Show that $\sum_{n=1}^{\infty} \sin \left(\frac{1}{n^2}\right)$ converges.

You can use the fact that $\sin x<x$ when $x>0$, so that $\sin\dfrac 1 {n^2} < \dfrac 1 {n^2}$. Then cite the "comparison test".

Your conclusion that $\displaystyle \sum_{n=1}^\infty \sin \frac 1 {n^2} = \sum_{n=1}^\infty \frac 1 {n^2}$ is too much: it should say "less than" rather than "equal to".

Use $\sin(x)\le x$ for $x\ge 0$.