A statistics problem about sum of random variables

on average how many random variables between (0,1) needed to have a sum > 1:

P(sum of 1 random variables < 1) = 1
P(sum of 2 random variables < 1) = 1/2
P(sum of 3 random variables < 1) = 1/3
P(sum of 4 random variables < 1) = 1/4
...
P(sum of n random variables < 1) = 1/n

They are all independent events

P (until nth random variable, all sum < 1) = 1/n!
p1 = P (before and include nth random variable, any sum > 1) = 1 - 1/n!
p2 = P (before and include (n - 1)th random variable, any sum > 1) = 1 - 1/(n - 1)!
p (x = n) = p1 - p2 = (1–1/n!) — (1–1/(n — 1)!) = (n — 1)/n!
E(x) = sum_{n*(n -1)/n!} = sum_{n/n!} = e