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  "body": "\n## Definition\n\nLet $(a_n)_{n \\ge 1}$ be a sequence of complex numbers.\nThe *Lambert series* of $(a_n)$ is the [[series|0NM626FH6Q]]\n$$L(q) = \\sum_{n=1}^{\\infty} a_n \\frac{q^n}{1 - q^n},$$\nconsidered as a function of the complex variable $q$.\nOnly the points $q$ with $\\abs{q} \u003c 1$ where the series converges are considered.\n\n## Remarks\n\nIf $\\abs{q} \u003c 1$ and $\\sum_n a_n q^n$ [[converges absolutely|0YBDSJARWN]], then the Lambert series converges absolutely and\n$$L(q) = \\sum_{m=1}^{\\infty} \\Big( \\sum_{d \\mid m} a_d \\Big) q^m.$$\nCheck: $\\frac{q^n}{1 - q^n} = \\sum_{k \\ge 1} q^{nk}$, and $\\sum_{n, k} \\abs{a_n} \\abs{q}^{nk} \\le \\frac{1}{1 - \\abs{q}} \\sum_n \\abs{a_n} \\abs{q}^n \u003c \\infty$, so the double series may be regrouped by $m = nk$.\nExample: for $a_n = 1$, the coefficient of $q^m$ is the [[number of divisors|0BKJT6WG96]] $d(m)$, so $\\sum_n \\frac{q^n}{1 - q^n} = \\sum_m d(m) q^m$.\nExample: for $a_n = \\mu(n)$, the [[Möbius function|0C8QENHYZW]], the divisor sums vanish for $m \u003e 1$, so $\\sum_n \\mu(n) \\frac{q^n}{1 - q^n} = q$.\nExample: for $a_n = \\varphi(n)$, [[Euler's totient function|0VV04M41AN]], the [[divisor sum of the totient|09JXE6XMCC]] is $m$, so the Lambert series is $\\sum_m m q^m = q/(1-q)^2$.\nNon-example of convergence: for $a_n = n!$ and $0 \u003c \\abs{q} \u003c 1$, the terms satisfy $\\abs{a_n q^n / (1 - q^n)} \\ge n! \\abs{q}^n / 2$, which tends to $\\infty$, so the series diverges.\n",
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