TITLE: Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$
QUESTION [18 upvotes]: 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 $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$

REPLY [5 votes]: 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 $1/2$.