Infinite Product

Let us define the general term for this product.

$T_n = \frac {(n+1)^2 - 1}{(n+1)^2} = \frac {n(n+2)}{(n+1)^2}$

Now, Product to n terms,

$P_n = \frac{( 1 \times 2 \times ...\times n ) \times ( 3 \times 4 \times ... \times (n+2))} {2^2 \times 3^2 \times ... \times (n+1)^2}$

or,

$P_n = \frac{n! (n+2)!}{2 ((n+1)!)^2 } = 2 \times \frac{n!}{(n+1)!} \times \frac{(n+2)!}{(n+1)!} = \frac{1}{2} \times \frac{1}{n+1} \times (n+2)$

Thus,

$P_n = \frac{n+2}{2n + 2}$

So, product of infinite terms, $P = \lim_{n \to \infty} P_n = \lim_{n \to \infty} \frac{n+2}{2n + 2}$

Dividing both numerator and denominator by $n \neq 0$ , we get $P = \lim_{n \to \infty} \frac{ 1 + \frac{2}{n}}{2 + \frac{2}{n}} = \frac{1}{2} = \boxed{0.5}$

We see that each term can be factored with difference of squares: (1- $\frac{1}{2}$ )(1+ $\frac{1}{2}$ )(1- $\frac{1}{3}$ )(1+ $\frac{1}{3}$ ) ...

Evaluating each term results in: ( $\frac{1}{2}$ )( $\frac{3}{2}$ )( $\frac{2}{3}$ )( $\frac{4}{3}$ )( $\frac{3}{4}$ )...

Aside from the first term, all other terms cancel out, leaving us with $\frac{1}{2}$

The expression:

$\begin{aligned} \prod_{k=2}^\infty {\frac{k^2-1}{k^2}} & = \prod_{k=2}^\infty {\frac{(k-1)(k+1)}{k^2}} \\ & = \frac{1\dot{}3}{2\dot{}2} \dot{} \frac{2\dot{}4}{3\dot{}3} \dot{} \frac{3\dot{}5}{4\dot{}4} \dot{} \frac{4\dot{}6}{5\dot{}5} \dot{} ... \\ & = \frac{1\dot{}\color{#D61F06}{\not{3}}}{2\dot{}\color{#3D99F6}{\not{2}}} \dot{} \frac{\color{#3D99F6}{\not{2}} \dot{}\color{#20A900}{\not{4}}}{\color{#D61F06}{\not{3}}\dot{}\color{#D61F06}{\not{3}}} \dot{} \frac{\color{#D61F06}{\not{3}}\dot{}\color{#3D99F6}{\not{5}}}{\color{#20A900}{\not{4}}\dot{}\color{#20A900}{\not{4}}} \dot{} \frac{\color{#20A900}{\not{4}}\dot{}\color{#D61F06}{\not{6}}}{\color{#3D99F6}{\not{5}}\dot{}\color{#3D99F6}{\not{5}}} \dot{} ... \\ & = \boxed{\frac{1}{2}} \end{aligned}$