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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $a_0, a_1, \\dots, a_n$ be complex numbers such that no division by zero occurs on the right-hand side below.\nThen\n$$a_0 + a_0 a_1 + a_0 a_1 a_2 + \\cdots + a_0 a_1 \\cdots a_n = \\cfrac{a_0}{1 - \\cfrac{a_1}{1 + a_1 - \\cfrac{a_2}{1 + a_2 - \\cfrac{\\cdots}{\\cdots + a_{n-1} - \\cfrac{a_n}{1 + a_n}}}}}.$$\nConsequently, let $(a_k)$ be an infinite sequence, and consider the infinite [[continued fraction|0BK6JBK58W]] of the same shape.\nIts $n$-th [[convergent|0SRWNR42BQ]] equals the $n$-th [[partial sum|0NB643K60Q]] of the [[series|0JH4MN4K1C]] $\\sum_{k \\ge 0} a_0 a_1 \\cdots a_k$.\nSo the continued fraction converges exactly when the series does, and to the same value.\n\n## Remarks\n\nProof idea: by induction on $n$.\nReplacing $a_n$ by $a_n(1 + a_{n+1})$ turns the left side for $n$ into the left side for $n + 1$.\nOn the right side the same replacement turns the last fraction $\\frac{a_n}{1 + a_n}$ into $\\frac{a_n}{1 + a_n - \\frac{a_{n+1}}{1 + a_{n+1}}}$, because both equal $\\frac{a_n(1 + a_{n+1})}{1 + a_n + a_n a_{n+1}}$.\nCheck for $n = 1$: $\\cfrac{a_0}{1 - \\frac{a_1}{1 + a_1}} = a_0(1 + a_1)$.\nExample: with $a_0 = 1$ and $a_k = x$ for $k \\ge 1$, the series is the [[geometric series|01X5CVV1Q5]], and for $\\abs{x} \u003c 1$\n$$\\frac{1}{1 - x} = \\cfrac{1}{1 - \\cfrac{x}{1 + x - \\cfrac{x}{1 + x - \\cdots}}}.$$\n\n## Explanations\n\n### L1 human\n\nAny series whose terms are running products $a_0$, $a_0 a_1$, $a_0 a_1 a_2$, and so on, can be rewritten as a continued fraction with the same partial values.\nSo results about series transfer directly to continued fractions of this shape, and back.\n\n### L2 human\n\nPower series have this form with $a_k = c_k x / c_{k-1}$ when the coefficients $c_k$ are nonzero.\nIn this way the formula turns the series of $e^x$, $\\log(1 + x)$ or $\\arctan x$ into continued fractions.\nThese continued fractions converge at the same rate as the series; other transformations are needed to obtain faster convergence.\n",
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