The series ∑n=1∞sin2nn2\sum_{n=1}^{\infty} \frac{\sin^2 n}{n^2}∑n=1∞n2sin2n converges. Which statement best justifies this convergence?
By Direct Comparison: sin2nn2≤1n2\frac{\sin^2 n}{n^2} \leq \frac{1}{n^2}n2sin2n≤n21 for all nnn, and ∑1n2\sum \frac{1}{n^2}∑n21 converges
By the Alternating Series Test: the series has alternating signs
By the Root Test: limn→∞sin2nn2n=0\lim_{n \to \infty} \sqrt[n]{\frac{\sin^2 n}{n^2}} = 0limn→∞nn2sin2n=0
By the nth-Term Test: limn→∞sin2nn2=0\lim_{n \to \infty} \frac{\sin^2 n}{n^2} = 0limn→∞n2sin2n=0 guarantees convergence