Every Sunday, Bob calls Liz to see if she will play tennis with him on that day. If Liz has not played tennis with Bob since i Sundays ago, the probability that she will say yes to him is i/k, k ≥ 2,...

Every Sunday, Bob calls Liz to see if she will play tennis with him on that day. If Liz has not played tennis with Bob since i Sundays ago, the probability that she will say yes to him is i/k, k ≥ 2, i = 1, 2, ... , k. Therefore, if, for example, Liz does not play tennis with Bob for k − 1 consecutive Sundays, then she will play with him the next Sunday with probability 1. On the nth Sunday, after Bob calls Liz, the number of times Liz has said no to Bob since they last played tennis is denoted by X_n. For 0 ≤ i ≤ k − 1, find the long-run probability that Liz says no to Bob for i consecutive Sundays. (Note that the answer is not 1/k.)