Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$ 
Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$

I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$
 A: Your way works nice if you employ the Euler's infinite product for the sine function. Then 
$$\lim_{n \to \infty}\sum_{k=2}^{n} \ln\left(1-\frac{1}{k^2}\right)=\lim_{x\to\pi}\ln\left(\frac{\pi^2\sin x}{x(\pi^2-x^2)}\right)=\lim_{y\to0}\ln\left(\frac{\pi^2\sin y}{y(2\pi-y)(\pi-y)}\right)=\ln(1/2)$$
Thus, your product is $1/2$.
A: The identity $k^2-1=(k+1)(k-1)$ shows that
$$
\prod_{k=2}^n\left(1-\frac{1}{k^2}\right)=\prod_{k=2}^n\frac{k^2-1}{k^2}=\prod_{k=2}^n\frac{k-1}k\cdot\prod_{k=2}^n\frac{k+1}k=\frac1n\cdot\frac{n+1}2,
$$
and the value of limit should follow.
A: The $r$th term is $\frac{r^2-1}{r^2}=\frac{(r-1)}r\frac{(r+1)}r$
So, the product of $n$ terms is $$\frac{3.1}{2^2}\frac{4.2}{3^2}\frac{5.3}{4^2}\cdots \frac{(n-1)(n-3)}{(n-2)^2}\frac{(n-2)n}{(n-1)^2}\frac{(n-1)(n+1)}{n^2}$$
$$=\frac12\frac32\frac23\frac43\cdots\frac{n-2}{n-1}\frac n{n-1}\frac{n-1}n\frac{n+1}n=\frac12\frac{(n+1)}n$$
as the 1st half of any term is cancelled by the last half of the previous term except for the 1st & the last term.
