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.
 A: Note that, by induction, we have $$F(K)=\prod _{n=3}^K \left(1-\frac{4}{n^2}\right) = \frac{(K+1) (K+2)}{6 K (K-1)}$$ for $K\geq3$.
It's them simple to see that $$\lim_{K\to\infty}F(K) = \frac{1}{6}.$$
A: A different solution. You can use the fact that
$$\sin(\pi x)=\pi x \prod_1^\infty \left(1-\frac{x^2}{n^2} \right).$$
So your product is e qual to
$$\lim_{x \to 2} \frac{\sin(\pi x)}{\pi x (1-x^2)(1-x^2/4)}=1/6$$
A: $$ \prod_{n=3}^{\infty} \left( 1 - \frac{4}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{(n-2)(n+2)}{n^2} \right) = \prod_{n=3}^{\infty} \left( \frac{n-2}{n} \right)\prod_{n=3}^{\infty} \left( \frac{n+2}{n} \right) = \lim_{n\rightarrow \infty} \frac{2^2}{4!} \frac{(n-2)!}{n!} \frac{(n+2)!}{n!} $$
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.
