Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$
How to start the proof by induction? Is there any way to show this?

Let $$P = \prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) =\prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \left(1+\frac{1}{r}\right) $$
So $$\prod^{n}_{r=2}\left(1-\frac{1}{r^2}\right) = \prod^{n}_{r=2}\left(1-\frac{1}{r}\right)\cdot \prod^{n}_{r=2}\left(1+\frac{1}{r}\right)$$
Now Open These two products.
