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+1. It is not every day that I see the Cauchy root test cited to justify convergence of a geometric series. – 2011-02-12
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0@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. – 2011-02-12
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0Sivaram, 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. – 2011-02-12
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0@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. – 2011-02-12
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0@Sivaram, I'm sure I'm being dense, but where in your argument does it explicitly show that the limit is *strictly* positive? – 2011-02-13
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0@cardinal: Good point. I didn't notice it. But the convergence of the sum $|a_n|$ also implies the product is non-zero, since for large $n$, $|log(1+a_n)| < |a_n|$, and hence the sum of the logs dont blow up to $-\infty$ either. I will clear it up in the post. Thanks – 2011-02-13
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0@Sivaram, I hope you don't mind. I've provided an alternate argument and explicitly calculated a bound, similar to the method you suggest. – 2011-02-13
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0@cardinal: No problem at all. – 2011-02-13
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0@Sivaram, also note that for $-1 < x < 0$, it is never true that $|\log(1+x)| < |x|$. – 2011-02-13
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0@cardinal: Yup. I missed it in the previous comment. Your argument carries through. – 2011-02-13
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0@Sivaram, sorry to be so pedantic. But I don't believe $\sum |a_n| < \infty$ is necessary for convergence here. Indeed, let $a_n = 1/(n+1)$, or more extreme yet, let it be $a_n = 1/n$. – 2011-02-13
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0@cardinal: If I am right, if none of the terms in the product are zero and if $\sum a_n$ converges, then the product also converges to a strictly positive number. – 2011-02-13
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0@Sivaram, *that* does sound likely to me, provided, of course that $a_n < 1$. The version that I have normally seen of the "theorem" given in your answer takes $a_n \leq 0$. – 2011-02-13
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0@cardinal: I don't completely understand what you are saying. Isn't that what I have proved in the above argument? $\sum{\log{(1+a_n)}}$ converges also implies the product doesn't go to zero? – 2011-02-13
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0@Sivaram, I think part of the confusion we may be having in communicating is that in one case we consider $\prod (1-a_n)$ and in another $\prod (1+a_n)$. I've been thinking in terms of the former of the two. :) – 2011-02-13
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0@cardinal: Oh Ok. But in that case still, replace $a_n$ by $-a_n$ and the argument will stil carry through right? since $\sum a_n$ converges, then so does $\sum (-a_n)$ – 2011-02-13
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0@Sivaram, my point, stated rather ineloquently, was that it is not true in general that $\prod (1+a_n)$ converges if and only if $\sum a_n$ converges. A previous comment provides a counterexample. However, if $-1 < a_n \leq 0$, then $\prod (1+a_n)$ has a nonzero (i.e., strictly positive) limit if and only if $\sum a_n$ has a finite limit. The latter *does* follow essentially from your argument. – 2011-02-13
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0@cardinal: If I am right, your argument is if $a_n = \frac{1}{n+1}$, then the product $\prod(1-a_n)$ is $\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} \times \ldots = 0$, whereas $\sum \frac{1}{n}$ diverges. (Note $\prod(1+a_n)$ diverges.) So If I were to state it as follows: If none of the terms in the product are zero, then $\sum a_n$ converges iff $\prod (1+a_n)$ converges to a $\textbf{strictly positive number}$. Sorry if I sound confusing. – 2011-02-13
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0@Sivaram, no. I believe the signs of the $a_n$ are (also) important. For example, let $a_{2n-1} = e^{1/\log (n+1)} - 1$ and $a_{2n} = e^{-1/\log (n+1)} - 1$. Then $\prod (1+a_n) = 1$ and $\sum a_n$ does not converge since $\sum_{k=1}^{2n} a_k = \sum_{k=2}^{n+1} (e^{1/\log k} + e^{-1 / \log k} - 2) \geq n (e^{1/\log(n+1)} + e^{-1/\log(n+1)} - 2) \to \infty$. Correct? – 2011-02-13
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0@cardinal: Ok. I see. But where is the flaw in my proof and how do I correct it. Feel free to edit my post. – 2011-02-14
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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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0You mean "bounded above" in the your answer, right? – 2011-02-13
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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