Limit of the sequence For all natural numbers $n$
$$\begin{align}
a_n &= 2^{1/2}\cdot 2^{1/4}\cdot 2^{1/8}\cdot \dotsc\cdot 2^{1/{2^n}}\\
b_n &= \cos(x/2)\cdot\cos(x/4)\cdot\dotsc \cdot\cos(x/2n)
\end{align}$$
Find limit of $a_n$ and $b_n$ when $n$ approach to infinity.
I know that $\lim a_n$ is $2$, but I don't know how to show it.
 A: For $a_n$, you have $\lim_{n \to \infty} a_n = 2^{\sum_{n=1}^\infty 1/2^n}$ and the infinite sum in the exponent is equal to $1$, so $\lim_{n \to \infty} a_n = 2^1 = 2$.
A: For the first one, note that
$$2^{1/2} \cdot 2^{1/4} \cdot 2^{1/8} \cdots 2^{1/2^n} = 2 ^{1/2+1/4+\cdots1/2^n} = 2^{1-1/2^n}$$
For the second one, I assume you have
$$b_n = \cos(x/2)\cos(x/4) \cdots\cos(x/2^n)$$
If $x = (2n+1) \pi \cdot 2^{k-1}$, where $k \in \mathbb{Z}^+$, then $b_k$ onwards is zero and the limit is zero. If not, note that
\begin{align}
b_n & = \cos(x/2)\cos(x/4) \cdots\cos(x/2^n) = \dfrac{\cos(x/2)\cos(x/4) \cdots\cos(x/2^n) \cdot \sin(x/2^n)}{\sin(x/2^n)}\\
& = \dfrac{\cos(x/2)\cos(x/4) \cdots\cos(x/2^{n-1}) \sin(x/2^{n-1})}{2\sin(x/2^n)}\\
& = \dfrac{\sin(x/2^{n-1})}{2 \sin(x/2^n)} b_{n-1} = \dfrac{\sin(x/2^{n-1})}{2 \sin(x/2^n)} \dfrac{\sin(x/2^{n-2})}{2 \sin(x/2^{n-1})} b_{n-2} = \dfrac{\sin(x/2^{n-2})}{2^2 \sin(x/2^n)} b_{n-2}\\
& = \dfrac{\sin(x)}{2^n \sin(x/2^n)}
\end{align}
A: Hint: 
First problem:  Calculate the first few "partial" products.
The first is $2^{\frac{1}{2}}$.
The second is $2^{\frac{1}{2}+\frac{1}{4}}$.
The third is $2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}$.
And so on. Now you certainly know what $\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^n}$ approaches as $n\to\infty$. Then use the fact that $2^x$ is continuous to finish the argument.
Second problem: Treat the case $x=0$ separately. 
We will assume that the last entry is supposed to be $\cos(x/2^n)$.
If $x\ne 0$, and $n$ is large enough to ensure that $\sin(x/2^n)\ne 0$,  multiply on the right by $\sin(x/2^n)$, and use repeatedly $\sin\theta\cos\theta=\frac{1}{2}\sin 2\theta$.
Doing this for the product $\cos(x/2)\cos(x/4)\cos(x/8)$ will be enough for the pattern to become clear. After multiplying by $\sin(x/8)$ on the right, things collapse to $\frac{1}{2^3}\sin x$. 
If we call our product $f_n(x)$, then 
$$f_n(x)=\sin x\frac{1/2^n}{\sin(x/2^n)}=\frac{\sin x}{x}\frac{x/2^n}{\sin (x/2^n)}.$$
Now we can find the limit using standard facts.
