 Stationary distribution of a regenerative countdown chain

For the geometric-jump countdown Markov chain, regeneration at zero givesA cycle visits positive state $j$ precisely when its jump reaches at least $j$, with probability $q^{j-1}$. Dividing this expected occupation by the mean cycle length proves the formula. Mean recurrence times are $1/{\pi }_j$, and the limiting probability away from zero is $1/(1+p)$.