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 $1/2$?

$$\displaystyle \prod_{k=2}^n f(k)$$ is a short hand notation for $$f(2) \times f(3) \times f(4) \times \cdots \times f(n-1) \times f(n)$$
For instance, $$\displaystyle \prod_{k=2}^{4} \left(1 - \dfrac1{k^2} \right) = \left(1 - \dfrac1{2^2} \right) \times \left(1 - \dfrac1{3^2} \right) \times \left(1 - \dfrac1{4^2} \right)$$
HINT
To solve the problem, note that $$\left(1 - \dfrac1{k^2} \right) = \dfrac{k-1}{k} \times \dfrac{k+1}k$$ Now write out the first few terms and see the cancellation.
