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Limit of an infinite sum.

  1. Mar 14, 2005 #1
    Sorry, i had already post a thread about this, but it was worng. As Dextercioby said,

    [tex] \lim_{n\rightarrow \infty}\left( \frac{1}{n}\cdot\sum_{i=1}^{n}\sqrt{1-\frac{i^{2}}{n^{2}}}\right)= \frac{\pi}{4} [/tex]

    So, how do I show this. Can I do it with an integral (as Dextercioby did)?
    This limit came out when I was triyng to derive the formula of the surface area of a sphere in the way Archimedes did.
  2. jcsd
  3. Mar 14, 2005 #2


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    It looks very much like a Riemann sum.
    If you split the interval [0,1] into n subintervals of width [itex]\Delta x = 1/n[/itex], then the right endpoint of the i-th subinterval is [itex]x_i = i/n[/itex].

    So the sum can be written:

    [tex]\sum_{i=1}^{n}\sqrt{1-x_i^2}\Delta x[/tex]
    So if we let [itex]f(x)=\sqrt(1-x^2)[/itex] then the limit is:

    [tex]\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(x_i)\Delta x= \int_0^1\sqrt{1-x^2}dx[/tex]
    which is the area of a quarter of the unit disc: [itex]\pi/4[/itex].
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