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Ok so the question is in the title, and I'm just trying to do some practice problems and this question has me stumped. I do realize that as n goes to infinity, the brackets at the end will always be 1, but I have a hunch that the limit is 0, I just don't know how to prove it. Any help appreciated, thanks.

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    $\begingroup$ $$1-\frac{1}{k^2} = \frac{k^2-1}{k^2} = \frac{(k-1)(k+1)}{k^2}$$ $\endgroup$ Nov 2, 2014 at 19:10

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Rewrite each $1-\frac{1}{k^2}$ as $\frac{(k-1)(k+1)}{k^2}$ and observe the mass cancellations. It will be useful to do this explicitly for say the product of the first $5$ terms.

$$\left(\frac{3}{2}\frac{4}{3}\frac{5}{4}\cdots\frac{n}{n-1}\frac{n+1}{n}\right)\cdot\left(\frac{1}{2}\frac{2}{3}\frac{3}{4}\cdots\frac{n-2}{n-1}\frac{n-1}{n}\right)$$

After mass cancellations, pull the $$\frac{n+1}{2}\text{ and }\frac{1}{n}$$

$$\frac{n+1}{2}\cdot\frac{1}{n}$$

Limit of this function tending to infinity $= 1/2$.

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