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I know from my intuition that the sequence

$$x_n=\left(1-\cfrac{1}{3}\right)^2 \left(1-\cfrac{1}{6}\right)^2 \left(1-\cfrac{1}{10}\right)^2\cdots \cdots\left(1-\cfrac{1}{\cfrac{n\left(n+1\right)}{2}}\right)^2,\quad n\geq2$$

is convergent. But i don't know how to prove it.I almost try to apply every theorem I know (for eg ratio test ,monotone convergence theorem,...). Help me to prove this.

Proof or idea is needed.Where does the sequence converge to?

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  • $\begingroup$ are you sure sir $\endgroup$ – Sathasivam K Jul 10 '17 at 13:53
  • $\begingroup$ hm let me Control this $\endgroup$ – Dr. Sonnhard Graubner Jul 10 '17 at 13:54
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    $\begingroup$ no i was wrong sorry $\endgroup$ – Dr. Sonnhard Graubner Jul 10 '17 at 13:59
  • $\begingroup$ i think it would be zero or some simple fraction like 1/3 $\endgroup$ – Sathasivam K Jul 10 '17 at 13:59
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$$\prod_{k=2}^n\left(1-\frac{2}{k(k+1)}\right)=\prod_{k=2}^n\frac{(k+2)(k-1)}{k(k+1)}=$$ $$\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot...\cdot\frac{n(n-3)}{(n-2)(n-1)}\cdot\frac{(n+1)(n-2)}{(n-1)n}\cdot\frac{(n+2)(n-1)}{n(n+1)}=\frac{n+2}{3n}$$

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    $\begingroup$ i think a square is missing $\endgroup$ – Dr. Sonnhard Graubner Jul 10 '17 at 14:10
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    $\begingroup$ Squaring it's for you, dear doctor. $\endgroup$ – Michael Rozenberg Jul 10 '17 at 14:11
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    $\begingroup$ i have already squared $\endgroup$ – Dr. Sonnhard Graubner Jul 10 '17 at 14:11
  • $\begingroup$ @Michael Rozenberg very thanks.....now i am clear with telescoping series $\endgroup$ – Sathasivam K Jul 10 '17 at 14:23
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$$ \begin{align} \prod_{k=2}^\infty\left(1-\frac2{k^2+k}\right) &=\lim_{n\to\infty}\prod_{k=2}^n\frac{\color{#C00}{(k-1)}\color{#090}{(k+2)}}{\color{#00F}{k}\color{#C90}{(k+1)}}\\ &=\lim_{n\to\infty}\frac{\color{#C00}{(n-1)!}\,\color{#090}{(n+2)!/3!}}{\color{#00F}{n!/1!}\,\color{#C90}{(n+1)!/2!}}\\[3pt] &=\frac13\lim_{n\to\infty}\frac{n+2}{n}\\[3pt] &=\frac13 \end{align} $$ Your product is just the square of this one.

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HINT:

$$1-\dfrac2{n(n+1)}=\dfrac{n^2+n-2}{n(n+1)}=\dfrac{(n+2)(n-1)}{n(n+1)}=\dfrac{\dfrac{n-1}n}{\dfrac{n+1}{n+2}}=\dfrac{f(n-1)}{f(n+1)}$$

where $f(m)=\dfrac m{m+1}$

See Telescoping series

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1
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prove by induction that for your sum is hold $$\frac{(n+2)^2}{9n^2}$$

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  • $\begingroup$ sir please give some more hint sir....i'm approaching $\endgroup$ – Sathasivam K Jul 10 '17 at 14:05
  • $\begingroup$ yes sir the limit should be 1/9...but please give a one more small hint...I will finish it sir $\endgroup$ – Sathasivam K Jul 10 '17 at 14:08
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    $\begingroup$ i think you can find enough hints $\endgroup$ – Dr. Sonnhard Graubner Jul 10 '17 at 14:13

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