>>797
蛇足だけど
マクローリン展開
log(1+x) = ∫[0,x] 1/(1+y) dy = x - (1/2)x^2 + (1/3)x^3 - ････,
で　x = e^{t'/2n} -1　とおいて
t'/2n = [e^{t'/2n} -1] - (1/2)[e^{t'/2n} -1]^2 + ････,
2n+t' = 2n (e^{t'/2n} - (1/2)[e^{t'/2n} -1]^2 + ････),
(2n+t')^n
　≒ (2n)^n (e^{t'/2} - (n/2)[e^{(n+1)t'/2n} - 2e^{t'/2} + e^{(n-1)t'/2n}] ),
(2n+t')^n e^{-t'}
　≒ (2n)^n (e^{-t'/2} - (n/2)[e^{-(n-1)t'/2n} - 2e^{-t'/2} + e^{-(n+1)t'/2n}] ),
∫[2n,∞] (t^n) e^{-t} dt
　= e^{-2n}∫[0,∞] (2n+t')^n e^{-t'} dt'
　≒ e^{-2n} (2n)^n {2 - n[n/(n-1) -2 +n/(n+1)]}
　= (2/e)^n (n/e)^n {2 - 2n/(nn-1)}
　≒ (2/e)^n n! √(2/πn) {1 - n/(nn-1)},