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Please help me with this limit.

  1. Mar 12, 2005 #1
    Please show that:

    limit when n goes to infinity of the sum from i=1 to n of 1/n * sqrt(1-i^2/n^2) equals to zero.

    Sorry, i havent learned yet to use that Tex thing.
     
  2. jcsd
  3. Mar 12, 2005 #2
    Not sure of the formal way, but here is what I get.

    The inside of the sqrt goes to zero because n^2 goes to infinity, making the interior go to zero. 0/n as n -> infinity is still 0.
     
  4. Mar 12, 2005 #3

    dextercioby

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    Yes,but it's an infinite sume of "zero-s"...Are us sure you're referring to

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

    My maple says it's [tex] \frac{\pi}{4} [/tex]...

    Daniel.
     
  5. Mar 12, 2005 #4

    dextercioby

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    It is.I'm a genius.That limit (together with the sum) can be put in connection to the Riemann sum of the integral (it actually is)

    [tex] I=:\int_{0}^{1}\sqrt{1-x^{2}} \ dx =\frac{\pi}{4} [/tex]...

    You must be talking about something else...

    Daniel.
     
  6. Mar 13, 2005 #5
    I think the 1/n is inside the summation, but I am not totally sure. Can we get clarity from the thread starter.
     
  7. Mar 13, 2005 #6
    You can take the [tex]\frac{1}{n}[/tex] out of the summation because it doesn't depend on [tex]i[/tex]. As stated above by Daniel,

    [tex]\lim_{n \rightarrow \infty} \left(\sum_{i=1}^n \left[\frac{1}{n} \sqrt{1 - \frac{i^2}{n^2}}\right]\right) = \int_0^1 \sqrt{1-x^2} \ dx = \frac{\pi}{4}[/tex]

    Here [tex]\frac{1}{n}[/tex] is the width [tex]\Delta x[/tex] of the intervals in the Riemann sum, and [tex]\frac{i}{n}[/tex] is the [tex]x_i^*[/tex] (and you will note that it is indeed always in the appropriate interval).
     
    Last edited: Mar 13, 2005
  8. Mar 13, 2005 #7
    Ah, my bad.

    I don't really deal with limits like that, so I was basing what I said on what I know.

    I was wrong though.
     
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