$\prod\limits_{k = 0}^\infty {(1 - {p_k})} = 0$ if and only if $\sum\limits_{k = 0}^\infty {{p_k}} $ diverges Let $p_0,p_1,p_2,...$ be real numbers in $(0,1)$. I am trying to prove that 
$\prod\limits_{k = 0}^\infty  {(1 - {p_k})}  = 0$ if and only if the series 
$\sum\limits_{k = 0}^\infty  {{p_k}} $ diverges.
By taking logarithms in the product we have
$\log \left( {\prod\limits_{k = 0}^\infty  {(1 - {p_k})} } \right) = \sum\limits_{k = 0}^\infty  {\log \left( {1 - {p_k}} \right)} $
and the latter series should diverge to $-\infty$ if the product were to be $0$.
Equivalently, $\sum\limits_{k = 0}^\infty  {\left( { - \log \left( {1 - {p_k}} \right)} \right)} $ should diverge to $+\infty$ and the conclusion may follow by approximating $-\log(1-p_k)$ by $p_k$ near $0$.
Any help on how to write this last step formally? How can we prove that $\sum\limits_{k = 0}^\infty  {\left( { - \log \left( {1 - {p_k}} \right)} \right)} = +\infty$ if and only if 
$\sum\limits_{k = 0}^\infty  {{p_k}} = +\infty$ ?
 A: You can try this: if $p_k \not \to 0$ then clearly both $\sum u_k = \infty$ and $\prod_k(1-p_k)=0$. Assume now $p_k \to 0$. Then we have 
$$\log\frac{1}{1-p_k} \simeq p_k$$
so the series 
$\sum \log\frac{1}{1-p_k}$ and $\sum p_k$ are simultaneously convergent.
Obs: You can also throw in the product $\prod_k (1+p_k)$
A: As $1-x_i>0$            it is obvious that $$\lim_{n \to \infty}\prod_{n \ge i \ge 1} (1-x_i)=0$$ is equivalent to $$\lim_{n \to \infty}\prod_{n \ge i \ge 1} \frac{1}{(1-x_i)} = \infty.$$ At first let us transform the latter product into something that has the form of the first (namely $\prod (1+a_i)$) by writing $\frac{1}{1-x_i} = 1+ \frac{x_i}{1-x_i}$. Let $y_i=\frac{x_i}{1-x_i}$. Now all we need to do is show that $\prod(1+y_i) $ and $\sum x_i$ are of the same nature. The result follows directly from the two following facts :
Lemma 1.
For any positive sequence $(a_i)_{i>0}$, $$S_n = \sum_{n \ge i \ge 1} a_i$$ and $$P_n= \prod_{n \ge i \ge 1} (1+a_i)$$are both of the same nature (they either both converge or both diverge).
Lemma 2.
For $0<x_i<1$ $$A_n= \sum_{n \ge i \ge 1} x_i $$ and $$B_n=\sum_{n \ge i \ge 1} \frac{x_i}{1-x_i}$$ are of the same nature.

Now just set $a_n = x_n$ and note that $B_n \sim A_n=S_n \sim P_n$ where $ \sim$ means "is of the same nature as" which is an equivalence relation.

Proof of Lemma 1. By the inequality $\exp(x) \ge x+1$ and induction (or alternatively by Newton's inequalities) we can see that $$\exp(\sum_{n \ge i \ge 1} a_i) \ge \prod_{n \ge i \ge 1} (1+a_i) $$
On the other hand, by Bernoulli's inequality we have $$\prod_{n \ge i \ge 1} (1+a_i) \ge 1+ \sum_{n \ge i \ge 1}a_i$$ hence $S_n$ and $P_n$ are really of the same nature. 
Proof of Lemma 2.
Note that we are dealing with positive numbers here and $0<1-x_i \to 1$. Therefore there exists an integer $n_0$ such that for all $i>n_0$ we have $1>(1-x_i)> 0,5$  and therefore $x_i<\frac{x_i}{1-x_i}< 2 x_i$. From here we get
$$\sum_{n \ge i >n_0} x_i \le \sum_{n \ge i >n_0}\frac{x_i}{1-x_i} \le 2\sum_{n \ge i >n_0} x_i$$
which implies the result.
