Abel's theorem and generating function

Question

Corollary:

Let $P_{ij}\left ( s \right )=\sum_{n=0}^{\infty}P_{ij}s^{n}, \forall \left | s \right | < 1$, be the generating function of $\left \{ p_{ij}^{n} \right \}$.

and

$F_{ij}\left ( s \right ) =\sum_{n=0}^{\infty}f_{ij}^{n}s^{n}$, for $\left | s \right | < 1$ be the generating function of $\left \{ f_{ij} \right \}^{n}$

Then, $P_{ij}\left ( s \right )=\delta _{ij}+F_{ij}\left ( s \right )P_{jj}\left ( s \right )$

Abel's theorem:

Let $\left \{ a_{n} \right \}$ be a sequence of real numbers such that $0 \leq a_{n} \leq 1, n=0,1, \cdot \cdot \cdot$ Let $A\left ( z \right )=\sum_{n=0}^{\infty}a_{n}z^{n}$ be the generating function of $\left \{ a_{n} \right \}$

If $\sum_{n=0}^{\infty}a_{n}=a<\infty$, then $\lim_{z\rightarrow 1^{-}}A\left ( z \right )=a$.

$\space$

Theorem: i is recurrent IFF $\sum_{n=0}^{\infty}P_{ii}^{n}<\infty$

Proof from author:

Suppose i is recurrent and that $f_{ii}=1$ where $f_{ii} \equiv \sum_{n=0}^{\infty}f_{ii}^{n}.$

By Abel's theorem:

$\lim_{s\rightarrow 1^{-}}F_{ii}\left ( s \right )=1\Rightarrow \lim_{s\rightarrow 1^{-}}1-F_{ii}\left ( s \right )=0$

At this point, I see an issue: if the author has defined $f_{ii}^{n}$ to be the coefficient of the generating function, then the term $s^{n}$ in the generating function $\sum_{n=0}^{\infty}f_{ii}s^{n}$ must have $\left | s \right | \leq 1$, contrary to $\left | s \right | < 1$ as mentioned by the author.

Next, using the fact then that $P_{ij}\left ( s \right )=\delta _{ij}+F_{ij}\left ( s \right )P_{jj}\left ( s \right )$,

for j=i, we have

$P_{ii}\left ( s \right )=\delta _{ii}+F_{ii}\left ( s \right )P_{ii}\left ( s \right )=P_{ii}\left ( s \right )=1+F_{ij}\left ( s \right )P_{jj}\left ( s \right )$

$\Rightarrow P_{ii}\left ( s \right )=\frac{1}{1-F_{ii}\left ( s \right )}$

Then, $\lim_{s\rightarrow 1^{-}}P_{ii}\left ( s \right )=\infty.$

Again, how does the above tends to infinity as $s\rightarrow 1^{-}$? It can only be true that $P_{ii}\left ( s \right )\rightarrow \infty$ as $F_{ii}\left(s\right) \rightarrow 1$

Does he mean $\lim_{F_{ii}\rightarrow 1^{-}}P_{ii}\left ( s \right )=\infty$?.

Answer 1

If $i$ is recurrent then $\sum f_{ii}^{n} =1$. Also $\sum f_{ij}^{n} <\infty $ always. Hence there is no difficulty in defining F for $|s| \leq 1$. I think $|s|<1$ in the definition is only a typo. Since $F_{ii}(s) \to \sum f_{ii}^{n} =1$ as $s\to 1-$ we get $\lim_{s \to 1-} P_{ii}(s)=\infty$.