Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$

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:

$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$

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.

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).