[proofplan] We write the probability [generating function](/page/Generating%20Function) as a [power series](/page/Power%20Series) with coefficients $p_n=\mathbb P(X=n)$. For $s<1$, the left difference quotient can be rewritten using the finite geometric identity as a nonnegative series whose $n$th summand is $p_n(1+s+\dots+s^{n-1})$. As $s\uparrow 1$, these summands increase to $n p_n$, so the [monotone convergence theorem](/theorems/509) identifies the limit with $\sum_{n=1}^{\infty} n p_n$. Finally, we verify separately that this sum is exactly $\mathbb E[X]$. [/proofplan] [step:Expand the probability generating function by its point masses] For each $n\in\{0,1,2,\dots\}$, define \begin{align*} p_n:=\mathbb P(\{\omega\in\Omega:X(\omega)=n\}). \end{align*} Since $X$ takes values in $\{0,1,2,\dots\}$, the events $\{X=n\}$ are pairwise disjoint and their union is $\Omega$. Hence \begin{align*} \sum_{n=0}^{\infty}p_n=1. \end{align*} For every $n\in\{0,1,2,\dots\}$, let $\mathbb 1_{\{X=n\}}:\Omega\to\{0,1\}$ denote the indicator function of the event $\{\omega\in\Omega:X(\omega)=n\}$. For every $s\in[0,1]$, the [random variable](/page/Random%20Variable) $s^X:\Omega\to[0,1]$ is nonnegative and bounded, and the pointwise partition of $\Omega$ by the events $\{X=n\}$ gives \begin{align*} s^X=\sum_{n=0}^{\infty}s^n\mathbb 1_{\{X=n\}}. \end{align*} The partial sums are nonnegative and increase pointwise to $s^X$, so the monotone convergence theorem gives \begin{align*} G_X(s)=\mathbb E[s^X]=\sum_{n=0}^{\infty}s^n\mathbb E[\mathbb 1_{\{X=n\}}]=\sum_{n=0}^{\infty}p_n s^n. \end{align*} In particular, \begin{align*} G_X(1)=\sum_{n=0}^{\infty}p_n=1. \end{align*} [/step] [step:Rewrite the left difference quotient as a nonnegative series] Let $s\in[0,1)$. Since $G_X(1)=1$, the difference quotient in the statement satisfies \begin{align*} \frac{G_X(s)-G_X(1)}{s-1}=\frac{1-G_X(s)}{1-s}. \end{align*} Using the series expression for $G_X(s)$ and the identity $\sum_{n=0}^{\infty}p_n=1$, we obtain \begin{align*} \frac{1-G_X(s)}{1-s}=\sum_{n=0}^{\infty}p_n\frac{1-s^n}{1-s}. \end{align*} The term with $n=0$ is zero. For $n\ge 1$, the finite geometric identity gives \begin{align*} \frac{1-s^n}{1-s}=1+s+\dots+s^{n-1}. \end{align*} Therefore \begin{align*} \frac{G_X(s)-G_X(1)}{s-1}=\sum_{n=1}^{\infty}p_n(1+s+\dots+s^{n-1}). \end{align*} [guided] The sign in the left derivative is worth handling carefully. Since $s<1$, both $s-1$ and $G_X(s)-G_X(1)$ are nonpositive, so it is cleaner to rewrite the quotient as \begin{align*} \frac{G_X(s)-G_X(1)}{s-1}=\frac{1-G_X(s)}{1-s}. \end{align*} Now insert the probability generating function expansion: \begin{align*} G_X(s)=\sum_{n=0}^{\infty}p_n s^n. \end{align*} Because $\sum_{n=0}^{\infty}p_n=1$, we may write \begin{align*} 1-G_X(s)=\sum_{n=0}^{\infty}p_n-\sum_{n=0}^{\infty}p_n s^n. \end{align*} For $s\in[0,1]$, each summand $p_n(1-s^n)$ is nonnegative, so this subtraction is legitimate as an identity of