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  "body": "\n## Definition\n\nLet $\\sum_{k \\ge 0} a_k$ be a [[series|0JH4MN4K1C]] of complex numbers, and let $q \\ge 0$ be real.\nThe series is *Euler summable* with parameter $q$, or $(E, q)$ summable, to $s$ if the series\n$$\\sum_{p \\ge 0} \\frac{1}{(1 + q)^{p + 1}} \\sum_{k=0}^{p} \\binom{p}{k} q^{p - k} a_k$$\nconverges and its sum is $s$.\nHere $\\binom{p}{k}$ is the [[binomial coefficient|092GEF38YD]].\nThe case $q = 1$ is called Euler summation without qualification:\n$$\\sum_{p \\ge 0} \\frac{1}{2^{p + 1}} \\sum_{k=0}^{p} \\binom{p}{k} a_k = s.$$\n\n## Remarks\n\nFor $q = 0$ the inner sum is $a_p$ alone, and $(E, 0)$ is ordinary convergence.\n\nExample: take $a_k = (-1)^k (k + 1)$, that is $1 - 2 + 3 - 4 + \\cdots$, and $q = 1$.\nThe identities $\\sum_{k=0}^{p} \\binom{p}{k} (-1)^k = 0$ for $p \\ge 1$ and $\\sum_{k=0}^{p} \\binom{p}{k} (-1)^k k = 0$ for $p \\ge 2$ show that only $p = 0$ and $p = 1$ contribute.\nThese give $\\frac12 \\cdot 1 + \\frac14 \\cdot (1 - 2) = \\frac14$, so the series is $(E, 1)$ summable to $1/4$.\n\nNon-example: for $1 + 1 + 1 + \\cdots$ and $q = 1$, the inner sum is $2^p$, and every term of the new series is $\\frac12$.\nThat series diverges, so $1 + 1 + 1 + \\cdots$ is not $(E, 1)$ summable.\n",
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