Is $\prod_{n=1}^\infty\delta_n=0$ for $\{\delta_n\}_{n\in\mathbb{N}}\subset(0,1)$ decreasing to $0$?

Question

Is $\prod_{n=1}^\infty\delta_n=0$ for $\{\delta_n\}_{n\in\mathbb{N}}\subset(0,1)$ decreasing to $0$?

I used the following approach: Let $g(n)=\Pi_{i=1}^n\delta_i$. We know that $g(n)$ is strictly decreasing in $n$ given $\delta_i\in(0,1)$. We also know that the product of any number of positive reals is positive. Hence $\{g(n)\}_{n\in\mathbb{N}}$ is a strictly decreasing sequence bounded below by $0$. Such s sequence converges to $\inf \{g_n\}=0.$

Is this proof correct?

Answer 1

If $\delta_i$ are decreasing, then $\delta_i < r$ for some $0 < r < 1$ and all $i$, thus

$$ 0 \le \prod_{i=1}^{\infty} \delta_i \le \prod_{i=1}^{\infty} r = \lim_{n\to\infty}r^n \le 0.$$

By squeeze theorem, the product is zero. In fact, you don't need that they're decreasing at all, just that they stay away from $1$ and the above works. This is of course a sufficient condition, but infinite series and products can have extremely bizarre behavior. I doubt there's a good way to further classify this to catch those products whose $\delta_i$ do not stay away from $1$ (some work, some don't).

Answer 2

As pointed out in the comments, your proof has a gap. And it is also needlessly complicated.

You assume that $(\delta_n)_{n\in\Bbb N}\subset(0,1)$ strictly decreases to $0$. In particular,

$$\delta_1\in (0,1), \quad \delta_n\in (0,\delta_1] \; \forall n\ge 1,$$

which yields

$$g(N)=\prod_{n=1}^N\delta_n\le \delta_1^N\xrightarrow{N\to\infty} 0.$$