How to show $\prod_{i=1}^n (1-p_i) = \exp{(-\sum_{i=1}^n p_i)} + O(\sum_{i=1}^n p_i^2)$ I have a (seemingly) simple question. How can I see (rigorously) that
\begin{equation}
\prod_{i=1}^n (1-p_i) = \exp{\left(-\sum_{i=1}^n p_i\right)} + O\left(\sum_{i=1}^n p_i^2\right)
\end{equation}
for $\max_{i}p_i\rightarrow 0$ and $p_i\geq 0,\, \forall i$. For $n=1$, the above is a just a Taylor approximation. But, for $n>1$, I have trouble combining the individual error terms (for the individual $i$'s) in the product to obtain the $O(\sum p_i^2)$ error term. Any ideas? Every hint/help is much appreciated. Thanks in advance!
 A: Consider 
$$ \log \prod_{i=1}^n (1-p_i) = \sum_{i=1}^n \log(1-p_i).$$
Since $\max_i p_i \to 0$, use the Taylor series expansion of $\log(1-p_i)$, then it follows that
$$ \log \prod_{i=1}^n (1-p_i) = -\sum_{i=1}^n p_i + O\left( \sum_{i=1}^n p_i^2 \right). $$
Take exponents of both sides and use $e^x=1+ O(x)$, $x$ around $0$, to obtain
$$  \prod_{i=1}^n (1-p_i)= exp(-\sum_{i=1}^n p_i) + 1 + O\left( \sum_{i=1}^n p_i^2 \right). $$
A: I think that the problem is now correct. 
Suppose $\forall i,\;|1-p_i| \leq  1$ (for instance if $\forall i,\;0\leq p_i \leq 1$). We have
$$
\left|\prod_{i=1}^n (1-p_i) - \prod_{i=1}^n e^{-p_i}\right| \leq \sum_{i=1}^n |1 - p_i - e^{-p_i}| \leq \frac{1}{2} \sum_{i=1}^n p_i^2.
$$
Hence,
$$
\prod_{i=1}^n (1-p_i) = \exp\left(-\sum_{i=1}^n p_i\right) + O\!\left(\sum_{i=1}^n p_i^2\right).
$$
The first inequality comes from the fact that if $a_1,\dots,a_n$ and $b_1,\dots,b_n$ are complex numbers with modulus lesser than $1$, then $|\prod a_i - \prod b_i| \leq \sum |a_i - b_i|$ (this follows from an easy recurrence).
