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$P_n= \prod^n_k_=_2 \frac{k^2-1}{k^2}$ for $n \ge 2$ I calculated $P_1 to P_3$ . I have been trying to come up with a formula but I can't really see any pattern. $P_2 = \frac{3}{4} , P_3 = \frac{2}{3}, P_4 = \frac{5}{8}$

user60334
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    The key is to (i) Calculate somewhat longer and (ii) do not really calculate, just write down the product, making only the simplification $k^2-1=(k-1)(k+1)$, and look. – André Nicolas Feb 23 '13 at 04:54

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$$P_n = \prod_{k=2}^n \dfrac{k^2-1}{k^2} = \prod_{k=2}^n \dfrac{(k-1)(k+1)}{k^2}$$ Hence, $$P_n = \dfrac{1 \times \color{red}3}{2 \times \color{blue}2} \cdot \dfrac{\color{blue}2 \times \color{green}4}{\color{red}3 \times \color{orange}3} \cdot \dfrac{\color{orange}3 \times \color{magenta}5}{\color{green}4 \times \color{brown}4} \cdot \dfrac{\color{brown}4 \times 6}{\color{magenta}5 \times 5} \cdots \dfrac{(n-2) \times \color{darkgreen}n}{(n-1) \times \color{purple}{(n-1)}} \cdot \dfrac{\color{purple}{(n-1)} \times (n+1)}{\color{darkgreen}n \times n}$$ Rearranging we get, $$P_n = \dfrac12 \cdot \dfrac{3 \times 2}{2 \times 3} \cdot \dfrac{4 \times 3}{3 \times 4} \cdot \dfrac{5 \times 4}{4 \times 5} \cdots \dfrac{n \times (n-1)}{(n-1) \times n} \cdot \dfrac{n+1}n$$ All the terms in the middle are $1$, except the two end terms. Hence, we get that $$P_n = \dfrac12 \cdot \dfrac{n+1}{n}$$