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anonymous
 4 years ago
how does sum sin(pi/n) n>infinity diverge?
anonymous
 4 years ago
how does sum sin(pi/n) n>infinity diverge?

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JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1For small \( x \) \[ \sin x \approx x \] Hence intuitively at least, \[ \sum_n \sin(\pi/n) \approx \sum_n \pi/n \] and that second sum diverges. Now that's not a proof, but it does suggest something. Try and bound \[ \sin(\pi/n) \] below by something that does diverge also.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that was my patter of thought, but doesn't that only work when x>0 not infinity?

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.1As n > infty, 1/n > 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does make sense sort of. thanks!
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