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Evaluate the limit of a series with an integral

  1. Aug 9, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]\lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n}\sqrt{\frac{4i}{n}}[/itex]


    3. The attempt at a solution

    This seems to give the right answer, 16/3, but I can't figure out why:

    [itex]\lim_{n \to \infty}\int_{1}^{n}\frac{4}{n}\sqrt{\frac{4x}{n}}dx[/itex]

     
  2. jcsd
  3. Aug 9, 2013 #2

    vanhees71

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    I'd rather think that you have to interpret the sum as an approximation to the (Riemann) integral
    [tex]I=\int_0^{4} \mathrm{d} x \sqrt{x},[/tex]
    where the interval is divided in [itex]n[/itex] subintervals of equal size.
     
  4. Aug 9, 2013 #3

    HallsofIvy

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    This is more than just approximating. If you have the function [itex]f(x)= \sqrt{x}[/itex], on the interval [0, 4], the Riemann sum, dividing [0, 4] into n equal subintervals, so that each subinterval has length 4/n and x= 4i/n, gives [tex]\sum{i= 1}^n \frac{4}{n}\sqrt{\frac{4i}{n}}[/tex]. As n goes to infinity, we are dividing the interval into more and more smaller and smaller intervals and the limit is the Riemann integral.
     
  5. Aug 9, 2013 #4

    haruspex

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    vanhees71 said the sum approximates the Riemann integral; you're saying the limit of the sum is the Riemann integral. No contradiction there.
     
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