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 November 2nd, 2011, 10:33 AM #1 Newbie   Joined: Oct 2011 Posts: 8 Thanks: 0 What is the limit of The sequence $r_n= \prod_{k=1}^{n} (1-\frac{1}{2^n})$ is bounded and monotone and therefore it converges. What is the limit: $lim_{n\to\infty}r_n$
 November 2nd, 2011, 02:06 PM #2 Senior Member   Joined: May 2011 Posts: 501 Thanks: 6 Re: What is the limit of This infinite product does not have a closed form value. But, you can try to get closer and closer to what it converges to. By changing a product to a sum often helps. This can be done by using the log. $ln(L)=ln\left(\Pi_{k=1}^{\infty}\left(1-\frac{1}{2^{k}}\right)\right)$ $ln(L)=\sum_{k=1}^{\infty}ln(1-\frac{1}{2^{k}})$ $=\sum_{k=1}^{\infty}\left[ln(2^{k}-1)-kln(2)\right]$ $\approx -1.24206$ $e^{-1.24206}=.288788...$ Just playin' around. Perhaps someone has a more elegant method.
 November 2nd, 2011, 02:13 PM #3 Newbie   Joined: Oct 2011 Posts: 8 Thanks: 0 Re: What is the limit of Thanks!! So to find approximately the limit I need to take the logarithm of the product. Is there any simpler way to show that the limit is not 0?
 November 2nd, 2011, 02:55 PM #4 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: What is the limit of [color=#000000]How is it possible for this product to be equal to 0? Let $a_{n}=1-\frac{1}{2^n}$, $\prod_{k=1}^{n}\left(1-\frac{1}{2^n}\right)=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{2^3}\right)\ldots\left(1-\frac{1}{2^n}\right)$ in order for this product to be equal to 0, at least one term $a_{k}$ has to be zero for $1\leq k\leq n$ , but the sequence $a_n$ is strictly increasing $a_1=\frac{1}{2}$ and $\lim_{n\to+\infty}a_{n}=1$, so $\frac{1}{2}\leq a_{n}< 1$ and there exists no term of the product equal to zero, hence $\prod_{k=1}^{\infty}a_k\neq 0$.[/color]
November 3rd, 2011, 06:56 AM   #5
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Re: What is the limit of

Quote:
 How is it possible for this product to be equal to 0?
Dear ZardoZ,
What you say is true for finite products, but when the number of multipliers $n\to \infty$ it is not true.

 November 3rd, 2011, 09:10 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: What is the limit of [color=#000000]Ok then we will have to prove two theorems, which have to be answered in the complex analysis section! My proof is obvious for finite products though! I'll post them later or tomorrow.[/color]
 November 4th, 2011, 07:09 AM #7 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: What is the limit of [color=#000000]Using the second theorem which can be found here, since $\sum_{n=1}^{\infty}\frac{1}{2^n}=\sum_{n=0}^{\inft y}\left(\frac{1}{2}\right)^n-1=\frac{1}{1-\frac{1}{2}}-1=1<+\infty=$ we conclude that $\prod_{n=1}^{\infty}\left(1-\frac{1}{2^n}\right)>0=$.[/color]

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