Prove that $\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$ prove the $$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$ for all n greater or equal to 2.
$\pi$ should be a big pi from $i=2$ to $n$ for $(1-1/i^2)$. I'm really confused about the $\prod$ function.
UPDATE:
Suppose$$\prod_{i=2}^n (1-1/i^2) = {n+1\over 2n}$$
then $$\prod_{i=2}^{n+1} (1-1/i^2) = {n+2\over 2(n+1)}$$
so $$\prod_{i=2}^{n+1} (1-1/i^2)  \times  \left(1-{1\over (n+1)} \right)$$
then $${n+1\over 2n} \left(1-{1\over (n+1)} \right)$$
but this equals $1/2$?

$\prod$ means multiply the things together, starting with $i=2$, then $i=3$, $i=4$, and so on until you reach $i=n$. Try it for some small values of $n$, and see if you don't see a pattern in the answers. Then use induction to prove the pattern continues. 
