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? 

The result follows from the Addition Law for Cosines. 
We have $\cos(x+y)=\cos x\cos y-\sin x\sin y$ and $\cos(x-y)=\cos x\cos y+\sin x\sin y$. Subtract. We get 
$$\cos(x-y)-\cos(x+y)=2\sin x\sin y.$$
Let $x-y=\frac{2}{n}$ and $x+y=\frac{4}{n}$. Solve for $x$ and $y$. We get $x=\frac{3}{n}$ and $y=\frac{1}{n}$.
