<s>[INST] Determine convergence of $\sum_{n=1}^{\infty} (\cos{\frac{2}{n}}-\cos{\frac{4}{n}})$ Determine convergence of $$\sum_{n=1}^{\infty} \left(\cos{\frac{2}{n}}-\cos{\frac{4}{n}}\right)$$
In the answer, it says 
$$\cos{\frac{2}{n}}-\cos{\frac{4}{n}} = 2\sin{\frac{3}{n}}\sin{\frac{1}{n}} \le 2\cdot \frac{3}{n} \cdot \frac{1}{n} = \frac{6}{n^2}$$
But how do I get the above trig substitution? I guess removing the fractions, I will get $\cos{x}-\cos{2x}=2\sin{(2x-1)}\sin{(x-1)}$ ... probably this is wrong, but how do I get that? 
 [/INST] I suggest another way:
$\cos\left(\frac{2}{n}\right)-\cos\left(\frac{4}{n}\right)=-\left(1-\cos\left(\frac{2}{n}\right)\right)+1-\cos\left(\frac{4}{n}\right)\sim -\frac{4}{2n^2}+\frac{16}{2n^2}=\frac{6}{n^2}$ and $\sum \frac{6}{n^2}$ converges.
</s>