<s>[INST] Evaluate this infinite product: $\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)$ 
$$\prod_{n=3}^{\infty} \left(\;1-\frac{4}{n^2}\;\right)\;=\;\text{???}$$

I took the LCM and split the numerator as $(n+2)(n-2)$ and then took the product of the numerator and the denominator separately but I was not able to get the answer from that so can you please help me in what to do next.
 [/INST] $$ \prod_{n=3}^{\infty} \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{(n-2)(n+2)}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{n-2}{n} \right)\prod_{n=3}^{\infty} \left( \frac{n+2}{n} \right) = \lim_{n\rightarrow \infty} \frac{2^2}{4!} \frac{(n-2)!}{n!} \frac{(n+2)!}{n!} $$
So now we pull out Stirling's Approximation 
$$ \frac{1}{3!} \lim_{n \rightarrow \infty} \frac{2 \pi \sqrt{n^2-4}\left( \frac{n-2}{e} \right)^{n-2}\left( \frac{n+2}{e} \right)^{n+2}  }{2 \pi n \left( \frac{n}{e} \right)^{2n}} $$
Dividing out common terms we have:
$$ \frac{1}{3!} \lim_{n \rightarrow \infty} \frac{\sqrt{n^2-4}\left( \frac{n-2}{e} \right)^{n-2}\left( \frac{n+2}{e} \right)^{n+2}  }{n\left( \frac{n}{e} \right)^{2n}} $$
We can take a natural logarithm and then exponentiate our answer to recover the original. So now consider:
$$ \lim_{n \rightarrow \infty} \frac{1}{2} \ln \left(\frac{n^2 - 4}{n}  \right) + n \ln \left( \frac{(n+2)^2}{n^{n}(n-2)^2} \right)+n - \ln(3!) $$
Our problem then simplifies to:
$$ \lim_{n \rightarrow \infty}   n \ln \left( \frac{(n+2)^2}{(n-2)^2} \right)+n - \ln(3!) -  n \ln(n)$$
Which at limit yields:
$$ \lim_{n \rightarrow \infty} O(n) - O(n \ln n)$$
And so obviously tends to $-\infty$. This means that the original tends to 0. 
So we conclude that the product tends to 0 (using our approximation).

An easier method.
Note that
$$ \frac{1 \times 2 ... \times n-2}{3 \times 4 ... \times n} = \frac{2}{(n-1)n} $$
On the other hand we have:
$$ \frac{5 \times 6 ... \times n+2}{3 \times 4 ... \times n} = \frac{(n+1)(n+2)}{3 \times 4} $$
So their combined product is
$$ \frac{2(n+1)(n+2)}{12(n-1)n} = \frac{1}{6} \frac{(n+1)(n+2)}{n(n-1)}$$
Now it's easy to prove $\lim_{n\rightarrow \infty} 
 \frac{(n+1)(n+2)}{n(n-1)} = 1$ and you get the desired result.
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