I'm struggling with the following theorem:
If $s$ is a recurrent aperiodic state in a Markov Chain $X$, then $\forall s'\in[s]$ i can say that $$\lim_{n\to\infty}p^{(n)}_{s',s}=f_{s',s}\frac{1}{\mu_s}$$
where $p^{(n)}_{s',s}$ is the probability to go from the state $s'$ to the state $s$ in exactly $n$ steps, $f_{s',s}$ is the probability to go from the state $s'$ to the state $s'$ and it's defined as follows $$f_{s',s}=\sum_{j=1}^\infty f_{s',s}^{(j)}=\sum_{j=1}^\infty\mathbb{P}(X_1\neq s,X_2\neq s,...,X_{j-1}\neq s,X_n=s|X_0=s')$$ and $\mu_s$ is the invariant distribution.
The proof i have use the renewal theorem in this way: $$p^{(n)}_{s',s}=\sum_{j=1}^n f_{s',s}^{(j)}p^{(n-j)}_{s,s}$$ but after that i'm completely lost and my notes get really confused. Can anyone help me?
Edit: I think I found a solution with the problem i was having, but I'm not completely certain about a passage, so I'll ask confirmation here.
As said before I'll use the renewal theorem so I'll get
$$p^{(n)}_{s',s}=\sum_{j=1}^n f_{s',s}^{(j)}p^{(n-j)}_{s,s}$$
Now I'm not completely sure about what I'm about to say, but since $p_{s,s}^{(n-k)}$ is the probability to get from the state $s$ to itself in $n-k$ steps, can I say that it's equal to the probability of $n-k$ being a renewal time?
In that way I can say that
$$p_{s,s}^{(n-k)}=\mathbb{P}(\exists m |\tau^{(s)}_m =n-k )$$
that for the low of large number aproaches the expected value of the return time.
Just to be clearer about notation, which I know it's confusing and it's probably why I'm having all this troubles the $(\tau^{(s)}_i)_{i\in\mathbb{N}}$ are the time in which my markov chain gets back to $s$ starting in $s$ at time $0$.
The return time is defined as follows:
$$T_1=\tau^{(s)}_1,\;\;T_i=\tau^{(s)}_{i}-\tau^{(s)}_{i-1}\;\operatorname{for}\;i\geq 2$$