## anonymous 5 years ago how does sum sin(pi/n) n->infinity diverge?

1. 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.

2. anonymous

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

3. JamesJ

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

4. anonymous

does make sense sort of. thanks!