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lim n --> oo sin(pi/n)/[sin(pi/n)] = 1 <>0. Since a necesssary condition for Sum a_n to converge is that lim a_n = 0, it follows your series diverges. It goes to oo.

The sequence ((pi/n)) /(sin(pi/n)) is neither decreasing nor incresing, its sign changes infinitely many times.

2007-12-03 07:18:54 · answer #1 · answered by Steiner 7 · 0 0

I won't do this for you. But I will tell you how to do it.

pi/n obviously goes to zero as n gets large.

Take the derivative of x/sin(x) to see if it is decreasing for positive numbers that are close to zero. That will tell you if the sequence is increaseing or decreasing.

The series cannot converge if the sequence does not go to zero as x gets large. I think it does (but haven't looked at it). To show it converges -- you need to find a sequence that is larger than this (for large values on n) that you know converges. To show that is does not converge -- you need to find a sequence that is less than this (for large values on n) that you know does not converge.

You may want to start by looking at the Taylor Series expansion od sin.

My instinct tells me that this does not converge. I could be wrong.

2007-12-01 20:45:50 · answer #2 · answered by Ranto 7 · 0 0

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