# 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 $\frac{1}{2}$?

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 $a_n<0$.

• The terms of the sequence must be in the open interval $(0,1)$ Jul 10 '12 at 18:39
• @Hassan Take $b=1/2$ for instance. Jul 10 '12 at 19:17
• You can write that product for $b=1/2$ in an intuitive (but not very compact) form: $\sqrt{1/2}\sqrt{\sqrt{1/2}}\sqrt{\sqrt{\sqrt{1/2}}}\cdots$. The sequence of factors however has a very nice recursive definition: $x_1=\sqrt{1/2}$, $x_{n+1}=\sqrt{x_n}$ Jul 10 '12 at 19:27
• @HassanMuhammad : If $b>1$, then $\log_b(1/2)<0$, and $b^{a_n}$ will be in $(0,1)$ if $a_n<0$. Jul 10 '12 at 22:03
• @MichaelHardy: Thanks for your answer (+1) Jul 11 '12 at 7:31

$$x_n=2^{-1/2^n}\qquad (n\geqslant1)$$

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.

• I don't think $x_{n}=\frac{n^2-1}{n^2}$ will do the job because if $n=1$, $x_{n}=0$ which is not in $(0,1)$, and of course $\prod_{n=1}^{\infty}=0$ which is not my goal. Jul 10 '12 at 18:25
• @HassanMuhammad: I believe the product starts at $n=2$. Jul 10 '12 at 18:26
• @HassanMuhammad Just shift the sequence if you want it to start at $1$ or $0$, but it doesn't matter. Jul 10 '12 at 18:33
• @oen: But that is not my question. I wrote the product of all its term, so n can be 1 Jul 10 '12 at 18:35
• @HassanMuhammad: As Davide Giraudo mentions, you can just shift the sequence. Let $m = n-1$. Jul 10 '12 at 18:38

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 $1/2$.

• To see that the terms are in the desired range, it helps to write a formula for $x_n$ in terms of $n$. Jul 10 '12 at 21:30
• I like the telescoping product, too. Jul 11 '12 at 5:17

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 $$\prod_{n=1}^\infty x_n=1/2.$$

• Are you sure that for a natural number n, $n\gt 1$, $x_{n}$ is in $(0,1)$? Jul 10 '12 at 19:09
• As long as the sequence $\pi_n$ is positive and strictly decreasing. Jul 10 '12 at 19:32

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 $\dfrac{\pi}{4}$.

• This also gives $$\prod_{n\ge2}\left(1-\frac{1}{n^2}\right)=\frac{1}{2}.$$ (Though it is telescoping and thus easier..)
– anon
Jul 10 '12 at 23:09

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.