Determine whether the series $\sum\limits_{k=1}^{\infty }\dfrac{\sin k}{k^{2}}=\dfrac{\sin 1}{1^{2}}+\dfrac{\sin \,2}{2^{2}}+\dfrac{\sin 3}{3^{2}}+\cdots$ converges.

Solution This series has both positive and negative terms, but it is not an alternating series. Use the Absolute Convergence Test to investigate the series $\sum\limits_{k=1}^{\infty }\left\vert \dfrac{\sin k}{k^{2}}\right\vert$. Since $$\left\vert \frac{\sin n}{n^{2}}\right\vert \leq \frac{1}{n^{2}}$$

for all $n$, and since $\sum\limits_{k=1}^{\infty}\dfrac{1}{k^{2}}$ is a convergent $p$-series, then by the Comparison Test for Convergence, the series $\sum\limits_{k=1}^{\infty }\left\vert \dfrac{\sin k}{k^{2}}\right\vert$ converges. Since $\sum\limits_{k=1}^{\infty }\dfrac{\sin k}{k^{2}}$ is absolutely convergent, it follows that $\sum\limits_{k=1}^{\infty }\dfrac{\sin k}{k^{2}}$ is convergent.