Solution
= Solution
{c}
For $t\geq0$, $|\sin t|\leq t$, so
$$
0\leq\frac{n\sin^2(1/n)}{(\log n)^2}\leq\frac1{n(\log n)^2}.
$$
The <integral test> applies to the decreasing positive comparison function, and
$$
\int_2^\infty\frac{dx}{x(\log x)^2}=\frac1{\log2}<\infty.
$$
The <comparison test> proves that the <series> \b[converges]; since all terms are nonnegative, this is also <absolute convergence>.