My lecture notes state the following theorem:
Theorem 2. Let $(X_n)$ be an irreducible, aperiodic, null-recurrent Markov chain. Then $$\forall i,j \in S : p_{ij}^{(n)}\to 0 \text{ and } \sum_n p_{ij}^{(n)} = \infty$$
And the proof starts like this
The divergence of the series $\color{red}{\text{follows}}$ from the Theorem 1, so we just need to show $p_{ij}^{(n)}\to 0$.
Okay, I go to the theorem 1 and see roughly the following:
Theorem 1. Let $(X_n)$ be a Markov chain.
If $i\in S$ is recurrent then $\sum_n p_{ii}^{(n)} = \infty$.
If $i\in S$ is transient then $\sum_n p_{ii}^{(n)} < \infty$.
I tried to figure out why would it follow by trying to come up with an inequality like $$p_{ii}^{(n+m+k)}\geq p_{ij}^{(n)} p_{jj}^{(m)} p_{ji}^{(k)}$$ yet it didn't help.
In other sources, it seems to be a non-trivial result which is proven with Abel's lemma.
So my question is: can the divergence of the series in theorem 2 be followed from theorem 1?
Since the Markov chain is irreducible, it is possible to get from state $i$ to state $j$, so $p_{ij}^{(k)} > 0$ for some $k$. Then $p_{ij}^{(n+k)} \ge p_{ii}^{(n)} p_{ij}^{(k)}$, so $$\sum_{n\ge 0} p_{ij}^{(n)} \ge \sum_{n\ge k} p_{ij}^{(n)} = \sum_{n\ge 0} p_{ij}^{(n+k)}\ge p_{ij}^{(k)} \sum_{n \ge 0} p_{ii}^{(n)} = \infty$$