Vietnam Undergraduate Mathematics Competition 2003 B-4

[Namdung]

Let xk = 1/2! + 2/3! + 3/4! + ... + k/(k+1)!. Find lim (x1n + x2n + ... + x2003n )1/n as n→∞.

Source VUMC 2003

Namdung

[harazi]

I suppose that x_1n means (x_1)^n. Let's proceed. It's trivial to prove that x_k=1-1/(k+1)!. Thus (sum (x_k)^n)^1/n=((1-1/2!)^n+(1-1/3!)^n+...+(1-1/(n+1)!)^n)^1/n, which obviously tends to 1, since it is greater than 1-1/(n+1)! and smaller than (n*(1-1/(n+1)!)^n)^1/n, and these sequences tend to 1.