Solution
= Solution
Adding $\varepsilon I$ leaves every invariant distribution unchanged. On the unique closed communicating class it also supplies a self-loop, making the chain aperiodic; transient states do not affect the <limit>. Therefore
$$
P(X_n^{(\varepsilon)}=0)\longrightarrow\pi_0
=\frac1{E[T_1]}=\frac1{\sum_{j=1}^Njq_j},
$$
independently of $\varepsilon\in(0,1)$.