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Find $\frac{2^2}{2^2-1} \cdot \frac{3^2}{3^2-1} \cdot \frac{4^2}{4^2-1} \cdot \dots \cdot \frac{2006^2}{2006^2-1}.$

Guest Feb 15, 2020

Guest · Answer 1 · 2020-02-15T07:14:21

∏[n^2 / (n^2 - 1), n, 2, 2006] = 1.999003488