Does there exist an example of converging infinite product when all elements are below 1?

Question

I consider the infinite product $$p = \prod_{k=1}^\infty z(k),$$ where $z:\mathbb{N}\to[0,1).\ $ I am wondering if there exists $z$ such that the product converges to a nonzero value? Obviously, we thus need that $z(k)\to 1$ as $k\to\infty$.

Answer 1

You can fit a sequence $z(n)$ so that the infinite product above converges to any number $b\in(0,1)$ you want.

Here is how. Take any strictly decreasing sequence $b_n$ for $n\ge 1$ such that $0<b_n<1$ and $\lim_{n\to\infty}b_n = b>0$ - a positive number. For example, $b_n=\frac{1}{n+1}+\frac{1}{4}$ (with $b=\frac{1}{4}$).

Now set $z(1)=b_1$ and $z(n)=\frac{b_n}{b_{n-1}}$ for $n>1$. What you get is:

$$\prod_{i=1}^n z(i)=z(1)z(2)z(3)\cdots z(n)=b_1\frac{b_2}{b_1}\frac{b_3}{b_2}\cdots\frac{b_n}{b_{n-1}}=b_n\to b\,(n\to\infty)$$

i.e. $\prod_{n=1}^\infty z(n)=b>0$.

Answer 2

Certainly there are strictly decreasing sequences of real numbers $(x_n)_{n\in\mathbb{N}}$ with $0<x_1<1$ and with greatest lower bound $l>0.$ Choose any such sequence and call it $(x_n)_{n\in\mathbb{N}}.$

Now let $z(1) = x_1.$ The we can find $z(2) \in [0,1)$ such that $z(1)\cdot z(2) = x_2,\ $ we can find $z(1)\cdot z(2) \cdot z(3) = x_3,\ $ etc.

Defining the function $z$ in this way, we get $\displaystyle\prod_{k=1}^n z(k) = x_n, $ and taking limits as $n\to\infty,$ we get:

$\displaystyle\prod_{k=1}^{\infty} z(k) = \lim_{n\to\infty} x_n = l > 0.$