Markov chain that is $\phi$-irreducible but not $\phi$-recurrent In trying to understand the underlying mathematics of the Metropolis-Hasting algorithm, I came across this paper on continuous state Markov chains. On page 3, Example 2 gives a Markov chain which is claimed to be strongly $\phi$-irreducible but not $\phi$-recurrent (in the example $\pi$ is used instead of $\phi$)
For convenience I'm transcribing the relevant definitions and the example.
On page 1 the following definitions are given ($P$ is a probability kernel, and $(X, B)$ is a measurable space):


*

*$P$ is "$\phi$-irreducible" iff for all $x \in X$ and all $A \in B$ with $\phi(A) > 0$, there is a positive integer $n = n_{xA}$ such that $P^n(x, A) > 0$


*$P$ is "strongly $\phi$-irreducible" iff for all $x \in X$ and all $A \in B$ with $\phi(A) > 0$, there is a positive integer $n = n_{xA}$ such that $P^m(x, A) > 0$ for all $m \geq n$


*$P$ is "$\phi$-recurrent" iff for all $x \in X$ and all $A \in B$ with $\phi(A) > 0$, a Markov chain which starts from $x$ at time $0$ hits $A$ at some positive time, a.s.; of course, this time is random

The example is the following:

Let $X_1$ be a finite set, and $P$ a stochastic matrix on $X_1$, with all entries strictly positive. Under the circumstances, there is a unique stationary probability $\pi$, and $P$ is $\pi$-recurrent. Adjoin a sequence of states 1, 2,..., each with $\pi$-probability $0$, and the following transition rules: $i \rightarrow  i + 1$ with probability $1/2^i$; with the remaining probability, $i$ goes to a point in $X_1$, chosen at random from $\pi$. The resulting kernel is strongly $\pi$-irreducible, but not $\pi$-recurrent, due to the adjoined states.

I don't see why the resulting kernel is not $\pi$-recurrent. Specifically, I take it that the definition of $\phi-recurrent$ is equivalent to: $P(\tau_A = \infty \: | \: starting \: at \: x) = 0$ for all $x \in X$ and all $A \in B$ with $\phi(A) > 0$, where $\tau_A = min\{n > 0, X_n \in A\}$ (in this case $P$ denotes probability, not the kernel)
The states we need to examine for this property are the adjoined states. Hence, if at time $0$ the chain begins at $i$ we have:
$$
P(\tau_A = \infty \: | \: starting \: at \: i) = P(i \rightarrow i + 1)P(\tau_A = \infty \: | \: starting \: at \: i+1) + (1-P(i \rightarrow i + 1))P(\tau_A = \infty \: | \: starting \: at \: x \in X_1)
$$
However, $P(\tau_A = \infty \: | \: starting \: at \: x \in X_1) = 0$, since the initial kernel is $\pi$-recurrent, so:
$$
P(\tau_A = \infty \: | \: starting \: at \: i) = P(i \rightarrow i + 1)P(\tau_A = \infty \: | \: starting \: at \: i+1)
$$
Inductively:
$$
P(\tau_A = \infty \: | \: starting \: at \: i) = \prod_{j=i}^\infty \frac{1}{2^j} = 0
$$
So why isn't the resulting kernel $\pi$-recurrent? Is there something I'm missing?
 A: This answer gives details on my comment:
The example in the paper should be modified to assuming
$$ P_{i,i+1} = 1 - \frac{1}{(i+1)^2} \quad \forall i \in \{1, 2, 3, ...\}$$
This is because
$$ \prod_{j=1}^{\infty} \left(1-\frac{1}{(i+1)^2}\right)= 1/2$$
and so if we start in state 1, with probability $1/2$ will will always transition one step forward (in the augmented states) and so we will never visit states in $X_1$.

It is also true that
$$ \prod_{j=1}^{\infty}   (1-2^{-j}) >0$$
and so this is why I think that example simply had a typo (they meant $1-2^{-i}$ rather than $2^{-i}$). Indeed it seems difficult to calculate $\prod_{j=1}^{\infty}(1-2^{-j})$ exactly, but we can show it is positive by defining
$$ y_n = \prod_{j=1}^{n} (1-2^{-j}) \quad \forall n \in \{2, 3, 4, ...\}$$
Then for all $n \in \{2, 3, 4, ...\}$ we get
\begin{align}
\log(y_n) &= \sum_{j=1}^n \log(1-2^{-j}) \\
&=\log(1-2^{-1}) + \sum_{j=2}^n\log(1-2^{-j})\\
&=\log(1/2) - \sum_{j=2}^n \log\left(1 + \frac{1}{2^j-1}\right)\\
&\overset{(a)}{\geq} \log(1/2) - \sum_{j=2}^n \frac{1}{2^j-1}\\
&\overset{(b)}{\geq} \log(1/2) - \sum_{j=2}^n \frac{1}{2^{j}-(1/2)2^{j}} \\
&\geq \log(1/2) - \sum_{j=2}^{\infty}\frac{1}{2^{j}-(1/2)2^{j}}\\
&= \log\left(\frac{1}{2e}\right) 
\end{align}
where (a) uses $\log(1+x)\leq x$ for all $x>-1$; (b) uses $\frac{1}{2^j-1}\leq \frac{1}{2^{j}-(1/2)2^{j}}$ for $j \geq 2$. Thus
$$ y_n \geq  \frac{1}{2e} \approx 0.1839 \quad \forall n \in \{2, 3, 4, ...\} $$
