Determining Whether a Series Converges

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.