# Summing up this series

1. Oct 14, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
Let An = $\sum_{r=1}^n arccos \frac{r}{n}$ and Bn=$\sum_{r=0}^{n-1} arccos \frac{r}{n}$ 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)$\lim_{n \to \infty} \dfrac{A_n}{n} = 1$

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. Oct 14, 2013

### Office_Shredder

Staff Emeritus
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. Oct 15, 2013

### utkarshakash

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?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted