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  "body": "\n## Proof\n\nWe rewrite $G$ with the partial sums by summation by parts.\n\n1. Let $s_n = \\sum_{k=0}^{n} a_k$ be the [[partial sums|0NB643K60Q]], and let $s = \\lim s_n$, which exists by hypothesis. Put $s_{-1} = 0$, so that $a_n = s_n - s_{n-1}$ for all $n \\ge 0$.\n2. The sequence $(s_n)$ converges, so it is [[bounded|0S4JF619V7]]: there is $K$ with $\\abs{s_n} \\le K$ for all $n$. Then $\\abs{a_n} \\le 2K$ by the [[triangle inequality|03HF5PMY5T]].\n3. Fix $x$ with $0 \\le x \u003c 1$. Then $\\abs{a_n x^n} \\le 2K x^n$ and $\\abs{s_n x^n} \\le K x^n$. The [[geometric series|01X5CVV1Q5]] $\\sum x^n$ converges.\n4. By the [[comparison test|04Z5G0THN6]], $\\sum a_n x^n$ and $\\sum s_n x^n$ converge absolutely. So both [[converge|0Y123X48GA]].\n5. For each $N$,\n$$\\sum_{n=0}^{N} a_n x^n = \\sum_{n=0}^{N} s_n x^n - \\sum_{n=1}^{N} s_{n-1} x^n = (1 - x) \\sum_{n=0}^{N} s_n x^n + s_N x^{N+1}.$$\n6. We have $\\abs{s_N x^{N+1}} \\le K x^{N+1}$, and [[$x^{N+1} \\to 0$|0Q7K4AXE53]]. So $s_N x^{N+1} \\to 0$.\n7. Let $N \\to \\infty$ in step 5. The [[sum rule|0RAM4E07NR]] and the [[product rule|0734RSVF2V]] for limits give $G(x) = (1 - x) \\sum_{n=0}^{\\infty} s_n x^n$.\n8. The geometric series gives $(1 - x) \\sum_{n=0}^{\\infty} x^n = 1$. So $s = (1 - x) \\sum_{n=0}^{\\infty} s x^n$.\n9. Subtract step 8 from step 7, using the limit rules on partial sums: $G(x) - s = (1 - x) \\sum_{n=0}^{\\infty} (s_n - s) x^n$.\n10. Let $\\varepsilon \u003e 0$. By the [[epsilon-N form|09G6KHS6GY]], there is $N$ with $\\abs{s_n - s} \u003c \\varepsilon / 2$ for all $n \\ge N$. Let $C = \\sum_{n=0}^{N-1} \\abs{s_n - s}$.\n11. Split the series of step 9 into the terms with $n \u003c N$ and the tail with $n \\ge N$. Since $x^n \\le 1$, the first part has absolute value at most $C$.\n12. The tail has terms of absolute value at most $(\\varepsilon/2) x^n$. By the [[bound for absolutely convergent series|0Y123X48GA]] and the comparison test, its absolute value is at most $(\\varepsilon/2) \\sum_{n \\ge N} x^n$.\n13. The geometric series gives $(1 - x) \\sum_{n \\ge N} x^n = x^N \\le 1$. With steps 9, 11 and 12, $\\abs{G(x) - s} \\le (1 - x) C + \\varepsilon/2$.\n14. Let $\\delta = \\min(1, \\varepsilon / (2C + 1))$. If $1 - \\delta \u003c x \u003c 1$, then $0 \\le x \u003c 1$, and $(1 - x) C \\le \\delta C \\le C \\varepsilon / (2C + 1) \u003c \\varepsilon / 2$.\n15. So $\\abs{G(x) - s} \u003c \\varepsilon$ for all $x$ with $1 - \\delta \u003c x \u003c 1$. By the definition of the [[left limit|0909PWZR62]], $G(x) \\to s = \\sum_{n=0}^{\\infty} a_n$ as $x \\to 1$ from the left.\n\n## Idea\n\nSummation by parts writes $G(x) - s$ as $(1 - x) \\sum (s_n - s) x^n$. The factor $1 - x$ kills the finitely many early terms as $x \\to 1$. The later terms have small coefficients $s_n - s$, and $(1 - x) \\sum x^n \\le 1$ keeps their total small.\n",
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