Why does (-1)^n(sin(pi/n)) converge when (sin(p/n)) diverges

[solour]

Homework Statement

I know that ∑n=1 to infinity (sin(p/n)) diverges due using comparison test with pi/n, despite it approaching 0 as n approaches infinity.

However, an alternating series with (-1)^n*sin(pi/n) converges. Which does not make sense because it consists of two diverging functions.

Is there any intuitive explanation for this? Or is it just a rule that I need to remember when treating alternating series

Thank you.

 

[John Park]

How about this: For large n, sin(π/n) is approximately π/n, so the series is approximately πΣ(1/π) which is known to diverge.

With alternating signs, the series approximates πΣ( (1/n) - 1/(n+1) ) which approaches π.ln(2). Actually I think I find it more persuasive to group pairs of successive terms:
(1/n) - 1/(n+1) = 1/(n(n+1)) = O(1/n²), but I gather that's not always a safe thing to do.

 

[vela]

However, an alternating series with (-1)^n*sin(pi/n) converges. Which does not make sense because it consists of two diverging functions.

The fact that $\sum (-1)^n$ and$\sum \sin(\pi/n)$ both diverge says absolutely nothing about the alternating series. If you don't believe that, you might consider $a_n = b_n = 1/n$. Neither $\sum a_n$ nor $\sum b_n$ converge, but $\sum a_nb_n = \sum 1/n^2$ converges.