Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$? Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$?
 A: How about this telescoping product? Let $a_n=2-1/n$ and then let $x_n=a_n/a_{n+1}$. Then
$$
\prod_{k=1}^n x_k = \left(\frac{a_1}{a_2}\right)\left(\frac{a_2}{a_3}\right)\cdots\left(\frac{a_n}{a_{n+1}}\right)=\frac{a_1}{a_{n+1}}=\frac{n+1}{2n+1}
$$
It's easy to verify that $0<x_n<1$ and the partial products obviously tend to $1/2$.
A: Take any decreasing sequence $(\pi_k)_{k\ge0}$ such that
$$\pi_0=1;\quad \forall k\gt0,\pi_k \lt \pi_{k-1};\quad\lim_{k\to\infty}\pi_k=1/2.$$  We simply set $(\pi_k)_{k\ge1}$ as a sequence of partial products,
$$\pi_k=\prod_{n=1}^k x_n,\text{ where }x_n=\pi_n/\pi_{n-1},$$
guaranteeing that $$\prod_{n=1}^\infty x_n=1/2.$$
A: Recall either of Euler's two famous expressions for $\sin x$:
$$\sin x=x\prod_{n=1}^\infty \cos\left(\frac{x}{2^n}\right),$$ 
or
$$\sin x=x\prod_{n=1}^\infty\left(1-\frac{x^2}{\pi^2n^2}\right).$$
Now let $x=\dfrac{\pi}{6}$.
Or else use the following formula of Viète
$$\frac{2}{\pi}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}\cdots,$$
and multiply both sides by $\dfrac{\pi}{4}$. 
A: If you take any sequence $a_1,a_2,a_3,\ldots$ whose sum is $\log_b (1/2)$, then $b^{a_1}, b^{a_2}, b^{a_3},\ldots$ is a sequence whose product is $1/2$.
Later note: Notice that $\frac 1 2 + \frac 1 4 + \frac 1 8 + \frac 1 {16} + \cdots = 1$.  If you multiply every term by $\log_b \frac 1 2$ then you get a series whose sum is $\log_b \frac 1 2$.
Still later note: If $b>1$, then $\log_b(1/2)<0$, and $b^{a_n}$ will be in $(0,1)$ if $a_n<0$.
A: $$x_n=2^{-1/2^n}\qquad (n\geqslant1)$$
A: Try to find a sequence such that $\frac{n+1}{2n}=\prod_{j=2}^nx_j$ (it will do the job). We have $x_2=3/4$ and 
$$x_{n+1}=\frac{\prod_{j=2}^{n+1}x_j}{\prod_{j=2}^nx_j}=\frac{n+2}{2(n+1)}\frac{2n}{n+1}=\frac{n(n+2)}{(n+1)^2}=\frac{n^2+2n}{(n+1)^2}<\frac{n^2+2n\color{red}{+1}}{(n+1)^2}=1.$$
So $x_n=\frac{n^2-1}{n^2}$ does the job. 
A: Here is another:
$$
\frac{1}{2}=(e^{\frac{1}{2}}-1)\prod_{k=1}^{\infty}\left(\frac{2}{e^{2^{-(k+1)}}+1} \right)
$$
See here.