nonnegative series: \begin{align*} 1-G_X(s)=\sum_{n=0}^{\infty}p_n(1-s^n). \end{align*} Dividing by $1-s>0$ gives \begin{align*} \frac{1-G_X(s)}{1-s}=\sum_{n=0}^{\infty}p_n\frac{1-s^n}{1-s}. \end{align*} The $n=0$ term vanishes because $1-s^0=0$. For $n\ge 1$, the finite geometric identity says \begin{align*} 1-s^n=(1-s)(1+s+\dots+s^{n-1}). \end{align*} Thus \begin{align*} \frac{G_X(s)-G_X(1)}{s-1}=\sum_{n=1}^{\infty}p_n(1+s+\dots+s^{n-1}). \end{align*} This is the key transformation: it replaces a boundary derivative by a monotone limit of nonnegative series. [/guided] [/step] [step:Pass to the limit by monotone convergence] Let $s\in[0,1)$ and define \begin{align*} Q(s):=\frac{G_X(s)-G_X(1)}{s-1}. \end{align*} For each $n\in\{1,2,\dots\}$, the function $s\mapsto p_n(1+s+\dots+s^{n-1})$ from $[0,1)$ to $[0,\infty)$ is nondecreasing. Hence, by the representation from the preceding step, $Q:[0,1)\to[0,\infty]$ is nondecreasing. Choose any sequence $(s_m)_{m=1}^{\infty}$ in $[0,1)$ such that $s_m\uparrow 1$. For each $m\in\{1,2,\dots\}$ and each $n\in\{1,2,\dots\}$, define \begin{align*} a_m(n):=p_n(1+s_m+\dots+s_m^{n-1}). \end{align*} For fixed $n$, the sequence $(a_m(n))_{m=1}^{\infty}$ is nondecreasing because $s_m\uparrow 1$ and all powers are nondecreasing on $[0,1]$. Moreover, \begin{align*} \lim_{m\to\infty}a_m(n)=n p_n. \end{align*} Applying the monotone convergence theorem to the counting measure on $\{1,2,\dots\}$ gives \begin{align*} \lim_{m\to\infty}\sum_{n=1}^{\infty}p_n(1+s_m+\dots+s_m^{n-1})=\sum_{n=1}^{\infty}n p_n. \end{align*} Therefore $Q(s_m)\to\sum_{n=1}^{\infty}n p_n$ along one increasing sequence $s_m\uparrow 1$. Since $Q$ is nondecreasing on $[0,1)$, for every $s\in[s_m,1)$ we have $Q(s_m)\le Q(s)\le\sum_{n=1}^{\infty}n p_n$, because each summand satisfies $1+s+\dots+s^{n-1}\le n$. Letting $m\to\infty$ proves \begin{align*} \lim_{s\uparrow 1}\frac{G_X(s)-G_X(1)}{s-1}=\sum_{n=1}^{\infty}n p_n. \end{align*} [/step] [step:Identify the limiting series with the expectation of $X$] For each $m\in\{1,2,\dots\}$, define the simple random variable $X_m:\Omega\to[0,\infty)$ by \begin{align*} X_m(\omega):=\sum_{n=0}^{m}n\,\mathbb 1_{\{X=n\}}(\omega). \end{align*} Then $0\le X_m\le X_{m+1}$ pointwise on $\Omega$, and $X_m(\omega)\to X(\omega)$ for every $\omega\in\Omega$. Applying the monotone convergence theorem to the [probability space](/page/Probability%20Space) $(\Omega,\mathcal F,\mathbb P)$ gives \begin{align*} \mathbb E[X]=\lim_{m\to\infty}\mathbb E[X_m]. \end{align*} For each $m$, \begin{align*} \mathbb E[X_m]=\sum_{n=0}^{m}n\,\mathbb P(X=n)=\sum_{n=0}^{m}n p_n. \end{align*} Therefore \begin{align*} \mathbb E[X]=\sum_{n=0}^{\infty}n p_n=\sum_{n=1}^{\infty}n p_n. \end{align*} The hypothesis $\mathbb E[X]<\infty$ ensures that this value is finite. Combining this identity with the preceding step yields \begin{align*} G_X'(1-)=\lim_{s\uparrow 1}\frac{G_X(s)-G_X(1)}{s-1}=\mathbb E[X]. \end{align*} [/step]