Michael K. answered 03/22/21
PhD professional for Math, Physics, and CS Tutoring and Martial Arts
We see that sin(n) will alternate in sign as n increases from 1 to infinity (although not aligned to points of π/2, we do know that sin(x) will oscillate between values of -1 and 1.
So to remove the alternating aspect, provide an upper bound for the sin(n)/2n ≤ |sin(n)|/2n ≤ 1/2n
Taking the infinite sum (lower bound = 1, upper bound = infinity) we can find the sum of 1/2n over this range of n as 1. Since the non-alternating sum is bounded, the alternating series in this case, is also bounded to be less than or equal to 1. This proves the original summation is bounded and is less than 1 (convergence). Furthermore, since the non-alternating series converges on its own, the alternating sum also converges (absolutely).
So we have absolute convergence (D)