<s>[INST] 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?
 [/INST] Here is an inductive proof:


*

*For $n=2$ we get $$1-\frac{1}{2^{2}}=\frac{2+1}{2*2}=\frac{3}{4}$$

*Let $n=k>2$ and $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})=\frac{k+1}{2*k}$$
and now the induction step:

*Let $n=k+1$, then $$(1-\frac{1}{2^{2}})(1-\frac{1}{3^{2}})...(1-\frac{1}{k^{2}})(1-\frac{1}{(k+1)^{2}})=(\frac{k+1}{2*k})(1-\frac{1}{(k+1)^{2}})= \\
=(\frac{k+1}{2*k})\frac{(k+1)^{2}-1}{(k+1)^{2}}=(\frac{1}{2*k})\frac{k(k+2)}{(k+1)}=\frac{1}{2}\frac{(k+2)}{(k+1)}=\frac{(k+1)+1}{2*(k+1)}$$
thus the proof is complete.

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