# How to prove $\lim_{n \to \infty}\prod_{k=1}^n(1 - ma_k) = 0$ if $\sum a_k = \infty$, $1/a_1 \geq m > 0$

As mentioned above, we have a sequence $$\{a_k\}$$, $$k=\{1, 2, \ldots \}$$, $$a_k \in (0, +\infty)$$ is decreasing and $$\sum a_k = \infty$$, $$\lim_{k\to \infty} a_k = 0$$.

$$\frac 1 {a_1} \geq m > 0.$$

Then, how to prove that

$$\lim_{n \to \infty}\prod_{k=1}^n(1 - ma_k) = 0$$

I think we can use log. Also, the condition $$\sum a_k = \infty$$ is necessary.

Taking the log on both sides, we wish to prove $$\sum_{k=1}^\infty\log(1-ma_k) = -\infty$$ Note that the choice of $$m$$ ensures that $$ma_k\le ma_1\le 1$$, so the logarithm is well-defined (except if $$ma_1=1$$, but that case is trivial).
We use the inequality $$\log x\le x-1$$: $$\sum_{k=1}^\infty\log(1-ma_k) \le \sum_{k=1}^\infty -ma_k = -m\sum_{k=1}^\infty a_k = -\infty$$
The assumption that $$a_k\to 0$$ is not actually necessary. In fact, if $$a_k\to c>0$$, then the claim is trivial by an argument similar to David Sillman's answer. (I.e. $$\prod (1-ma_k)$$ $$\le \prod (1-mc)$$ $$= (1-mc)^n$$ $$\to 0$$).
As stated, $$m\in(0,1/(a_1)]$$. Keep in mind, as well, that $$(a_k)\searrow$$. This means that the greatest value of $$ma_k$$ occurs at $$k=1$$. Due to the limit condition on $$a_k$$, we have that $$\inf (ma_k)=0$$ AND that $$\inf (ma_k) = 0\notin (ma_k)$$ because $$a_k$$ cannot take the value 0. Therefore, the product decays which can be shown using that $$\sup (ma_k) = 1$$ and $$\sup (ma_k) = 1 \notin (ma_k)$$ due to the limits on $$m$$. (Specifically, try comparing it to $$\prod (1-ma_1)$$, which is necessarily greater than or equal to our product, $$P$$, due to the decreasing property of $$(a_k)$$. HINT: $$(1-ma_1)\in(0,1)$$ via what we've shown above).
• That does not work. $\prod(1-ma_k) \ge \prod(1-ma_1) \to 0$, which tells us nothing useful. – Milten Nov 2 '19 at 22:33