If XXX is a discrete random variable such that P(X=k)=1k(k+1)P(X=k) = \frac{1}{k(k+1)}P(X=k)=k(k+1)1 for k=1,2,…k=1, 2, \dotsk=1,2,…, does E[X]E[X]E[X] exist?
Yes, E[X]=1E[X] = 1E[X]=1
Yes, E[X]=2E[X] = 2E[X]=2
No, the sum diverges.
Yes, E[X]=0E[X] = 0E[X]=0