Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$ 
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
 \frac{1}{2n+1}\right)$$

I promise this is the last one for today:
Using Simpson's rules:
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
 \frac{1}{2n+1}\right) = \sum_{n=1}^\infty 2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}$$
Now, $$\left|2\cos\frac{4n+1}{4n² + 2n}\sin\frac{1}{8n² + 4n}\right| \leq \frac{2}{8n² + 4n}$$
hence by the comparison test, the series converges absolutely, and hence it also converges. Is this correct?

As an alternative, since:
$$\sin \frac{1}{2n} - \sin
 \frac{1}{2n+1}=\frac{1}{2n}-\frac{1}{2n+1}+o\left(\frac{1}{n^2}\right)=\frac{1}{2n(2n+1)}+o\left(\frac{1}{n^2}\right)$$
the series converges by limit comparison with:
$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$
