I don't like this solution. The problem is reminiscent of the
Poisson Distribution. In short, this determines the probability that
k counts (called "arrivals") occurring if you expect
λ counts occurring. The Poisson Distribution's form is:
As shows it can be manipulated to looks like terms of this limit.
Some properties of the Poisson distribution is that it is asymptotic to zero as you reach infinity, which makes sense since the odds that you will have infinitely many things occurring when you expect a finite number is zero. The Poisson distribution has a peak probability at
λ, shown here [click image to enlarge]:
I've shown before that this problem's solution lies between 0 and 1, which is confirmed as probability is used. The proper way to compute the probability of a range of occurrences, is to add the individual probabilities together. For instance, if you average 10 people arriving per day in a travel office, and you want to know the probability of 9, 10, or 11 people arriving, then you:
Looking at this, if we find the probability of all possible occurrences, from none to infinitely many occurrences, then the probability will be 1.
My limit goes from zero to
n occurrences, with the expected value being
n. Since we are going from 0 to
n, the point of the peak, we are only analyzing the left range of the total values. Since
n is going to infinity, and the end
is already at infinity, we are analyzing only the left
half of all expected values. If we are only analyzing half the values, then the probability is 1/2.