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Homework Statement
*Find the distance between 1 and the various n-th roots of unity - denoted d(k)
*Find a formula for the sum of distances between 1 and each of the n-th roots of unity - denoted S(n)
*Find the limit as n->infinity of (1/n).S(n)
Homework Equations
*The n-th roots of unity are z[0], z[1], z[2],...,z[n-1] - ie. there are 5 5th roots of unity. z[0] is always 1.
*d(k) = abs{z(k)-1}
*S(n) = sum(from k=1 to k=n-1) of d(k) - so n=5 has four terms in the sum
The Attempt at a Solution
By vector subtraction and finding the magnitude, I calculated:
d(k)=2sin( k.pi / n )
Hence:
S(n) = 2.sum(from k=1 to k=n-1) of sin( k.pi / n )
But now I'm stuck on the limit question. I know the answer is about 1.27 from calculating the first dozen term by calculato, however I can't see how to prove this.
What I have is (where limit is as n goes to infinity):
lim(1/n)S[n] = (2/n){sin(pi/n)+sin(2pi/n)+sin(3pi/n)+sin(4pi/n)+...+sin((n-1)pi/n)}
There are n-1 terms in the braces. Every time I try solving it I end up with zero, which I know is not the case, since when I do a spreadsheet S[n]/n starts at 1 (the n=2 case) and increases with increasing n. I've tried decomposing sine into x-x^3/3!+x^5/5!+... but it doesn't seem to help.
Is anyone able to steer me in the right direction?