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I am stuck on the following problem:

Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$

I see that $x_n^{\frac{1}{2}}=\frac{2}{3}\frac{5}{6}\frac{9}{10} \ldots..\frac{(n-1)(n+2)}{n(n+1)}$. now not sure what to do next ?Any idea?

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It converges, as $\sum_{n\geq 1}\frac{1}{n(n+1)/2}$ converges. So you want the value of the limit, I assume. –  1015 May 2 '13 at 18:27
1  
If you consider $y_n = \log x_n$ and use $\log(ab) = \log a + \log b$, I think you will get a telescoping series. –  TMM May 2 '13 at 18:27
    
Converges to 1/3. –  gt6989b May 2 '13 at 18:32
    
@julien: You will get $y_n = 2 \sum \ln(n-1) + 2 \sum \ln(n+2) - 2 \sum \ln n - 2 \sum \ln(n+1)$, so except for a few initial terms and last terms, each number appears twice with a $+2$ and twice with a $-2$. –  TMM May 2 '13 at 18:32
    
@TMM Right, thanks. –  1015 May 2 '13 at 18:34

3 Answers 3

up vote 4 down vote accepted

Note you really have a telescoping product. To see this intuitively, write out a few terms. You start with $\frac{1 \cdot 4}{2 \cdot 3}$

$$\frac{1 \cdot 4}{2 \cdot 3} \times \frac{2 \cdot 5}{3 \cdot 4} = \frac{1 \cdot 5}{3 \cdot 3}$$

and adding another term

$$\frac{1 \cdot 5}{3 \cdot 3} \times \frac{3 \cdot 6}{4 \cdot 5} = \frac{1 \cdot 6}{3 \cdot 4}$$

and proceed by induction to prove that $x_n = \frac{1 \cdot (n+2)}{3 \cdot n}$ which in the limit should converge to 1/3.

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Typo: the first $4$ should be a $3$ on line $2$. +1. –  1015 May 2 '13 at 18:40
    
@julien Fixed thank you –  gt6989b May 2 '13 at 18:41
    
Thanks for the detailed clarification.But I have one question. It is a multiple choice question where both options $1/3$ and $1/9$ are included.So what should be the right choice? As $x_n^{\frac{1}{2}} \to 1/3$ ,should $x_n$ tends to $1/9$ or $1/3?$ –  learner May 2 '13 at 18:54
    
@learner: $\sqrt{x_n} \to 1/3$ so yes, $x_n \to (1/3)^2 = 1/9$. –  TMM May 2 '13 at 20:11
    
@TMM thank you very much –  gt6989b May 2 '13 at 20:42

Let $f(n)=\frac{n-1}{n+1}$. Then show $$1-\frac{1}{n(n+1)/2} = \frac{f(n)}{f(n+1)}$$

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$$ \begin{align} \prod_{j=2}^n\left(1-\frac2{j(j+1)}\right) &=\prod_{j=2}^n\left(\frac{(j-1)(j+2)}{j(j+1)}\right)\\ &=\color{#00A000}{\prod_{j=2}^n\frac{j-1}{j}} \color{#C00000}{\prod_{j=2}^n\frac{j+2}{j+1}}\\ &=\color{#00A000}{\frac1n}\color{#C00000}{\frac{n+2}{3}}\\ &=\frac13\frac{n+2}{n} \end{align} $$ Taking it to the limit, we get $$ \prod_{j=2}^\infty\left(1-\frac2{j(j+1)}\right)=\frac13 $$ The product of the squares is the square of the product (to get the answer to the modified question).

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nice approach.Thanks a lot sir. –  learner May 3 '13 at 11:21

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