令 t = n^(1/n) - 1 ,由 n^(1/n) > 1 ,可得：t > 0 ；则有：n = (1+t)^n = 1+nt+n(n+1)t^2/2+...+t^n > n(n+1)t^2/2 ,可得：t^2 < 2/(n+1) ；所以,0 < t < √[2/(n+1)] ,即有：0 < n^(1/n) - 1 < √[2/(n+1)] .已知,...