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?

$$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| = |cos(\xi)(\frac{1}{2n}-\frac{1}{2n+1})|$$
The above is followed by mean value theorem, then 
$$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| \leq \frac{1}{2n}-\frac{1}{2n+1}$$
So the series is absolutely convergent.
