I would like to know an idea on how to show this:
$\prod_{k=2}^{2n+1}$ $(1-{1\over k^2})$ = ${n+1\over 2n+1}$ $\forall$ $n\ge$ $2$.
I already checked for $2$ and tried it by induction but I didn't succeed.
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I would like to know an idea on how to show this: $\prod_{k=2}^{2n+1}$ $(1-{1\over k^2})$ = ${n+1\over 2n+1}$ $\forall$ $n\ge$ $2$. I already checked for $2$ and tried it by induction but I didn't succeed. |
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Two hints to start you off: 1) You do not need induction. The statement can be proved purely by algebraic manipulation of the given formula. 2) Write $$1 - \frac{1}{k^2} = \frac{k^2-1}{k^2}.$$ This can be put into the product easily. Think about factoring the terms here, and factorials in the product. Many terms will cancel. |
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