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Summing up this series

  1. Oct 14, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Let An = [itex]\sum_{r=1}^n arccos \frac{r}{n} [/itex] and Bn=[itex]\sum_{r=0}^{n-1} arccos \frac{r}{n} [/itex] for n=1,2,3....... then

    A)A(2010) < 2010 and B(2010)>2010
    B)A(2010) > 2010 and B(2010)<2010
    C)A(n) < B(n) for all n
    D)[itex]\lim_{n \to \infty} \dfrac{A_n}{n} = 1 [/itex]

    More than one option is correct.

    3. The attempt at a solution

    Finding A(2010) seems really challenging. I start by writing out all the terms manually.

    arccos(1/2010)+arccos(2/2010)+arccos(3/2010)+arccos(4/2010)+..... ..arccos(2010/2010)

    Atmost I can only simplify the last term to 0. But what about the rest of the terms? The formula arccosx + arccosy = arccos{xy − √[(1 − x2)(1 − y2)]} doesn't help. It only makes the terms more complicated.
     
  2. jcsd
  3. Oct 14, 2013 #2

    Office_Shredder

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    For A and B there is no way they expect you to evaluate that series directly - it's going to come from a more general statement like A(n) < n for all n, or A(n) > n for all n.

    As far as how to do that I recommend thinking about these series as Riemann sums (modulo a little algebra) if you have covered that.
     
  4. Oct 15, 2013 #3

    utkarshakash

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    So this is basically an integral where subintervals are of unit length and the function is arccos(x) and that instead of finding the sum I should find the integral. Is this what you mean?
     
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