Question

how does sum sin(pi/n) n->infinity diverge?

Answer 1

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.

Answer 2

that was my patter of thought, but doesn't that only work when x->0 not infinity?

Answer 3

As n --> infty, 1/n --> 0

Answer 4

does make sense sort of. thanks!