Summing up this series

  • #1
utkarshakash
Gold Member
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Homework Statement


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.

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.
 

Answers and Replies

  • #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.
 
  • #3
utkarshakash
Gold Member
855
13
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.

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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