limit of adjacent series Suppose $U_n = \prod_{k=2}^n \cos\left(\frac{\pi}{2^k}\right)$ and $V_n = U_n\cdot\cos\left(\frac{\pi}{2^n}\right)$.
How can I prove that $U_n$ and $V_n$ are adjacent?
What is the limit of $U_n$ and $v_n$?
Note: $W_n = U_n\cdot\sin\left(\frac{\pi}{2^n}\right)$ and $W_n$ is geometrical. 

By the double angle formula, we have that
$$\sin x=2\sin \frac x 2 \cos \frac x 2 $$
Using the double angle formula once more
$$\sin x=2\cdot 2 \sin \frac x 4\cos \frac x 4  \cos \frac x 2 $$
You should realize that by induction,
$$\sin x=2^n \sin \frac x {2^n} \prod_{k=1}^n \cos \frac x {2^k}   $$
This means that, for $x\neq 0$.
$$\frac{\sin x}x=\frac{2^n}x \sin \frac x {2^n} \prod_{k=0}^n \cos \frac x {2^k}   $$
$$\frac{\sin x}{x}= \frac{\sin \dfrac x {2^n}}{\dfrac x {2^n}} \prod_{k=0}^n \cos \frac x {2^k}   $$
Now let $x=\pi /2$. We get
$$\frac{{\sin \frac{\pi }{2}}}{{\frac{\pi }{2}}} = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 0}^n {\cos } \frac{\pi }{{{2^{k + 1}}}}$$
or
$$\frac{2}{\pi } = \frac{{\sin \frac{\pi }{{{2^{n + 1}}}}}}{{\frac{\pi }{{{2^{n + 1}}}}}}\prod\limits_{k = 2}^{n + 1} {\cos } \frac{\pi }{{{2^k}}}$$
