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  "body": "\n## Proof\n\nWe show that the series and $-\\operatorname{Log}(1 - z)$ have the same derivative and agree at $0$.\n\n1. Let $D = \\set{z : \\abs{z} \u003c 1}$ and $z \\in D$. Then $\\abs{z^n / n} \\le \\abs{z}^n$, and $\\sum \\abs{z}^n$ converges. By the [[comparison test|04Z5G0THN6]], $\\sum_{n \\ge 1} z^n / n$ converges absolutely.\n2. Consider the [[power series|0VARVQ9ZW1]] $\\sum_{n \\ge 1} z^n / n$. For $0 \\le s \u003c 1$, $s^n / n \\le 1$ for all $n \\ge 1$. So its [[radius of convergence|045EWAXY9P]] is at least $1$, and $D$ lies in its disc of convergence.\n3. Put $g(z) = \\sum_{n=1}^{\\infty} z^n / n$ for $z \\in D$. Since [[power series are holomorphic in the disc of convergence|0YMP521ZR5]], $g$ is holomorphic on $D$ and $g'(z) = \\sum_{n=1}^{\\infty} z^{n-1}$.\n4. By the [[finite geometric sum|05APWSX53S]], $\\sum_{n=1}^{N} z^{n-1} = \\frac{1 - z^N}{1 - z}$. Since $\\abs{z}^N \\to 0$ by [[the limit of $r^N$|0Q7K4AXE53]], we get $g'(z) = \\frac{1}{1 - z}$.\n5. For $z \\in D$, $\\operatorname{Re}(1 - z) \\ge 1 - \\abs{z} \u003e 0$. So $1 - z$ lies in the slit plane $S = \\C \\setminus (-\\infty, 0]$.\n6. Put $h(z) = \\operatorname{Log}(1 - z)$ for $z \\in D$, with $\\operatorname{Log}$ the [[principal branch|0YKY4M3KRQ]]. Fix $z_0 \\in D$, $\\zeta_0 = 1 - z_0$, and $k \\ne 0$ small. Then $\\frac{h(z_0 + k) - h(z_0)}{k} = - \\frac{\\operatorname{Log}(\\zeta_0 - k) - \\operatorname{Log} \\zeta_0}{-k}$.\n7. As $k \\to 0$, also $-k \\to 0$. Since [[$\\operatorname{Log}$ is holomorphic on $S$ with derivative $1/\\zeta$|0XVGJE7Q04]], the quotient in step 6 tends to $-1/\\zeta_0$. So $h$ is holomorphic on $D$ with $h'(z) = -\\frac{1}{1 - z}$.\n8. Put $F = g + h$ on $D$. By steps 3, 4 and 7, $F$ is holomorphic on $D$ and $F'(z) = 0$ for all $z \\in D$.\n9. Fix $z \\in D$ and let $\\gamma(t) = t z$ for $t \\in [0, 1]$, a path in $D$. By the [[fundamental theorem of contour integration|0YQ458D7PF]] for $F$ and the continuous function $0$, $F(z) - F(0) = \\int_\\gamma 0 \\, dw = 0$.\n10. We have $g(0) = 0$. Also $\\operatorname{Arg} 1 = 0$ by the [[principal argument|0SQSQP748R]], since $1 = \\cos 0 + i \\sin 0$. So $h(0) = \\operatorname{Log} 1 = \\ln 1 + 0 = 0$, and $F(0) = 0$.\n11. By steps 9 and 10, $g(z) + \\operatorname{Log}(1 - z) = 0$. So $-\\operatorname{Log}(1 - z) = \\sum_{n=1}^{\\infty} z^n / n$ for each $z \\in D$.\n\n## Idea\n\nDifferentiating the series term by term gives the geometric series $1/(1 - z)$, which is also the derivative of $-\\operatorname{Log}(1 - z)$. Two functions on a disc with the same derivative differ by a constant, and both vanish at $0$.\n",
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