TITLE: Does this series involving sine converge or diverge: $\sum\limits_{k=1}^\infty \frac{1}{k}\cdot \sin\frac{(-1)^k}{1+k^2}$?
QUESTION [12 upvotes]: I have been trying to show the convergence of this series but I can't seem to find a way to do it.
$$\sum_{k=1}^\infty \frac{1}{k}\cdot \sin\frac{(-1)^k}{1+k^2}$$

REPLY [13 votes]: Since for $k\geq 1$,
$$\left|\frac{1}{k}\cdot \sin\left(\frac{(-1)^k}{1+k^2}\right)\right|=\frac{1}{k}\cdot \sin\left(\frac{1}{1+k^2}\right)\sim \frac{1}{k}\cdot\frac{1}{1+k^2}\sim \frac{1}{k^3}$$
and $3>1$, then the series converges absolutely and therefore it is also convergent.