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This question already has an answer here:

I am having trouble figuring out how to solve this limit.

$$\lim _{x\to \infty }\left(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{x^2}\right)\right)$$

I understand that as '$x$' increases, the overall product becomes an even smaller number between $0$ and $1$ only because I tried plugging in numbers in hope to learn about the nature of the expression. But I can't seem to come to a pattern that will allow me to efficiently simplify it and solve the limit.

Help, anyone?

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marked as duplicate by Martin Sleziak, Jonas Meyer, Community Jul 28 '16 at 20:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I think the first factor is $1-\frac1{2^2}$ not $1-\frac1{1^2}$, the latter being equal to $0$. $\endgroup$ – Olivier Oloa Jul 28 '16 at 20:22
  • $\begingroup$ Should the first term 1 - 1/1 not be there? Since the whole expression is rendered 0 because of that. $\endgroup$ – zarathustra Jul 28 '16 at 20:23
  • $\begingroup$ yes you are right, I'll fix it right away, sorry. $\endgroup$ – intersomnium Jul 28 '16 at 20:23
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Overkill: $$\frac{\sin(\pi z)}{\pi z}=\prod_{n\geq 1}\left(1-\frac{z^2}{n^2}\right) \tag{1} $$ gives: $$ \prod_{n\geq 2}\left(1-\frac{1}{n^2}\right) = \lim_{z \to 1}\frac{\sin(\pi z)}{\pi z(1-z^2)} \stackrel{DH}{=}\lim_{z\to 1}\frac{\cos(\pi z)}{1-3z^2}=\color{red}{\frac{1}{2}}.\tag{2} $$ That stops being an overkill if the infinite product you want to compute is $\prod_{n\geq 1}\left(1-\frac{1}{4n^2}\right)=\frac{2}{\pi}$ (aka Wallis' product), for instance.

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  • $\begingroup$ What does the "DH" above the equals sign stand for? $\endgroup$ – Jonas Meyer Jul 28 '16 at 20:44
  • $\begingroup$ @JonasMeyer: De l'Hopital rule. $\endgroup$ – Jack D'Aurizio Jul 28 '16 at 20:45
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Hint. One may observe that $$ \prod_{n=2}^N\left(1-\frac1{n^2}\right)=\prod_{n=2}^N\frac{n^2-1}{n^2}=\prod_{n=2}^N\frac{n+1}{n}\cdot \prod_{n=2}^N\frac{n-1}{n} $$ then factors telescope.

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  • $\begingroup$ yeah,i also suspected it $\endgroup$ – haqnatural Jul 28 '16 at 20:23
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The logarithm of the limit is the limit of the logarithm, now:

$$\lim_{n \to \infty} \sum_{k=2}^{n} \log \left( 1 - \frac1{k^2}\right) = \lim_{n \to \infty} \left[ \sum_{k=2}^{n} (\log (k+1) - \log k) - \sum_{k=2}^n (\log k - \log (k-1)) \right] = \lim_{n \to \infty} \left[ \log\left( \frac{n+1}n \right) - \log 2 \right] = - \log 2$$

So the answer is $1/2$.

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