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  "body": "\n## Definition\n\nLet $\\sum_{k \\ge 0} a_k$ be a [[series|0JH4MN4K1C]] of complex numbers, and let $\\Gamma$ be the [[gamma function|0V3FZMVB4K]].\nFor $\\alpha \u003e 0$ put\n$$M(\\alpha) = \\sum_{k \\ge 0} \\frac{a_k}{\\Gamma(1 + \\alpha k)}$$\nwhenever this series converges.\nThe series $\\sum a_k$ is *Mittag-Leffler summable* to $s$ if $M(\\alpha)$ is defined for all small $\\alpha \u003e 0$ and the [[one-sided limit|0909PWZR62]] $\\lim_{\\alpha \\to 0^+} M(\\alpha)$ equals $s$.\n\n## Remarks\n\nFor $a_k = z^k$ one gets $M(\\alpha) = \\sum_{k \\ge 0} \\frac{z^k}{\\Gamma(1 + \\alpha k)}$, the Mittag-Leffler function $E_\\alpha(z)$.\n\nExample: let $\\sum a_k$ converge absolutely to $s$.\nSince $\\Gamma(x) \\ge \\frac12$ for all $x \\ge 1$, one has $\\abs{a_k / \\Gamma(1 + \\alpha k)} \\le 2 \\abs{a_k}$ for all $\\alpha$ and $k$.\nEach term tends to $a_k$ as $\\alpha \\to 0^+$, so by dominated convergence for series $M(\\alpha) \\to s$.\n\nNon-example: for $1 + 1 + 1 + \\cdots$, the terms $1/\\Gamma(1 + \\alpha k)$ are positive and tend to $1$ for each $k$.\nSo for each $N$, $M(\\alpha) \\ge \\sum_{k=0}^{N} \\frac{1}{\\Gamma(1 + \\alpha k)} \\to N + 1$ as $\\alpha \\to 0^+$, and $M(\\alpha)$ has no finite limit.\n",
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