{
  "aliases": [
    "mercator-series",
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  "body": "\n## Statement\n\nLet $\\operatorname{Log}$ be the [[principal branch|0YKY4M3KRQ]] of the logarithm.\nFor each $z \\in \\mathbb{C}$ with $|z| \u003c 1$, the series $\\sum_{n \\ge 1} z^n / n$ converges absolutely, and\n$$-\\operatorname{Log}(1 - z) = \\sum_{n=1}^{\\infty} \\frac{z^n}{n} = z + \\frac{z^2}{2} + \\frac{z^3}{3} + \\cdots.$$\n\n## Remarks\n\nReplacing $z$ by $-z$ gives $\\operatorname{Log}(1 + z) = \\sum_{n \\ge 1} (-1)^{n+1} z^n / n$ for $|z| \u003c 1$.\nThe [[radius of convergence|045EWAXY9P]] is $1$. For $|z| \u003c 1$, the number $1 - z$ has positive real part, so it is not on the branch cut.\nProof idea: both sides vanish at $0$, and both have the derivative $1/(1 - z) = \\sum_{n \\ge 0} z^n$ on the unit disc.\nAt $z = -1$ the series still converges, and gives $\\ln 2 = 1 - \\frac{1}{2} + \\frac{1}{3} - \\cdots$.\n\n## Explanations\n\n### L1 human\n\nIntegrate the geometric series $1 + z + z^2 + \\cdots = 1/(1 - z)$ term by term from $0$ to $z$.\nOn the left you get $z + z^2/2 + z^3/3 + \\cdots$, and on the right you get $-\\operatorname{Log}(1 - z)$.\n",
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    "aliases": [
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    "name": "Power series of the complex logarithm",
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  "msc": [
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  "name": "Power series of the complex logarithm",
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