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?

Base case:
For $n=2$,
$$LHS=(1-\frac{1}{4})=\frac{3}{4}$$
$$RHS=\frac{(2+1)}{2\cdot 2}=\frac{3}{4}$$
Hence for $n=2$, the equality holds.
Induction hypothesis:
Let it hold for some $n=k$.
Then,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right) = \frac{k + 1}{2k}$$
For $n=k+1$,
$$LHS=\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right)\left(1 - \frac{1}{(k+1)^2}\right)$$
Inductive step:
Using the induction hypothesis, for $n=k+1$,
$$LHS=\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right)\left(1 - \frac{1}{(k+1)^2}\right)=\frac{k + 1}{2k}\left(1 - \frac{1}{(k+1)^2}\right)$$
$$=\frac{(k+1)^2-1}{2k(k+1)}=\frac{(k^2+2k)}{2k(k+1)}=\frac{k+2}{2(k+1)}=\frac{(k+1)+1}{2(k+1)}=RHS$$
Thus, whenever the equality holds for $n=k\ge2$, it also hold for $n=k+1$.
Using the principle of mathematical induction, it holds for all $n\ge2$.
