How do I determine the convergence of $\sum_{n=1}^\infty (-1)^n n\sin(\frac{1}{n^2})$ I tried to prove whether $\sum_{n=1}^\infty (-1)^n n\sin(\frac{1}{n^2})$ converges or not for some long time.
Of course it is alternating, but the positive term is not monotonically decreasing. Liouville's theorem implies that $n\sin(\frac{1}{n^2})\leq \frac{A}{n}$ for some positive constant $A$. But this is not relevant to determine the convergence of the series. Even though it has a bound, it need not be monotonic.
How do I prove this? Thank you in advance.

Let us look at the behaviour of $n\sin(1/n^2)$ for large $n$. 
Let $f(t)=\frac{\sin(t^2)}{t}$.
We have 
$$f'(t)=\frac{2t^2\cos(t^2)-\sin(t^2)}{t^2}.$$
The numerator can be rewritten as $\frac{1}{t^2\cos^2 t}\left(2-\frac{\tan(t^2)}{t^2}\right)$. 
Since $\frac{\tan u}{u}\to 1$ as $u\to 0$, it follows that for small positive $t$ the function $f'(t)$ is positive. 
So for large enough $n$ the function $n\sin(1/n^2)$ is decreasing, and we can use the alternating series test.
