Evaluating the infinite product $\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)$

Note that $$1-\frac1{k^2}=\left(1-\frac1k\right)\left(1+\frac1k\right)=\frac{k-1}{k}\frac{k+1}{k}=\frac{a_k}{a_{k-1}}$$ with $a_k= \frac{k+1}k$, hence this is a telescoping product, i.e. $$ \prod_{k=2}^n\left(1-\frac1{k^2}\right)=\frac{a_2}{a_1}\frac{a_3}{a_2}\cdots\frac{a_n}{a_{n-1}}=\frac{a_n}{a_1}=\frac{n+1}{2n}.$$

Let $g(k)=\frac{k-1}{k}$. Then this product is:

$$\prod_{k=2}^n \frac{g(k)}{g(k+1)}$$

which is a telescoping product, and thus equal to $\frac{g(2)}{g(n+1)}$

$$ 1-\frac{1}{k^2} = \frac{(k-1)(k+1)}{k^2}$$ $$ \prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ) = \underbrace{\frac{2-1}{2}}_{\text{from first}} \times \underbrace{\frac{(n+1)}{n}}_{\text{from last}}$$