Question: Assume that the state space of Markov Chain $S=\mathbb{N}$, and if state $i$ is recurrent, then prove \begin{align*} \mathrm{lim}_{N\rightarrow\infty}\mathbb{P}(X_k\neq i \ \mathrm{for\ all\ integers}\ k\in[n+1,n+N]|X_0=i)=0 \end{align*} where $n\in \mathbb{N}^*$
Here is what I thought: \begin{align*} \mathbb{P}(X_k\neq i \ \mathrm{for\ all\ integers}\ k\in[n+1,n+N]|X_0=i)=\sum_{j=0}^{\infty}P_{ij}^{(n)}\sum_{l=N+1}^{\infty}f_{ji,n}^{(l)} \end{align*} where \begin{align*} f_{ji,n}^{(l)}=\mathbb{P}(X_{n+l}=i,X_m\neq i,m\in[N+1,N+l]|X_{n}=j) \end{align*} and my question is whether the $f_{ji,n}^{(l)}$ is equal to $f_{ji,0}^{(l)}$.
If the above statement is true, then I want to prove the remainder $\sum_{l=N+1}^{\infty}f_{ji,n}^{l}$ goes to $0$, when $N\rightarrow\infty$.
So, I think about whether or not the state $j$ is recurrent, if it's true then I can apply the theorem which is $\sum_{l=1}^{\infty}f_{ji}^{(l)}=1$, then the remainder will become zero, when $N\rightarrow\infty$. However, what if the state $j$ is not recurrent, how should I prove?
You can do this more simply. Let $\tau$ denote the hitting time of $i$ after at least $n$ steps. Then, $\Pr_i(\tau < \infty) = 1$ because $i$ is recurrent. Indeed, wherever $X$ is after $n$ steps, it will definition return.
So, given $\varepsilon > 0$, there exists $N_\varepsilon > 0$ such that $\Pr_i(\tau \le N_\varepsilon) \ge 1 - \varepsilon$. But, this exactly says that your long probability is at least $1 - \varepsilon$ for $N \ge N_\varepsilon$. Since $\varepsilon > 0$ was arbitrary, this proves that the limit is $0$.