<s>[INST] Sum of series $\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$ How can we get  sum of this series? $$\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$$
I think we must apply this theorem.

if in  $\sum_{n=1}^\infty a_{n}$, $a_n = b_{n+1}-b_n$ and $\lim_{n\to\infty} b_n = b$; $ \implies $ $\sum_{n=1}^\infty a_{n}=b-b_1$ 

But how? 
 [/INST] The identity $\sin x\cos 3x=\frac{1}{2}(\sin 4x-\sin 2x)$ gives$$\sum_{n\ge 1}\sin\frac{1}{2^n}\cos\frac{4}{2^n}=\frac{1}{2}\sum_{n\ge 1}(\sin\frac{4}{2^n}-\sin\frac{2}{2^n})=\frac{1}{2}\sin 2.$$
</s>