<s>[INST] 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?
 [/INST] Since $|\sin t|\le |t|$ we have 
 $$|\sin(a)-\sin(b)|= 2\left|\sin\left(\frac{a-b}{2}\right)\cos\left(\frac{a+b}{2}\right)\right|\le |a-b|=  \frac{1}{2n}-\frac{1}{2n+1}\sim \frac{1}{n^2}$$
with $a= \frac{1}{2n}$ and $b=\frac{1}{2n+1}$
Hence, the series is absolutely convergent.
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