Show that $\lim_{n \rightarrow \infty} \prod_{k=1}^{n} (1-e^{-ka})$ exists an is positive If $a$ is a positive constant,then show that $\displaystyle \lim_{n \rightarrow \infty} \prod_{k=1}^{n} (1-e^{-ka})$ exists and is strictly positive.
 A: I assume your question is a product of the form $\displaystyle \lim_{n \rightarrow \infty}\prod_{k=1}^n (1 - e^{-ka})$.
Any product of the form $\displaystyle \lim_{n \rightarrow \infty}\prod_{k=1}^n (1 - a_k)$ converges iff $\displaystyle \lim_{n \rightarrow \infty}\sum_{k=1}^n a_k$ converges.
In your cases, $a_k = e^{-ka}$.
$\displaystyle \lim_{n \rightarrow \infty}\sum_{k=1}^n  a_k = \lim_{n \rightarrow \infty}\sum_{k=1}^n e^{-ka}$ converges to $\frac{1}{e^a-1}$ $(\text{geometric series with }e^{-a}<1\text{ as }a>0)$.
Hence, the infinite product $\displaystyle \lim_{n \rightarrow \infty}\prod_{k=1}^n (1 - e^{-ka})$ converges.
EDIT
The equivalence of the convergence of $\displaystyle \prod_{n=1}^{\infty} (1 + a_n)$ and $\displaystyle \sum_{n=1}^{\infty} a_n$.
Assume that $\displaystyle \sum_{n=1}^{\infty} a_n$ converges. This implies $\displaystyle \lim_{n \rightarrow \infty} a_n = 0$. Hence, $\displaystyle \lim_{n \rightarrow \infty} \frac{\log(1+a_n)}{a_n} = 1$.
Now consider $b_n = \log(1+a_n)$. By limit comparison test, since $0<\frac{b_n}{a_n}<\infty$, we have that $\displaystyle \sum_{n=1}^{\infty} b_n$ converges. (Intuitively, what the limit comparison test means is that the tail sums differ only by a factor and hence the convergence of one implies the other.)
Hence, $\displaystyle \sum_{n=1}^{\infty} \log(1+a_n)$ converges which essentially means $\displaystyle \prod_{n=1}^{\infty} (1+a_n)$ converges.
Similarly, you can argue out that if $\displaystyle \prod_{n=1}^{\infty} (1+a_n)$ converges, then $\displaystyle \sum_{n=1}^{\infty} a_n$ converges.
A: Here's a slightly different argument than @Sivaram's answer. I'm not sure his makes explicit (or even proves) that the limit is strictly positive, as requested.
First note that if $a > 0$, then $0 < 1 - e^{-an} < 1$ for all $n$ and so the "partial product"
$$
P_n = \prod_{k=1}^n (1 - e^{-a k})
$$
is decreasing and bounded below. The fact that $\lim_n\, P_n$ exists follows immediately.
To show that the limit is strictly positive, note that for all $n \geq \frac{1}{a} \log 2$, we have that $e^{-a n} \leq \frac{1}{2}$. Let $N(a) \equiv N = \lceil \frac{1}{a} \log 2\rceil$. 
Since for $0 \leq x \leq 1/2$, $\log(1-x) \geq - x - x^2$, we get that
$$
\prod_{n=N}^\infty (1 - e^{-a n}) = \exp\big( \sum_{n=N}^\infty \log(1-e^{-an}) \big) \geq \exp( -\sum_{n=N}^\infty (e^{-an}+e^{-2an}) ) \geq \exp( - 2\sum_{n=N}^\infty e^{-an} ) .
$$
Now, $\sum_{n=N}^\infty \,e^{-a n} \leq \frac{1}{1-e^{-a}}$ and so
$$
\lim_n \, P_n = P_{N-1} \prod_{m=N}^\infty (1 - e^{-a m}) \geq e^{-2/(1-e^{-a})} P_{N-1} > 0 .
$$
