Does this series involving sine converge or diverge: $\sum\limits_{k=1}^\infty \frac{1}{k}\cdot \sin\frac{(-1)^k}{1+k^2}$?

Using the limit comparison test for the following:

$$a_k=\frac{\sin\frac1{k^2+1}}k\;\;,\;\;\;b_k:=\frac1{k^2+1}\implies \frac{a_k}{b_k}=\frac1k\frac{\sin\frac1{k^2+1}}{\frac1{k^2+1}}\xrightarrow[k\to\infty]{}0\cdot1=0$$

so $\;\sum a_k\;$ converges because $\;\sum b_k\;$ does, and thus your series converges absolutely ( since

$$\;\left|\sin\cfrac{(-1)^k}{k^2+1}\right|=\sin\cfrac1{k^2+1}\;)$$

and then it converges.

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.

Note that $\sin((-1)^k/(1+k^2))=\sin(1/(1+k^2))$ if $k$ is even and $-\sin(1/(1+k^2))$ is $k$ is odd. Further, as $k\to\infty$ we have $(1/k)\sin(1/(1+k^2))\downarrow0$. So you are in the shape of $\sum_na_n$ where $a_n$ are alternating, $a_n\to0$. Thus by Leibnitz test the series converges.