## Calculus (3rd Edition)

Given $$\sum_{k=1}^{\infty}\frac{\sin^2k }{k^2}$$ Compare with $\displaystyle\sum_{k=1}^{\infty}\frac{1}{k^{2}}$, which is a convergent series ( $p-$series with $p>1$) and for $k\geq 1$ \begin{align*} \frac{\sin^2k }{k^2} &\leq \frac{1 }{k^2} \end{align*} Then $\displaystyle\sum_{k=1}^{\infty}\frac{\sin^2k }{k^2}$ also converges.