 | 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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