Markov Chains - Proof that a finite chain has at least one recurrent state I am looking at the proof in Grimmett and Stirzaker's Probability and Random Processes of the statement:
If $S$ is finite, then at least one state is persistent
The proof given is to assume by contradiction that all states are persistent, and then take the limit as $n \rightarrow \infty$ of $1 = \sum_j p_{ij}(n)$, giving: $1=lim_{n \rightarrow \infty}\sum_j p_{ij}(n)=0$,
the second equality being derived from an earlier corollary that if $j$ is transient then $p_{ij}(n) \rightarrow 0$ as $n \rightarrow \infty$ for all $i$.
The proof makes sense, but what restricts it to the finite case? Intuitively I get that if you have an infinite number of states, that even when $n$ gets large, there is always another state to jump to - but my understanding is that the corollary and the summation in this proof are true for infinite state spaces too
 A: If the state space is infinite, then all states can be transient -- think of a chain on the positive integers that deterministically marches off to infinity. The cited proof fails in the infinite-state case because it it not always true that
$$
\lim_n\sum_j p_{ij}(n) = \sum_j \lim_n p_{ij}(n),
$$
the reason being that you are attempting to interchange the order of two limits (the sum over $j$ is also a limit), and you are not assured of the same result after interchanging.
A: A sequence of infinite sums, whose terms each tend to zero, need not converge (as a sequence) to $0$. Consider the sequence of sum $S_n=\sum_{k\geq 1}a_{n,k}$ where
$$a_{n,k}=\cases{\frac{1}{n}&$1\leq k\leq n$\\0&otherwise}$$
Then for each fixed $n\geq 1$, $S_n=1$. Nonetheless, $\lim_{n\to\infty}a_{n,k}=0$ for all $k\geq 1$. Essentially, this means that, in general, one cannot change the order of limits for infinite sums:
$$\lim_{n\to\infty}\sum_{k\geq 1}a_{n,k}\neq\sum_{k\geq1}\lim_{n\to\infty}a_{n,k}$$
A: This comes down to the fact that if you have a finite number of terms all tending individually to $0$, then their sum tends to $0$, but this isn't true for an infinite number of terms.
If there are only $k$ terms, say $p_1(n),\ldots p_k(n)$, and for each $i$ $p_i(n)\to\infty$ as $n\to\infty$, then for any $\delta$ and each $i$ there exists $n_i$ such that $|p_i(n)|<\delta/k$ for all $n>n_i$. Now if $n>\max_i n_i$, then each term is less than $\delta/k$, so $|\sum_{i=1}^kp_i(n)|<\delta$.
This doesn't work for infinitely many terms because there are infinitely many different $n_i$, and they might not be bounded. For example, let $p_i(n)=1$ if $n=i$ and $p_i(n)=0$ otherwise. Now for any fixed $i$ $p_i(n)\to 0$, but $\sum_{i=1}^{\infty}p_i(n)=1$ for every $n$. (This corresponds to the Markov chain on $\mathbb Z$ that takes one step to the right with probability $1$, which is transient.)
