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  "body": "\n## Definition\n\nLet $(a_n)_{n \\ge 0}$ be a sequence of real or complex numbers, and consider the [[alternating series|017BR1M98J]] $\\sum_{n \\ge 0} (-1)^n a_n$.\nThe forward difference is $\\Delta a_n = a_{n+1} - a_n$, and its iterates satisfy\n$$\\Delta^n a_0 = \\sum_{k=0}^{n} (-1)^{n-k} \\binom{n}{k} a_k,$$\nwith the [[binomial coefficients|092GEF38YD]] $\\binom{n}{k}$.\nThe *Euler transform* of the series is the [[series|0JH4MN4K1C]]\n$$\\sum_{n \\ge 0} \\frac{(-1)^n \\Delta^n a_0}{2^{n+1}}.$$\nIf $\\sum (-1)^n a_n$ converges, then its Euler transform converges to the same sum.\n\n## Remarks\n\nExample: for $a_n = \\frac{1}{n+1}$ one has $(-1)^n \\Delta^n a_0 = \\frac{1}{n+1}$, so the alternating harmonic series becomes $\\sum_{n \\ge 0} \\frac{1}{(n+1) 2^{n+1}}$, which equals $\\ln 2$.\nFor $a_n = 1$ the series $1 - 1 + 1 - \\cdots$ diverges, but $\\Delta^n a_0 = 0$ for $n \\ge 1$ and the transform is $\\frac12$; this is how Euler summation assigns the value $\\frac12$ to it.\nNon-example: for $a_n = 3^n$ one gets $\\Delta^n a_0 = 2^n$, and the transform $\\sum_{n \\ge 0} \\frac{(-1)^n}{2}$ diverges, so the transform does not always produce a convergent series.\n\n## Explanations\n\n### L1 human\n\nThe transform replaces the terms by weighted differences of the first terms.\nFor smooth terms the high differences are small, and the weights $2^{-n-1}$ make the new series converge fast.\n",
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