[proofplan] We write the coefficients $a_n$ as successive differences of the ordinary partial sums $A_n=\sum_{k=0}^{n}a_k$. This converts the [power series](/page/Power%20Series) into the Abel mean $(1-r)\sum_{n=0}^{\infty}A_n r^n$. Since $(A_n)$ converges to $s$, this Abel mean is forced to converge to the same limit: the finitely many early terms are killed by the factor $1-r$, and the tail is uniformly close to $s$. [/proofplan] [step:Express the power series through the partial sums] Let $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$ denote the scalar field containing all coefficients $a_n$ and the sum $s$. We use the convention $\mathbb{N}=\{1,2,3,\dots\}$. For each $n\in \mathbb{N}\cup\{0\}$, define the partial sum \begin{align*} A_n := \sum_{k=0}^{n} a_k \in \mathbb{K}. \end{align*} Since $\sum_{n=0}^{\infty}a_n$ converges to $s$, the sequence $(A_n)_{n=0}^{\infty}$ converges to $s$. In particular, there exists a constant $M>0$ such that $|A_n|\leq M$ for all $n\in \mathbb{N}\cup\{0\}$. Fix $r\in(0,1)$. For each $N\in\mathbb{N}$, using $a_0=A_0$ and $a_n=A_n-A_{n-1}$ for $1\leq n\leq N$, we compute \begin{align*} \sum_{n=0}^{N} a_n r^n &= A_0+\sum_{n=1}^{N}(A_n-A_{n-1})r^n \\ &= \sum_{n=0}^{N} A_n r^n-\sum_{n=0}^{N-1} A_n r^{n+1} \\ &= (1-r)\sum_{n=0}^{N-1}A_n r^n + A_N r^N. \end{align*} The scalar series $\sum_{n=0}^{\infty}A_n r^n$ converges absolutely because \begin{align*} \sum_{n=0}^{\infty}|A_n|r^n \leq M\sum_{n=0}^{\infty}r^n = \frac{M}{1-r}. \end{align*} Also $A_N r^N\to 0$ as $N\to\infty$, since $|A_N r^N|\leq Mr^N$ and $r^N\to0$. Taking $N\to\infty$ in the finite identity gives convergence of $\sum_{n=0}^{\infty}a_n r^n$ and the representation \begin{align*} f(r)=(1-r)\sum_{n=0}^{\infty}A_n r^n. \end{align*} [guided] The reason for introducing $A_n$ is that the original coefficients $a_n$ need not be absolutely summable, so direct comparison of $\sum a_n r^n$ with $\sum a_n$ is not available. The convergence of $\sum a_n$ tells us instead that its partial sums are bounded and approach $s$. Let $\mathbb{K}\in\{\mathbb{R},\mathbb{C}\}$ denote the scalar field containing all coefficients $a_n$ and the sum $s$. We use the convention $\mathbb{N}=\{1,2,3,\dots\}$. For each $n\in \mathbb{N}\cup\{0\}$, define \begin{align*} A_n := \sum_{k=0}^{n} a_k \in \mathbb{K}. \end{align*} Then $A_n\to s$ as $n\to\infty$. Every convergent sequence is bounded, so there exists $M>0$ such that $|A_n|\leq M$ for every $n\in\mathbb{N}\cup\{0\}$. Fix $r\in(0,1)$ and $N\in\mathbb{N}$. We rewrite $a_n$ as a difference of consecutive partial sums: \begin{align*} a_0=A_0,\qquad a_n=A_n-A_{n-1}\quad\text{for }1\leq n\leq N. \end{align*} Substituting this into the finite weighted sum gives \begin{align*} \sum_{n=0}^{N} a_n r^n &= A_0+\sum_{n=1}^{N}(A_n-A_{n-1})r^n \\ &= \sum_{n=0}^{N} A_n r^n-\sum_{n=0}^{N-1} A_n r^{n+1} \\ &= (1-r)\sum_{n=0}^{N-1}A_n r^n + A_N r^N. \end{align*} This is the finite summation-by-parts identity. It is useful because the sequence $(A_n)$ is bounded, and the weights $r^n$ are summable when $0<r<1$. Indeed, \begin{align*} \sum_{n=0}^{\infty}|A_n|r^n \leq M\sum_{n=0}^{\infty}r^n = \frac{M}{1-r}, \end{align*} so $\sum_{n=0}^{\infty}A_n r^n$ converges absolutely. The final boundary term vanishes because \begin{align*} |A_N r^N|\leq Mr^N\to0. \end{align*} Therefore, letting $N\to\infty$ in the finite identity proves that $\sum_{n=0}^{\infty}a_n r^n$ converges and that \begin{align*} f(r)=(1-r)\sum_{n=0}^{\infty}A_n r^n. \end{align*} [/guided] [/step] [step:Compare the Abel mean with the limiting partial sum] For $r\in(0,1)$, the geometric series identity gives \begin{align*} s=(1-r)\sum_{n=0}^{\infty}s r^n. \end{align*} Using the representation from the previous step, define $c_n:=A_n-s$ for $n\in\mathbb{N}\cup\{0\}$. Then $c_n\to0$ and \begin{align*} f(r)-s=(1-r)\sum_{n=0}^{\infty}c_n r^n. \end{align*} Let $\varepsilon>0$. Since $c_n\to0$, choose $N_0\in\mathbb{N}\cup\{0\}$ such that $|c_n|\leq\varepsilon$ for every $n\geq N_0$. Define the finite constant \begin{align*} C_{\varepsilon}:=\sum_{n=0}^{N_0-1}|c_n|, \end{align*} with the convention that $C_{\varepsilon}=0$ if $N_0=0$. Then, for every $r\in(0,1)$, \begin{align*} |f(r)-s| &\leq (1-r)\sum_{n=0}^{\infty}|c_n|r^n \\ &= (1-r)\sum_{n=0}^{N_0-1}|c_n|r^n +(1-r)\sum_{n=N_0}^{\infty}|c_n|r^n \\ &\leq (1-r)C_{\varepsilon}+\varepsilon(1-r)\sum_{n=N_0}^{\infty}r^n \\ &= (1-r)C_{\varepsilon}+\varepsilon r^{N_0} \\ &\leq (1-r)C_{\varepsilon}+\varepsilon. \end{align*} Taking $\limsup_{r\uparrow1}$ gives \begin{align*} \limsup_{r\uparrow1}|f(r)-s|\leq \varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, $\lim_{r\uparrow1}|f(r)-s|=0$, and hence \begin{align*} \lim_{r\uparrow1}f(r)=s. \end{align*} [guided] The identity from the previous step says that $f(r)$ is not mysterious: it is the weighted average of the partial sums $A_n$, with weights $(1-r)r^n$. Since \begin{align*} (1-r)\sum_{n=0}^{\infty}r^n=1, \end{align*} these weights form a probability-type averaging kernel on $\mathbb{N}\cup\{0\}$. As $r$ approaches $1$, the average places mass farther out in the sequence, where $A_n$ is close to its limit $s$. For $r\in(0,1)$, the geometric series identity gives \begin{align*} s=(1-r)\sum_{n=0}^{\infty}s r^n. \end{align*} Define the error sequence $c_n:=A_n-s$ for $n\in\mathbb{N}\cup\{0\}$. Since $A_n\to s$, we have $c_n\to0$. Subtracting the geometric representation of $s$ from the Abel representation of $f(r)$ gives \begin{align*} f(r)-s=(1-r)\sum_{n=0}^{\infty}c_n r^n. \end{align*} Now fix $\varepsilon>0$. Because $c_n\to0$, choose $N_0\in\mathbb{N}\cup\{0\}$ such that $|c_n|\leq\varepsilon$ whenever $n\geq N_0$. The finitely many earlier errors may be large, but there are only finitely many of them. Define \begin{align*} C_{\varepsilon}:=\sum_{n=0}^{N_0-1}|c_n|, \end{align*} with $C_{\varepsilon}=0$ if $N_0=0$. We split the weighted error into the early part and the tail: \begin{align*} |f(r)-s| &\leq (1-r)\sum_{n=0}^{\infty}|c_n|r^n \\ &= (1-r)\sum_{n=0}^{N_0-1}|c_n|r^n +(1-r)\sum_{n=N_0}^{\infty}|c_n|r^n. \end{align*} On the finite early part, we use $r^n\leq1$ and obtain the bound $(1-r)C_{\varepsilon}$. On the tail, we use $|c_n|\leq\varepsilon$ for all $n\geq N_0$, giving \begin{align*} (1-r)\sum_{n=N_0}^{\infty}|c_n|r^n \leq \varepsilon(1-r)\sum_{n=N_0}^{\infty}r^n = \varepsilon r^{N_0}. \end{align*} Therefore, \begin{align*} |f(r)-s| \leq (1-r)C_{\varepsilon}+\varepsilon r^{N_0} \leq (1-r)C_{\varepsilon}+\varepsilon. \end{align*} As $r\uparrow1$, the finite contribution $(1-r)C_{\varepsilon}$ tends to $0$, while the tail contribution is at most $\varepsilon$. Hence \begin{align*} \limsup_{r\uparrow1}|f(r)-s|\leq \varepsilon. \end{align*} Since $\varepsilon>0$ was arbitrary, this forces $\lim_{r\uparrow1}|f(r)-s|=0$, which is exactly \begin{align*} \lim_{r\uparrow1}f(r)=s. \end{align*} [/guided] [/step]