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  "body": "\n## Definition\n\nLet $\\sum_{n \\ge 1} a_n$ be a [[series|0JH4MN4K1C]] of complex numbers.\nThe series is *Lambert summable* to $s$ if the following two conditions hold.\n\n1. The series $\\sum_{n \\ge 1} n a_n \\frac{x^n}{1 - x^n}$ converges for every real $x$ with $0 \u003c x \u003c 1$.\n2. The [[one-sided limit|0909PWZR62]]\n$$\\lim_{x \\to 1^-} (1 - x) \\sum_{n \\ge 1} n a_n \\frac{x^n}{1 - x^n}$$\nexists and equals $s$.\n\n## Remarks\n\nExample: let $\\mu$ be the [[Möbius function|0C8QENHYZW]] and $a_n = \\mu(n)/n$.\nExpanding $\\frac{x^n}{1 - x^n} = \\sum_{m \\ge 1} x^{nm}$ and grouping by $N = nm$ gives\n$$\\sum_{n \\ge 1} \\mu(n) \\frac{x^n}{1 - x^n} = \\sum_{N \\ge 1} x^N \\sum_{d \\mid N} \\mu(d) = x,$$\nsince $\\sum_{d \\mid N} \\mu(d)$ is $1$ for $N = 1$ and $0$ for $N \u003e 1$.\nThe rearrangement is valid because the double series converges absolutely for $0 \u003c x \u003c 1$.\nSo $(1 - x) \\cdot x \\to 0$, and $\\sum \\mu(n)/n$ is Lambert summable to $0$.\n\nNon-example: for $1 + 1 + 1 + \\cdots$, the inequality $1 - x^n \\le n (1 - x)$ gives $n \\frac{x^n}{1 - x^n} \\ge \\frac{x^n}{1 - x}$.\nHence $(1 - x) \\sum_{n \\ge 1} n \\frac{x^n}{1 - x^n} \\ge \\sum_{n \\ge 1} x^n = \\frac{x}{1 - x} \\to +\\infty$, and condition 2 fails.\n",
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