For $n\in\Bbb N$ and $n\ge 2$, find and prove a formula for $\prod_{i=2}^n\left(1-\frac1{i^2}\right)$.
I can easily prove the formula using induction once I have the equivalent result but I'm having a bit of trouble actually finding said formula. I know I can separate the $1 - 1/i^2$ into two different large product notations but after that I've hit a wall. How do I proceed?
HINT: Note that
$$1-\frac1{i^2}=\frac{i^2-1}{i^2}=\frac{(i-1)(i+1)}{i^2}=\frac{i-1}i\cdot\frac{i+1}i\;.$$
Now write out the products for a few small values of $n$. For instance, for $n=4$ we get
$$\left(\frac12\cdot\frac32\right)\cdot\left(\frac23\cdot\frac43\right)\cdot\left(\frac34\cdot\frac54\right)=\frac12\cdot\left(\frac32\cdot\frac23\right)\cdot\left(\frac43\cdot\frac34\right)\cdot\frac54\;.$$
If you do this for $n=2,3,4,5$, and maybe $6$, I think that you’ll be able to spot the result.
We can write the product as
$$\begin{align} \prod_{i=2}^n\left(1-\frac1{i^2}\right)&=\frac{\prod_{i=2}^n(i^2-1)}{\prod_{i=2}^ni^2}\\\\ &=\frac{\prod_{i=2}^n(i+1)\prod_{i=2}^n(i-1)}{\prod_{i=2}^ni^2}\\\\ &=\frac{\frac12(n+1)!\,(n-1)!}{(n!)^2}\\\\ &=\frac{n+1}{2n} \end{align}$$
Yes, factorising is the way to go, but just two factors is not sufficient. $$\prod_{i=2}^n\Big(1-\frac 1{i^2}\Big) = \prod_{i=2}^n\dfrac{(i-1)\cdot(i+1)}{i\cdot i}$$
Can you finish from here?
