Considering e is the limit->+oo of (1+1/n)^n, then is e "what you get if you wait for the least gain, by waiting for the most amount of time"? Something like "e is the patience number".
You might wish to look at, for example, the actual limit of, for example, (1+1/(3n))^(2n) which, for every particular choice of "n" will have a less gain waited for for an even greater period of time than the one you happende to pick.