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?
 A: 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:
$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$
A: It's simpler still with equivalents:
$$\sin \frac{1}{2n} - \sin \frac{1}{2n+1}=2\sin\frac1{4n(2n+1)}\underbrace{\cos\frac{4n+1}{4n(2n+1)}}_{\substack{\downarrow\\\textstyle1}}l\sim_\infty2\,\frac1{4n(2n+1)}\cdot 1\sim_\infty\frac1{4n^2},$$
which converges.
A: It may be easier to prove a stronger result: Suppose $f$ is increasing on $[0,b].$ Let $b\ge b_1 > a_1 > b_2>a_2  > \cdots \to 0^+.$ Then
$$\sum_{n=1}^{\infty}(f(b_n)- f(a_n)) <\infty.$$
Proof: The $n$th partial sum is
$$S_n = f(b_1)-f(a_1) + f(b_2)-f(a_2)+ \cdots + f(b_n)-f(a_n)$$ $$\le\, f(b_1)-f(b_2) + f(b_2)-f(b_3)+ \cdots + f(b_n)-f(b_{n+1})$$ $$ = f(b_1)-f(b_{n+1}) \le f(b_1) - f(0).$$
Thus the sequence $S_n,$ which is increasing, is bounded above. Therefore $S_n$ converges. By definition, this implies the series converges (absolutely of course).
A: Another proof: The partial sums are the same as the even partial sums of the alternating series$$\sum (-1)^n\sin \frac{1}{n}$$ which converges by the Leibniz test.
A: 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.
A: $$|\sin(\frac{1}{2n})-\sin(\frac{1}{2n+1})| = |cos(\xi)(\frac{1}{2n}-\frac{1}{2n+1})|$$
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.
