I need to show via direct proof that:
$$ \prod_{i=2}^{n} \left (1- \frac{1}{i^2}\right) =\frac {n+1}{2n} $$
We first note that $$1-\frac{1}{i^2} = \frac{(i-1)(i+1)}{i\cdot i}.$$ Then \begin{align} \prod_{i=2}^{n} \left(1- \frac{1}{i^2}\right)&=\prod_{i=2}^{n} \frac{(i-1)(i+1)}{i\cdot i}\\ &=\frac {(n-1)!\cdot \frac{(n+1)!}{2} } {n!\cdot n!}\\ &=\frac {(n-1)!\cdot(n+1)!}{2\cdot n!\cdot n!}\\ &=\frac {(n+1)}{2n}. \end{align} Direct proof end.
Is that sufficient enough for a direct proof? Any critique would be helpful! Thanks!
That’s one way to do it, another way is to note that:
$$\begin{align}1-\frac1{i^2}&=\frac{(i-1)(i+1)}{i^2}\\ &=\dfrac{\frac{i-1}{i}}{\frac{i}{i+1}}\\&=\frac{f(i)}{f(i+1)} \end{align}$$ where $f(k)=\frac{k-1}k.$
So $$\begin{align}\prod_2^n \left(1-\frac1{i^2}\right)&=\frac{f(2)}{f(3)}\frac{f(3)}{f(4)}\cdots \frac{f(n)}{f(n+1)}\\&=\frac{f(2)}{f(n+1)}\\ &=\dfrac{\frac12}{\frac{n}{n+1}}\\&=\frac{n+1}{2n}\end{align}$$
(The simplest way to prove this equality is just by induction on $n,$ but your way and mine both show how to find the value, not just prove it is correct.)