lim n→∞ [(n!)^1/n]/n = 1/e Our advanced mathematics teacher uses the idea of LIM an = a, Lim BN = (a1 + A2 +... + an) / N = a

lim n→∞ [(n!)^1/n]/n = 1/e Our advanced mathematics teacher uses the idea of LIM an = a, Lim BN = (a1 + A2 +... + an) / N = a

We only need ln [(n!) ^ 1 / N] / N] to be ln [(n!) ^ 1 / N] / N] = 1 / N (ln1 + LN2 +... + lnn) - ln = 1 / n [ln (1 / N) + ln (2 / n) +... + ln (n / N)] - > ∫ (integral lower limit 0, upper limit 1) LNX DX (this is the defective integral, and 0 is the defect). By using the partial integral, it is not difficult to know that the value of the defective integral is - 1, and it is proved that