# Marriage proposals: value of $\lim_{n \to \infty} \prod_{i=1}^{n} (1-p_i)$ given $p_1>p_2...>p_n$ and $p_i \in (0,1)$

I was thinking about marriage proposals. If the probability of someone saying yes to the proposal is unvarying, say $$p$$, all one needs to do to get to a yes is to ask enough times. The probability of getting a no n times is $$(1-p)^n$$, which will surely converge to zero for any $$p>0$$.

I started to wonder, what if the successive probability of a 'yes' event decreases with each trial? In this case, the probability of getting a no n times is $$\prod_{i=1}^{n} (1-p_i)$$ and $$p_1>p_2...>p_n$$. This would perhaps be a better representation of reality.

What can be said about this series as n goes to infinity, or in general? I know the value must be between 0 and 1, but not much more than that.

Is any monotonic decrease in $$p_i$$ enough to make this value not approach to 0 as n approaches infinity? Or are there some conditions on the 'rate' of decrease in $$p_i$$ for this value to not be zero?

• It can certainly be non-zero, for example $\prod_{i=1}^{\infty}(1-\frac{1}{2^i})\approx 0.2887880951$
– Sil
Feb 22 '21 at 11:38
• The behavior is basically determined by whether $\sum_{i=1}^\infty p_i < \infty$. To see why, take the logarithm of the product and use that $\log(1-x)\sim -x$. Feb 22 '21 at 11:39
• en.wikipedia.org/wiki/Infinite_product#Convergence_criteria
– user
Feb 22 '21 at 11:57