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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")2012-10-17

2 Answers 2

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$\displaystyle \prod_{k=2}^n f(k)$ 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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    where did u get (n+1)^2?2012-10-17
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$\prod$ 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.