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  "body": "\n## Proof\n\nWe first treat $-1 \u003c x \u003c 1$ by comparing derivatives, then the endpoint $x = 1$ by Abel's theorem.\n\n1. Let $a_0 = 0$ and $a_n = \\frac{(-1)^{n+1}}{n}$ for $n \\ge 1$. Consider the power series $g(x) = \\sum_{n=0}^{\\infty} a_n x^n$.\n2. For $n \\ge 1$, $\\frac{|a_{n+1}|}{|a_n|} = \\frac{n}{n+1} \\to 1$. By the [[ratio formula|0JBMYAA9QH]], the radius of convergence of $g$ is $1$.\n3. By [[term-by-term differentiation|002X7EK91H]], $g$ is differentiable on $(-1, 1)$. Its derivative is $g'(x) = \\sum_{n=1}^{\\infty} n a_n x^{n-1} = \\sum_{n=1}^{\\infty} (-1)^{n+1} x^{n-1}$.\n4. Putting $m = n - 1$, this is $g'(x) = \\sum_{m=0}^{\\infty} (-x)^m$. Since $\\abs{-x} \u003c 1$, the [[geometric series|01X5CVV1Q5]] gives $g'(x) = \\frac{1}{1 + x}$.\n5. Let $h(x) = \\ln(1 + x)$ for $-1 \u003c x \u003c 1$, where $\\ln$ is the [[natural logarithm|0044CPHJBV]]. This is defined, since $1 + x \u003e 0$.\n6. The map $x \\mapsto 1 + x$ is a [[polynomial with derivative $1$|0MH1WK89PP]], and it sends $(-1, 1)$ into $(0, \\infty)$.\n7. By the [[derivative of the logarithm|07X1QA6BRW]] and the [[chain rule|0RXX2PT1MQ]], $h$ is differentiable on $(-1, 1)$ with $h'(x) = \\frac{1}{1 + x} \\cdot 1$.\n8. So $(g - h)'(x) = 0$ for every $x$ in the open interval $(-1, 1)$. By [[this result|003QDZ4T16]], $g - h$ is constant on $(-1, 1)$.\n9. At $x = 0$, $g(0) = a_0 = 0$ and $h(0) = \\ln 1 = 0$, since $e^0 = 1$ by the [[series of the exponential|0RMVS7RXZX]] and [[$\\ln(e^y) = y$|0RKW2X63YR]] for real $y$. So $g = h$ on $(-1, 1)$.\n10. This proves $\\ln(1 + x) = \\sum_{n=1}^{\\infty} \\frac{(-1)^{n+1}}{n} x^n$ for $-1 \u003c x \u003c 1$.\n11. Now let $x = 1$. Let $b_k = \\frac{1}{k+1}$ for $k \\ge 0$. Then $(b_k)$ is decreasing, and [[$b_k \\to 0$|08QAX4H1PF]].\n12. By the [[alternating series test|0PK8QH1W1V]], $\\sum_{k=0}^{\\infty} (-1)^k b_k$ converges. With $n = k + 1$, this is the series $\\sum_{n=1}^{\\infty} \\frac{(-1)^{n+1}}{n}$. Call its sum $S$.\n13. Then $\\sum_{n=0}^{\\infty} a_n = S$ converges, since $a_0 = 0$. By [[Abel's theorem|0S4EMDQS2F]], $g(x) \\to S$ as $x \\to 1$ from the left.\n14. By step 9, $g(x) = \\ln(1 + x)$ for $0 \u003c x \u003c 1$. Since [[$\\ln$ is continuous|0RKW2X63YR]] at $2$, $\\ln(1 + x) \\to \\ln 2$ as $x \\to 1$ from the left.\n15. Take any sequence $x_k \\in (0, 1)$ with $x_k \\to 1$. Then $g(x_k) \\to S$ and $g(x_k) \\to \\ln 2$. By [[uniqueness of limits|0ZXJ2455NJ]], $S = \\ln 2$.\n16. So $\\ln(1 + x) = \\sum_{n=1}^{\\infty} \\frac{(-1)^{n+1}}{n} x^n$ for every $x$ with $-1 \u003c x \\le 1$.\n\n## Idea\n\nInside the interval of convergence, the series and $\\ln(1 + x)$ have the same derivative $\\frac{1}{1 + x}$, by the geometric series. They also agree at $0$, so they are equal. At the endpoint $x = 1$ the series still converges, and Abel's theorem says its sum is the limit of the values inside, which is $\\ln 2$.\n",
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