| lim[n→∞] 1/n ({2n}P{n})^(1/n)
区分求積法を用います。 1/n ({2n}P{n})^(1/n) に log をとると
log( 1/n (2n(2n-1)...(n+2)(n+1))^(1/n) ) = log( ((1+n/n)(1+(n-1)/n)...(1+2/n)(1+1/n))^(1/n) ) = 1/n (log(1+n/n) + log(1+(n-1)/n) + ... + log(1+2/n) + log(1+1/n) ) → ∫[0,1] log(1 + x) dx, n → ∞ = [1/(1 + x)]_[0,1] = 1/2 - 1 = -1/2
したがって lim[n→∞] 1/n ({2n}P{n})^(1/n) = exp(-1/2).
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