>>275

a_n := (n!)^(1/n) / n
b_n := log(a_n)

とおく。

b_n
=
(1/n)*log(n!) - log(n) = (1/n)*(log(n!) - n*log(n)) = (1/n)*(log(1) + log(2) + … + log(n) - n*log(n))
=
(1/n)*( (log(1) - log(n)) + (log(2) - log(n)) + … + (log(n) - log(n)) )
=
(1/n)*( log(1/n) + log(2/n) + … + log(n/n) )

リーマン積分の定義より、

(1/n)*( log(1/n) + log(2/n) + … + log((n-1)/n) ) < [∫ log(x) dx from x = 1/n to x = 1] < (1/n)*( log(2/n) + log(3/n) + … + log(n/n) )が

が成り立つ。

式変形して、

[∫ log(x) dx from x = 1/n to x = 1] + (1/n)*log(1/n) < b_n < [∫ log(x) dx from x = 1/n to x = 1]

1/n - 1 < b_n < log(n)/n + 1/n - 1

1/n - 1 < b_n < log(n^(1/n)) + 1/n - 1

lim (1/n -1 ) = 0 - 1 = -1
lim (log(n^(1/n)) + 1/n - 1) = log(1) + 0 - 1 = -1
だから、はさみうちの原理により、

lim b_n = -1

lim a_n = lim exp(b_n) = exp(-1) = 1/e