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.
Show that $\lim_{n \rightarrow \infty} \prod_{k=1}^{n} (1-e^{-ka})$ exists an is positive
2 Answers
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.
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0@Sivaram, hmm, not sure I feel too comfortable about that. I'll think about it and if I can come up with a set of very minimal edits, then I'll do that. After "**EDIT**", mostly, I think you need to restrict the $a_n$ to fall in $(-1,0]$ and also add in the qualifier that the limit is strictly positive if and only if $\sum a_n$ is finite by your argument. (But, the fact that the product converges does not imply the sum does, since the product can converge to zero while the sum diverges.) – 2011-02-14
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 . $
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0@Billare, well $P_n$ is both bounded from above and bounded from below. **but** the important one in terms of quickly establishing that the limit exists is that it is bounded from below, as originally stated. Hope that helps. – 2011-02-13