I haven't seen that before and its not true in general. if its true for this distribution (and i haven't checked) then it might help but its probably just as easy to note the following:E(X)=sum(P(X>n)) where n € 0,1,2..... help me?

P(X>k) = P(X=k+1) + P(X=k+2) ....

if you write out the terms you will see a geometric progression. There is a standard formula for summing those.