# Bounding the product of a sequence

I am trying to find an upper bound for the following sequence:

$$(1-p_1)(1-(p_1+p_2))\cdots(1-(p_1+\cdots+p_n))$$

with $n$ groups to multiply. I have written it like this:

$$\prod_{i=1}^n \left({1 - \sum_{j=1}^i {p_j}}\right)$$

Where $p_i$ denotes a constant in the open interval $(0,1)$ and $\sum_{i=1}^n p_i = 1$.

Here's what I've attempted to do:

$$\prod_{i=1}^n \left({1 - \sum_{j=1}^{i}{p_j}}\right) = \prod_{i=1}^n \left(1 - O(1) \right) = \prod_{i=1}^n 1 - \prod_{i=1}^n O(1) = 1 - \prod_{i=1}^n O(1)$$

But I don't know how to bound $1 - \prod_{i=1}^n O(1)$.

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there is no $i$ in your expression. I suppose the sum goes from $j=1$ to $i$ –  Yimin Feb 17 '13 at 17:34
I will edit my post to clarify. –  proc-self-maps Feb 17 '13 at 17:36
Can $\sum_{k=1}^{n} p_k >1$? –  Alex Feb 17 '13 at 17:46
@Alex: No, I should have mentioned that. Thank you. More precisely, $\sum_{k=1}^n p_k = 1$. I'll also edit this into my question. –  proc-self-maps Feb 17 '13 at 17:48
if this is so, your last term is $0$ and so the producut equals $0$ –  Dominic Michaelis Feb 17 '13 at 17:51