Let $\{p_n(x)\}_{n=1}^{+\infty}$ be a set of functions such that for $\forall x \in \mathbb{R} \forall n \in \mathbb{N}: 0 < p_n(x) < 1$ and $\sum\limits_{n=1}^{+\infty} p_n(x) = x$. I wonder if it is always true that $1 - \prod\limits_{n=1}^{+\infty} (1-p_n(x)) \sim x$ as $x$ goes to $0$?
So can someone prove the equivalence in general or provide a counterexample? If a counterexample exists, I'm interested in finding $\{p_n(x)\}_{n=1}^{+\infty}$ that minimizes the following limit:
$$\lim\limits_{x \to 0}\dfrac{1- \prod\limits_{n=1}^{+\infty} (1-p_n(x))}{x}$$
If the limit above can be equal to $0$ for some $\{p_n(x)\}_{n=1}^{+\infty}$, I would be interested to see any example of such sequence.
Any ideas, suggestions, hints and references related to the problem would be greatly appreciated.
Since $$ \prod_{n=1}^{\infty} (1-p_n(x)) \leq \exp(-\sum_{n=1}^{\infty} p_n(x)), $$ you get $$ \frac{1 - \prod_{n=1}^{\infty} (1-p_n(x))}{x} \geq \frac{1 - e^{-x}}{x}, \qquad x\in (0,1), $$ hence $$ \liminf_{x\to 0} \frac{1 - \prod_{n=1}^{\infty} (1-p_n(x))}{x} \geq 1. $$ On the other hand, since $0< p_n(x) < x$ and, for every $x\in (0,1)$, $\log(1-t) \geq -t/(1-x)$ for every $t\in [0, x]$, we get $$ \frac{1 - \prod_{n=1}^{\infty} (1-p_n(x))}{x} \leq \frac{1-\exp(-\frac{x}{1-x})}{x} $$ hence $$ \limsup_{x\to 0} \frac{1 - \prod_{n=1}^{\infty} (1-p_n(x))}{x} \leq 1. $$