Consider the series

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series is conditionally convergent.

*Proof.* First, we recall the formula for the sine of a sum,

Therefore, we have

(Here we use that for all and for all .) This series is convergent since

and it is monotonically decreasing for .

This convergence is conditional since

But then, since and is decreasing for we have

for all . We know diverges since it is asymptotically equivalent to (i.e., ). Therefore, by the comparison test

diverges. Therefore, the convergence of the series in question is conditional