Does sin^2(1/n) converge?

By comparison, though it's a bit tricky. First, show that for x in [0,1], sin x ≤ x. You can do this by showing that the difference f(x) = sin x - x has one critical number at zero, and so the maximum occurs at either 0 or 1, the other endpoint. sin 0 - 0 = 0, sin 1 - 1 < 0, so the max value of the difference is zero. Since 0 < 1/n ≤ 1 for any positive integer n, it then follows that sin(1/n) ≤ 1/n. Hence sin²(1/n) ≤ 1/n². Both are positive, and since ∑1/n² is known to converge (p-series with p>1), it then follows that ∑sin²(1/n) converges.