Q5: Let PK = 1/(2*k(k + 1)), k = 1, 2, . .., and po = 1 - Ek>] PK. Notice that 2hipk = 1. k=1 So, if we let X1, X2, ... be i.i.d. with P(Xn = -1) = Po and P(Xn = 2k - 1) = Pk, 'IZ YA then EXn = 0. Let Sn = Xi+ ... + Xn. Show that Sn/ (n/ log2 n) - -1, in probability