This is self learning and it is stats.
$P_{ij}(s)=\sum^\infty_{n=0}p_{ij}^ns^n$ (which you'll probably recognise is a generating function)
and
$F_{ij}(s)=\sum^\infty_{n=1}f_{ij}^ns^n$ (note n=1 here)
I wish to show (I know this result, but I'd like to do it as if I did not however)
$P_{ij}(s)=p_{ij}^0+F_{ij}(S)P_{jj}(s)$
$P_{ij}(s)=p_{ij}^0+\sum^\infty_{n=1}p_{ij}^ns^n$
Now using a previous result: $p_{ij}^n=\sum^n_{k=1}f_{ij}^kp_{jj}^{n-k}$
We get:
$P_{ij}(s)=p_{ij}^0+\sum^\infty_{n=1}s^n\sum^n_{k=1}f_{ij}^kp_{jj}^{n-k}$
One can easily flip that around:
$P_{ij}(s)=p_{ij}^0+\sum^\infty_{n=1}s^n\sum^{n-1}_{k=0}f_{ij}^{n-k}p_{jj}^k$
and get close:
$P_{ij}(s)=p_{ij}^0+\sum^\infty_{n=1}\sum^{n-1}_{k=0}s^{n-k}f_{ij}^{n-k}s^kp_{jj}^k$
Then suddenly! Magic!
$=p_{ij}^0+\sum^\infty_{k=0}\sum^\infty_{n=k+1}s^{n-k}f_{ij}^{n-k}s^kp_{jj}^k$
The rest seems like something I can do.
So I looked at a simpler sum, $\sum^L_{i=1}\sum^i_{j=1}f(j)$ and noticed this is:
$\ \ \ f(1)$
$+f(1)+f(2)$
$+f(1)+f(2)+f(3)$
$...$
Which seems important, but I daren't make the jump to infinity.
The question adds at the end "for |s|$<$1" now isn't this a bit pointless, generating functions considered for their parameter being 1 is when they are actually useful!
Definitions (if they are needed / help)
$f_{ij}^n$ denotes $\mathbb{P}[X_n=j,X_k\ne j \forall 1\le j\le n-1|X_0=i]$
or in English: the probability that starting from state i we end up at state j taking a path that doesn't visit state j in exactly n steps.
$p_{ij}^n$ denotes $\mathbb{P}[X_n=j|X_0=i]$
It is the "probability that starting from i you end up at (via any path) at j in exactly n steps".
Note that why I have used the notation of powers, $p_{ij}^k\ne (p_{ij})^k$
(That's not true, but it'd be a special case where they are equal, like the identity matrix of state changes.... it doesn't denote power basically!)