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  "body": "\n## Definition\n\nLet $\\sum_{j \\ge 0} a_j$ be a [[series|0JH4MN4K1C]] with [[partial sums|0NB643K60Q]] $s_n = \\sum_{j=0}^{n} a_j$.\nLet $(\\omega_n)$ be a sequence of nonzero numbers, the remainder estimates, let $\\beta \u003e 0$, and let $k \\ge 1$ be an integer.\nThe *Levin transformation* is\n$$\\mathcal{L}_k^{(n)} = \\frac{\\displaystyle\\sum_{j=0}^{k} (-1)^j \\binom{k}{j} (\\beta + n + j)^{k-1} \\frac{s_{n+j}}{\\omega_{n+j}}}{\\displaystyle\\sum_{j=0}^{k} (-1)^j \\binom{k}{j} (\\beta + n + j)^{k-1} \\frac{1}{\\omega_{n+j}}},$$\ndefined when the denominator is not $0$.\nThe choice $\\omega_n = a_n$ gives Levin's $t$-transformation, and $\\omega_n = (\\beta + n) a_n$ gives Levin's $u$-transformation.\nTransformations of the same construction with other choices of remainder estimates or of the factors $(\\beta + n + j)^{k-1}$ are called *Levin-type transformations*.\n\n## Remarks\n\nThe formula is exact for the model sequences\n$$s_n = s + \\omega_n \\sum_{i=0}^{k-1} \\frac{c_i}{(\\beta + n)^i}.$$\nIndeed, $(\\beta + n)^{k-1}(s_n - s)/\\omega_n$ is then a polynomial in $n$ of degree at most $k - 1$, so its $k$-th difference vanishes.\nSolving this for $s$ gives $\\mathcal{L}_k^{(n)} = s$.\nExample: for $a_j = x^j$ with $x \\ne 0, 1$, one has $s_n - \\frac{1}{1-x} = a_n \\cdot \\frac{-x}{1 - x}$, so the $t$-transformation with $k = 1$ returns $\\frac{1}{1-x}$ for every $n$.\nNon-example: if some term $a_{n+j}$ is $0$, the remainder estimate $\\omega_{n+j} = a_{n+j}$ is $0$ and the $t$-transformation is not defined there.\n\n## Explanations\n\n### L1 human\n\nThe transformation divides out a guess of the error, removes a polynomial correction by taking differences, and reads off the sum.\n",
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