \to \infty} \left( \frac{\sqrt{1}}{n^{\frac{3}{2}}} + \frac{\sqrt{2}}{n^{\frac{3}{2}}} + \cdots + \frac{(n-2)^{\frac{1}{2}}}{n^{\frac{3}{2}}} + \frac{(n-1)^{\frac{1}{2}}}{n^{\frac{3}{2}}}\right)

    You can see that since the degrees of the denominators is greater than that of the numerators, they dominate (for each term).
    Yes, but the number of terms in the series is n, which can be tricky...

    Let  \sum_{i=1}^n \sqrt{i-1} = S_n and observe that

    \int_2^n \sqrt{x-2} \, dx< S_n < \int_1^n \sqrt{x-1} \, dx
    (2/3) (n-2)^{3/2} < S_n < (2/3) (n-1)^{3/2}
    \frac{(2/3) (n-2)^{3/2}}{n^{3/2}} < \frac{S_n}{n^{3/2} }< \frac{(2/3) (n-1)^{3/2}}{n^{3/2}}

    so, taking the limit,

    2/3 \leq \lim_{n \to \infty} \frac{S_n}{n^{3/2}} \leq 2/3
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  4. #4
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    why did you take Sn less than "Integrate of (n-1)" ?

    Shouldnt it be equal to Integrate of (n-1) itself?
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    Quote Originally Posted by champrock View Post
    why did you take Sn less than "Integrate of (n-1)" ?

    Shouldnt it be equal to Integrate of (n-1) itself?
    No, the sum is not equal to the integral.

    Try this: Draw x and y coordinate axes. Sketch rectangles of height \sqrt{n-1} for n = 1, 2, 3, 4, 5 (say), where the base of each rectangle is the line segment from (n, 0) to (n+1, 0). Now sketch the curves y = \sqrt{x-1} and y = \sqrt{x-2}. S_n is the total area of the first n rectangles. Can you "see" the inequality now?
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  6. #6
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    Your limit is equal to a Riemann sum  \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sqrt{n-i}}{n^{\frac{3}{2}}} = \lim_{n \to \infty} \sum_{i=1}^{n} \sqrt{1-\frac{i}{n}} \cdot \frac{1}{n} = \int_0^1 \sqrt{1-x}\,dx
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