# 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 $1/2$?

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"pi" is for "product" (like "sigma" for "sum") –  yoyo Oct 17 '12 at 0:37

$$\displaystyle \prod_{k=2}^n f(k)$$ 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.

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should i prove this by induction or just by manipulation? –  Jack F Oct 17 '12 at 0:36
To prove that a statement is true for all natural numbers always need induction. –  user17762 Oct 17 '12 at 0:38
so i get to the point of ((n+1)/2n) x (1-1/n+1) –  Jack F Oct 17 '12 at 0:43
$$\dfrac{n+1}{2n} \times \left(1 - \dfrac1{(n+1)^2} \right) = \dfrac{n+1}{2n} \times \dfrac{(n+1)^2-1}{(n+1)^2} = \dfrac{n+1}{2n} \times \dfrac{n(n+2)}{(n+1)^2} = \dfrac{n+2}{2(n+1)}$$ –  user17762 Oct 17 '12 at 0:55
where did u get (n+1)^2? –  Jack F Oct 17 '12 at 0:57

$\prod$ 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.

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