Proving $\prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right) = \frac{n+1}{2n}$ by induction So I have to prove the following using induction.
${\displaystyle \prod_{i=2}^{i=n} \left(1-\frac{1}{i^2}\right)} = \frac{n+1}{2n}$
I showed the basis step that if $n=i=2$, then the two functions are equal $\frac{3}{4}$, and I know that the induction step involves simplifying the function where $n=n+1$. But I'm not sure how to do it with the following algorithm. I am quite confused. Can someone show me how I can prove this in the induction step?
 A: Induction Hypothesis:
${\displaystyle \prod_{i=2}^{i=n} (1-\frac{1}{i^2})} = \frac{n+1}{2n}$
Now, we need to prove it for $n+1$ case.
${\displaystyle \prod_{i=2}^{i=n+1} (1-\frac{1}{i^2})} = (1-\frac{1}{(n+1)^2}) *{\displaystyle \prod_{i=2}^{i=n} (1-\frac{1}{i^2})} = (1-\frac{1}{(n+1)^2})*\frac{n+1}{2n} = \frac{n^2 + 2n}{(n+1)(2n)} = \frac{n+2}{2(n+1)}$
Hence Proved
A: Without really using induction, you can use telescoping:
$$\prod_{i=2}^n \left(1-\frac{1}{i^2}\right)=\prod_{i=2}^n\frac{i^2-1}{i^2}=\prod_{i=2}^n\frac{(i-1)(i+1)}{i^2}$$
$$=\frac{1\cdot 3}{2\cdot 2}\cdot \frac{2\cdot 4}{3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\frac{1\cdot \not3}{2\cdot \not 2}\cdot \frac{\not 2\cdot 4}{\not 3\cdot 3}\cdot \frac{3\cdot 5}{4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\frac{1\cdot \Box}{2\cdot \Box}\cdot \frac{\Box\cdot \not 4}{\Box\cdot \not 3}\cdot \frac{\not 3\cdot 5}{\not 4\cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\frac{1\cdot \Box}{2\cdot \Box}\cdot \frac{\Box\cdot \Box }{\Box\cdot \Box }\cdot \frac{\Box \cdot 5}{\Box \cdot 4}\cdots \frac{(n-1)(n+1)}{n\cdot n}$$
$$=\cdots =\frac{1}{2}\cdot \frac{n+1}{n}$$
