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$$\lim_{n\to\infty}{\bigg(1-\cfrac{1}{2^2}\bigg)\bigg(1-\cfrac{1}{3^2}\bigg) \cdots \bigg(1-\cfrac{1}{n^2}\bigg)}$$

This simplifies to $\prod_{n=1}^{\infty}{\cfrac{n(n+2)}{(n+1)^2}}$. Besides partial fractions and telescope, how else can we solve? Thank you!

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Using the fact $x^2-y^2=(x-y)(x+y)$ –  Jlamprong Feb 3 '14 at 11:00
$\prod_{n=2}^{\infty}{\cfrac{(n-1)(n+1)}{n^2}}$. A further hint? –  Daniel C Feb 3 '14 at 11:04
I suppose there are typo's in your product formula. –  Claude Leibovici Feb 3 '14 at 11:08
Yes. Expanding the product it's the key to observe that they simplify each other. Thank you! –  Daniel C Feb 3 '14 at 11:15
See Basel problem. –  Lucian Feb 3 '14 at 13:02

1 Answer 1

up vote 4 down vote accepted

\begin{align} L&=\lim_{n\to\infty}{\bigg(1-\cfrac{1}{2^2}\bigg)\bigg(1-\cfrac{1}{3^2}\bigg) \cdots \bigg(1-\cfrac{1}{n^2}\bigg)}\\ &=\lim_{n\to\infty}{\bigg(1-\cfrac{1}{2}\bigg)\bigg(1-\cfrac{1}{3}\bigg) \cdots \bigg(1-\cfrac{1}{n}\bigg)\bigg(1+\cfrac{1}{2}\bigg)\bigg(1+\cfrac{1}{3}\bigg) \cdots \bigg(1+\cfrac{1}{n}\bigg)}\\ &=\lim_{n\to\infty}\frac1n\frac{n+1}2\\ &=\frac12 \end{align}

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