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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.

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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.


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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+1. It is not every day that I see the Cauchy root test cited to justify convergence of a geometric series. – Jonas Meyer Feb 12 '11 at 18:11
@Jonas: oops... I was stuck with Cauchy root test since I used it to prove couple of convergences yesterday. Interesting how mind sticks on to a particular mode of thinking. I have now deleted it since Cauchy root test is derived from geometric series and hence I think using the geometric series would be a cleaner argument. – user17762 Feb 12 '11 at 18:22
Sivaram, could you go a little bit further into depth with line #2 please? I understand your reasoning, intuitively, that if those sums converge, then summing up the permuted products of things going rapidly to zero will doubtless converge, but I don't think I understand why the reverse implication holds and it would help me to see some manipulation. – Uticensis Feb 12 '11 at 18:24
@Billare: The reason why the convergence of the product is equivalent to the convergence of the sum is due to the limit comparison test. I shall update my post to explain this in detail. – user17762 Feb 12 '11 at 18:27
@Sivaram, I'm sure I'm being dense, but where in your argument does it explicitly show that the limit is strictly positive? – cardinal Feb 13 '11 at 1:57

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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You mean "bounded above" in the your answer, right? – Uticensis Feb 13 '11 at 12:34
@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. – cardinal Feb 13 '11 at 15:20

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