OK, with n growing without bound, and the ratio between n and pi being transcendental, the value of sin(2n) basically bounces around randomly. cos(n) is also e

OK, with n growing without bound, and the ratio between n and pi being transcendental, the value of sin(2n) basically bounces around randomly. cos(n) is also effectively random, and there's no real relationship between the sin and cos because everything's "random." So I'd bet it doesn't converge. In fact, for any epsilon we can find N such that |sin(2.pi.N)|<epsilon/4. Call the sum up to that point S - it will not be zero, and then look at the sum up to 2N. That will be very nearly 2S. Take one step further and the sum will be roughly 2S+0.7754. I think that's enough to show it doesn't converge, but I'm typing this between other tasks and don't have time to be more formal. If anyone really cares then I'll work on it later. But no, it doesn't converge.