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Limits playing hard to get

  1. May 14, 2010 #1
    By what tools can we evaluate limits like this:
    Lim (sin((n+1)/n^2) + sin((n+2)/n^2) + ... + sin((2n)/n^2)) as n->infinity

    I've played around with Maclaurin polynomials, squeeze and Stolz–Cesàro theorems...
  2. jcsd
  3. May 14, 2010 #2
    I would say that the limit is 0 and would reach this conclusion by looking at the value of each term as n --> ∞. I don't know if theres a specific method for this kind of limit.
  4. May 14, 2010 #3


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    Each term goes to zero but the number of terms goes to infinity, so that's not good enough. On the other hand this problem doesn't look all that subtle. Just estimate an upper bound for the size of the nth sum.
    Last edited: May 14, 2010
  5. May 15, 2010 #4
    the best estimate I was able to get is 1<=limit<=2
  6. May 15, 2010 #5


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    Well, that is not so bad, since the limit equals 3/2 :)

    Using [tex]\sin x = x + \mathcal{O}(x^3)[/tex], we get

    [tex]\sum_{i=0}^n\sin\left(\frac{n+i}{n^2}\right)=\sum_{i=0}^n \frac{n+i}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)=\frac{3(n+1)}{2n}+\mathcal{O}\left(\frac{1}{n^3}\right)\to\frac{3}{2}[/tex].
  7. May 15, 2010 #6
    I can't understand why this true?
  8. May 15, 2010 #7


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    It is not true, but that's not what he wrote. Don't forget that big O added to that term. What he has done is expand the sine using its power series. If you do that you will notice that the first term is (n+i)/n^2 while the rest is of order n^-3 or smaller.
  9. May 15, 2010 #8
  10. May 15, 2010 #9


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    You should quote correctly ;)
  11. May 15, 2010 #10
    I still have hard time understanding this =( (I'm new to taylor series)
    By the way I trying to calculate series limit...
  12. May 15, 2010 #11


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    Well, tell us what you don't understand? Do you understand sin x = x + O(x^3), i.e. do you know the taylor series of the sine?
  13. May 15, 2010 #12
    I Don't understand these points:
    1. [tex] \sum_ {i=0}^n \frac{n+i}{n^2}+\mathcal{O}\left(\frac{1}{n^3}\right)=\frac{3(n+1)}{2n}+\mathcal{O}\left(\frac{1}{n^ 3}\right)[/tex]
    2. How can we approximate series using Taylor polynomials? (they don't have derivative)

    Last edited: May 15, 2010
  14. May 15, 2010 #13


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    The only approximation that is made, is sin x = x +O(x^3). The rest is just substituting in the sum. I used that

    [tex]\sum_ {i=0}^n \frac{n+i}{n^2}=\frac{3(n+1)}{2n}.[/tex]

    This is easily calculated if you know that

    [tex]\sum_ {i=0}^n i=\frac{n(n+1)}{2}.[/tex]
  15. May 15, 2010 #14
    Oh thanks,
    but can we use this directly on a series?
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