anonymous
  • anonymous
how does sum sin(pi/n) n->infinity diverge?
Mathematics
schrodinger
  • schrodinger
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JamesJ
  • JamesJ
For 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
  • anonymous
that was my patter of thought, but doesn't that only work when x->0 not infinity?
JamesJ
  • JamesJ
As n --> infty, 1/n --> 0

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anonymous
  • anonymous
does make sense sort of. thanks!

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