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  "body": "\n## Definition\n\nLet $\\sum_{n \\ge 0} a_n$ be a [[series|0JH4MN4K1C]] of complex numbers.\nFor $h \u003e 0$ put\n$$L(h) = a_0 + \\sum_{n \\ge 1} n^{-hn} a_n$$\nwhenever this series converges.\nThe series $\\sum a_n$ is *Lindelöf summable* to $s$ if $L(h)$ is defined for every $h \u003e 0$ and the [[one-sided limit|0909PWZR62]] $\\lim_{h \\to 0^+} L(h)$ equals $s$.\n\n## Remarks\n\nExample: let $\\sum a_n$ converge to $s$.\nFor fixed $h \u003e 0$ the factors $n^{-hn} = e^{-hn \\ln n}$ decrease in $n \\ge 1$ and lie in $(0, 1]$.\nBy Abel's partial summation, $L(h)$ converges, and the tails of $L(h)$ are bounded by the tails of $\\sum a_n$ uniformly in $h$.\nSince each factor tends to $1$ as $h \\to 0^+$, one gets $L(h) \\to s$, so the series is Lindelöf summable to its sum.\n\nNon-example: for $1 + 1 + 1 + \\cdots$, $L(h) = 1 + \\sum_{n \\ge 1} n^{-hn}$ is defined for every $h \u003e 0$.\nFor each $N$, $L(h) \\ge 1 + \\sum_{n=1}^{N} n^{-hn} \\to N + 1$ as $h \\to 0^+$.\nSo $L(h) \\to +\\infty$, and the second condition fails.\n",
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