question (starting from Deduce that...) is trickier than it looks ! In order toprovide the correct answer, it might be useful, at some point, to use Stirling's formula, i.e. n ! \sim \sqrt{2 n \pi}\left(\frac{n}{e}\right)^{n} to find an equivalent of I(n, n), as n → +o. I(p, q)=\frac{p}{q+1} I(p-1, q+1) I(p, q)=\frac{p ! q !}{(p+q+1) !}

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