For ergodic Markov chains, when does $\lim_{N\to\infty} \mathbb{E}[\sum_{n=1}^{N}f(X_n)] - N\mu(f)$ exist For an ergodic Markov chain (process would be even better) $X_{n}$ with stationary distribution $\mu$, under which conditions does
$$
L:=\lim_{N\to\infty} \mathbb{E}[\sum_{n=1}^{N}f(X_n)] - N\mu(f)
$$
exist? In the other direction, do you know of examples where this limit does not exist?

Ergodicity says 
$$
\mathbb{E}[\frac{1}{N}\sum_{n=1}^{N}f(X_n)] - \mu(f) \to 0.
$$
If the limit $L$ above exists, this could be refined to 
$$
\mathbb{E}[\frac{1}{N}\sum_{n=1}^{N}f(X_n)] - \mu(f) = \frac{L}{N} + \mathcal{o}(\frac{1}{N}).
$$
If we denote by $\mu_{n}$ the distribution of $X_n$, we have 
$$
\mathbb{E}[\sum_{n=1}^{N}f(X_n)] = N \mu(f) + \sum_{n=1}^{N} (\mu-\mu_n)(f)
$$
and my question can be rephrased as

Under which conditions on the Markov chain $(X_n)$ and the function
  $f$ is $(\mu-\mu_n)(f)$ summable? In the other direction, do you know of examples where it is not summable?


I know that for many Markov chains we have $\mu_n =\mu+ \mathcal{O}(e^{-cn})$. I'm interested in theory for cases where the convergence is not geometric but still manageable. This question is therefore maybe a reference request or maybe can just be answered by a simple class of examples with easily tunable convergence properties.
 A: An important class of examples is the following:
Let $X$ be a Markov chain with state space $\mathcal{X}$ and $P$ its Markov operator. Suppose that $X$ is aperiodic and $f$-regular, i.e. there are functions $f\geq 1$ and $V\geq 0$, a Borel set $C$ and a constant $b<\infty$ such that 
$$\Delta V:=PV-V\leq -f+b\mathbb{I}_C,$$
where $\mathbb{I}$ denotes the indicator function. This is called drift condition (V3) in [1].
Suppose that $X$ is aperiodic and that $\mu(V)<\infty$. Then for any function $g$ satisfying $|g|\leq f$ we have some $R>0$ such that
$$\sum_{k=0}^{\infty}\left|(Pg)(x)-\mu(g)\right|\leq R(V(x)+1), \quad x\in\mathcal{X}.$$
Thus the sum
$$\hat g(x)=\sum_{k=0}^{\infty}\left((Pg)(x)-\mu(g)\right)$$
is absolutely convergent and satisfies $|\hat g|\leq R(V+1)$. You can therefore also get a bound on $\mu_0(\hat g)=E[\hat g(X_0)]$.
See Section 17.4 in [1] for generalizations and further details. 
[1] Meyn, S., Tweedie, R. L., & Glynn, P. W. (2009). Markov Chains and Stochastic Stability. Cambridge University Press.
