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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.

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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. $\prod (1-ma_k)$ $\le \prod (1-mc)$ $= (1-mc)^n$ $\to 0$).

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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).

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  • $\begingroup$ That does not work. $\prod(1-ma_k) \ge \prod(1-ma_1) \to 0$, which tells us nothing useful. $\endgroup$ – Milten Nov 2 '19 at 22:33

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