Do you remember the formula?

Let $\displaystyle S_n= \sum _{k=1}^{n} 2\cos \frac{3\pi}{2^{k+1}} \sin \frac{\pi}{2^{k+1}}$ .

From Product-to-Sum Formula,

$\displaystyle2\cos \frac{3\pi}{2^{k+1}} \sin \frac{\pi}{2^{k+1}} =2 \cdot \frac{1}{2} [\sin(\frac{3\pi}{2^{k+1}} + \frac{\pi}{2^{k+1}}) \sin(\frac{3\pi}{2^{k+1}} - \frac{\pi}{2^{k+1}})] =\sin \frac{4\pi}{2^{k+1}} - \sin\frac{2\pi}{2^{k+1}} =\sin \frac{\pi}{2^{k-1}} - \sin\frac{\pi}{2^k}$

Therefore,

$\displaystyle S_n=\sum _{k=1}^{n} (\sin\frac{\pi}{2^{2k-1}}-\sin\frac{\pi}{2^k}) =(\sin\pi -\sin\frac{\pi}{2}) +(\sin\frac{\pi}{2}-\sin \frac{\pi}{2^2}) +\cdots + (\sin\frac{\pi}{2^{n-1}}-\sin\frac{\pi}{2^n}) =\sin \pi-\sin \frac{\pi}{2^n} =-\sin\frac{\pi}{2^n}$

Thus,

$\displaystyle\lim_{n\to\infty} S_n=-\sin 0=0$

Therefore, $\displaystyle\sum _{k=1}^{\infty} 2\cos \frac{3\pi}{2^{k+1}} \sin \frac{\pi}{2^{k+1}} =0$