If $i$ is recurrent then expected recurrence time is the limit of $\frac 1 {(1-s)p_{ii}(s)}$

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Theorem: i is recurrent then $\mu_{ii}=\sum_{n=1}^\infty nf_{ii}^n = \lim_{s\rightarrow 1^{-}} \frac 1 {(1-s)p_{ii}(s)}$

Proof:

Let $i$ be recurrent.

Then, note:

\begin{align} & \lim_{s\rightarrow 1^{-}} F_{ii} (s))=\lim_{s\rightarrow 1^{-}} \sum_{n=0}^\infty f_{ii}^n s^n \\[10pt] & \Longrightarrow F_{ii}(\approx 1^{-})=\sum_{n=0}^\infty f_{ii}^n=1 \\[10pt] & \Longrightarrow \lim_{s\rightarrow 1^{-}} ( 1-F_{ii} (s)) =0 \end{align}

Again, observe:

\begin{align} & F_{ii} (s)=\sum_{n=0}^\infty f_{ii}^n s^n \\[10pt] \Longrightarrow & F_{ii}'(s)=\sum_{n=0}^\infty nf_{ii}^n s^{n-1} \end{align}

so,

$$F_{ii}'(\approx 1^{-})=\sum_{n=0}^\infty nf_{ii}^n \approx \lim_{s\rightarrow 1^{-}} F_{ii}' (s)=\lim_{s\rightarrow 1^{-}} \sum_{n=0}^\infty nf_{ii}^n s^{n-1} $$

How to I then conclude that $F_{ii}'(s)=\lim_{s\rightarrow 1^{-}} \frac{1-F_{ii} (s)}{1-s}?$

Vital hints are appreciated.