<s>[INST] 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.
 [/INST] 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}
</s>