Let
- $E$ be an at most countable Polish space and $\mathcal E$ be the Borel $\sigma$-algebra on $E$
- $X=(X_n)_{n\in\mathbb N_0}$ be a Markov chain with values $(E,\mathcal E)$, distributions $(\operatorname P_x)_{x\in E}$ and transition matrix $$p=\left(p(x,y)\right)_{x,y\in E}=\left(\operatorname P_x\left[X_1=y\right]\right)_{x,y\in E}$$
- $\tau_x^0:=0$ and $$\tau_x^k:=\inf\left\{n>\tau_x^{k-1}:X_n=x\right\}$$ for $x\in E$ and $k\in\mathbb N$ and $$\varrho(x,y):=\operatorname P_x\left[\tau_y^1<\infty\right]\color{blue}{=\operatorname P_x\left[\exists n\in\mathbb N:X_n=y\right]}$$
- $N(y)=\sum_{n\in\mathbb N_0}1_{\left\{X_n=y\right\}}$ be the number of visits of $X$ in $y\in E$
$x\in E$ is called
- recurrent $:\Leftrightarrow$ $\varrho(x,x)=1$
- absorbing $:\Leftrightarrow$ $p(x,x)=1$
Now, let $$G(x,y):=\operatorname E_x\left[N(y)\right]\;\;\;\text{for }x,y\in E$$ be the Green's function of $X$.
I want to show, that for a non-absorbing $x\in E$ $$x\text{ is recurrent }\Leftrightarrow\; G(x,x)=\infty$$
Using that $$\operatorname P_x\left[\tau_y^k<\infty\right]=\varrho(x,y)\varrho(y,y)^{k-1}\;\;\;\text{for all }x,y\in E\text{ and }k\in\mathbb N$$ we obtain
\begin{equation} \begin{split} G(x,y)&=&\sum_{k\in\mathbb N}\operatorname P_x\left[N(y)\ge k\right]\\ &=&\sum_{k\in\mathbb N}\operatorname P_x\left[\tau_y^k<\infty\right]+1_{\left\{x=y\right\}}\\ &=&\varrho(x,y)\sum_{k\in\mathbb N_0}\varrho(y,y)^k+1_{\left\{x=y\right\}} &=&\frac{\varrho(x,y)}{1-\varrho(y,y)}+1_{\left\{x=y\right\}} \end{split} \end{equation} where $1/0:=\infty$, $0/0:=0$ and $0\cdot\infty:=0$. Especially, we've got $$G(y,y)=\frac 1{1-\varrho(y,y)}\tag 1$$
Question:$\;\;\;$The statement follows immediately from the definitions and $(1)$. But why do we need to claim, that this statement only holds if $x$ is non-absorbing?
If $x$ is recurrent, then $(1)$ and $1/0=\infty$ yield $G(x,x)=\infty$. The same argument seems to yield the recurrence of $x$, whenever $G(x,x)=\infty$. So, what am I missing?