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  "body": "\n## Statement\n\nLet $0 \u003c \\varepsilon \u003c \\pi/2$ and let $a_1, a_2, \\dots$ be complex numbers with $\\abs{\\arg a_n} \u003c \\pi/2 - \\varepsilon$ for every $n$.\nLet $f_n$ be the $n$-th [[convergent|0SRWNR42BQ]] of the [[continued fraction|0BK6JBK58W]]\n$$f = \\cfrac{1}{a_1 + \\cfrac{1}{a_2 + \\cfrac{1}{a_3 + \\cdots}}}.$$\nThen:\n\n1. every $f_n$ is finite and satisfies $\\abs{\\arg f_n} \u003c \\pi/2 - \\varepsilon$;\n2. the sequences $(f_{2n})$ and $(f_{2n+1})$ of even and odd convergents both converge;\n3. the continued fraction converges, that is, $(f_n)$ has a finite limit, if and only if $\\sum_{n \\ge 1} \\abs{a_n}$ diverges.\n\n## Remarks\n\nItem 1 follows by induction from the tail: the open sector $S = \\set{z \\ne 0 : \\abs{\\arg z} \u003c \\pi/2 - \\varepsilon}$ is closed under addition, and $z \\mapsto 1/z$ maps $S$ onto itself.\nHere $\\arg$ is the [[argument|070PFSYJJR]] with values in $(-\\pi, \\pi]$.\nExample: $a_n = 1$ for all $n$ lies in the sector and $\\sum \\abs{a_n} = \\infty$, so the continued fraction converges; its value is $\\frac{\\sqrt5 - 1}{2}$.\nIf $a_n = 2^{-n}$, the sum $\\sum \\abs{a_n}$ is finite, and by item 3 the even and odd convergents tend to different limits.\nFor positive real $a_n$ the theorem contains the classical criterion that $\\cfrac{1}{a_1 + \\cfrac{1}{a_2 + \\cdots}}$ converges exactly when $\\sum a_n$ diverges.\n\n## Explanations\n\n### L1 human\n\nThe theorem treats continued fractions whose partial denominators all point into a fixed wedge around the positive real axis.\nThen the fraction converges exactly when these denominators are not too small in total, that is, when the sum of their sizes is infinite.\n\n### L2 human\n\nThe sector condition keeps all convergents away from $0$ and $\\infty$, so only the oscillation between even and odd convergents can prevent convergence.\nThe divergence of $\\sum \\abs{a_n}$ is what forces the even and odd limits to meet.\nCompare the [[Śleszyński–Pringsheim theorem|NEC48RHNN0]], where large partial denominators give convergence.\n",
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