Finding the limit of $\prod_{i=1}^\infty p_i$ with $p_i<1$ If we are considering $$\prod_{i=0}^\infty p_i$$ for $0<p_i<1$ is it always the case that $\prod_{i=1}^\infty p_i=0$?.
I mean this is clear to me if the $p_i$ are bounded above by some $c<1$ but what if $\lim_{i\rightarrow \infty} p_i=1$ so $\sup\{p_i\}=1$ which means there is no such $c$?
Cheers Guys
 A: The easy case to consider is
$$
\prod_{n=2}^{\infty} \left(1-\frac{1}{n^2}\right)
$$
In this case, the partial product$$
\prod_{n=2}^{K} \left(1-\frac{1}{n^2}\right)
$$
can be computed in closed form and the limit computed.  The limit is positive.  Try it!
A: Of course not. Think of $p_i = \mathbb{e}^{-\frac 1 {n^2}}$. If the product were $0$, then its logarithm would be $\log 0_+ = -\infty$. But in this example its logarithm is $\sum - \frac 1 {n^2}$ wich is known to be convergent.
A: In fact $\prod_{n=1}^\infty p_n > 0 \iff \sum_{n=1}^{\infty}(1-p_n) < \infty.$
A: You are asking whether $\lim_{n \to \infty} \prod_{i=1}^n p_i = 0$, which is equivalent to asking whether $-\lim_{n \to \infty} \sum_{i=1}^n \log p_i =\infty$.
To see this, suppose $\lim_{n \to \infty} \prod_{i=1}^n p_i = 0$. Fix $M > 0$. There exists some $N$ such that for all $n \geq N$, $\prod_{i=1}^n p_i < e^{-M}$. This implies $\sum_{i=1}^n \log p_i < -M$ for all $n \geq N$. Thus $-\lim_{n \to \infty} \sum_{i=1}^n \log p_i =\infty$.
Now suppose $-\lim_{n \to \infty} \sum_{i=1}^n \log p_i =\infty$. Fix $\epsilon > 0$. There exists some $N$ such that for all $n \geq N$ we have $ -\sum_{i=1}^n \log p_i > -\log \epsilon$. But this implies $\prod_{i=1}^n p_i < \epsilon$ for all $n \geq N$. 
Now $-\sum_{i=1}^n \log p_i$ is a monotonically increasing series, and therefore attains a limit. There are many tests to determine whether or not such a series converges to a finite limit, which in turn will imply that $\prod_{i=1}^n p_i \neq 0$. For example, if $-\log p_i = \frac{1}{i}$ then the series will diverge. This implies $\prod_{i=1}^\infty e^{-\frac{1}{i}} = 0$. On the other hand, setting $-\log p_i = \frac{1}{i^2}$ will yield a convergent series. This implies $\prod_{i=1}^\infty e^{-\frac{1}{i^2}} \neq 0$.
