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I am trying to find if

$$\prod_{j=0}^{k}(1-a_{k-j})$$

can be lower bounded by some constant $c$, with $0 \leq a_n \leq 1$. I have seen in other post that

$$\prod \limits_{j=0}^{k}(1-a_{k-j}) \geq1- \sum\limits_{j=k}^{n}a_{k-j}$$

but I do not know how to proceed. Given $\lim_{m\to \infty} a_m \to 0$ can we find such a lower bound $c$?

Thoth
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  • In the post you linked, there is the following statement: $$1+\sum_{i=1}^n a_i \le \prod_{i=1}^n(1+a_i)$$ and the proof, I think, was based on the nonnegativity of $a_n$'s. So, you cannot just apply this here – VIVID Feb 27 '21 at 17:00
  • @VIVID thanks for the comment, $a_m$ are non-negative in my case. – Thoth Feb 27 '21 at 17:08
  • Anyway you have the minus sign. – VIVID Feb 27 '21 at 17:09
  • I think $a_m \to 0$ is not enough to say something since the initial terms (as you have in your product) of this kind of sequence can be arbitrary. – VIVID Feb 27 '21 at 17:12

1 Answers1

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We cannot bound this product with some $c>0$. For example, let us take a sequence that satisfy the given conditions: $$a_n = \begin{cases} 2, & n = 0 \\ 1/2, & n = \overline{1,k} \\ 0, & n > k \end{cases}$$ Definitely, $a_n \to 0$ and $a_n \ge 0$. However, $$\prod_{j=0}^{k}(1-a_{k-j}) = \prod_{j=0}^k(1-a_j) = (1-2)\left(1 - \frac{1}{2}\right)^k < 0$$ Note that by choosing $a_0$ appropriately, we can make this product be less than any real number.

Therefore, with no constraints on $a_n$, we cannot lower bound the product.

EDIT: After the additional condition $0 \le a_n \le 1$, we have $$1-a_0 \ge 1-1 = 0 \\ 1-a_1 \ge 1-1 = 0 \\ ... \\ 1-a_k \ge 1-1 = 0 $$ Multiplying all inequalities we get $$\prod_{j=0}^k(1-a_j) \ge 0$$

VIVID
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  • Thanks, indeed the lower bound I need in my proof must be negative. I joust realized it. Sorry. Can we make other assumptions so as $c=-1$ (I will make the appropriate edits in the original post). – Thoth Feb 27 '21 at 17:29
  • @Thoth Pay attention to the last sentence of the answer. Just choose $a_0$ large enough to make the product as less as you want – VIVID Feb 27 '21 at 17:32
  • Sorry for editing the link in my post, I think it more correct now and more close to what I need in my proof. Does this change sign of the lower bound? Can we say something more precise for its value now? – Thoth Feb 27 '21 at 18:01
  • @Thoth $0\le a_n \le 1$ gives something. See the edit. – VIVID Feb 27 '21 at 18:09