Lambert series

Remarks

If $\abs{q} < 1$ and $\sum_n a_n q^n$ converges absolutely, then the Lambert series converges absolutely and $$L(q) = \sum_{m=1}^{\infty} \Big( \sum_{d \mid m} a_d \Big) q^m.$$ Check: $\frac{q^n}{1 - q^n} = \sum_{k \ge 1} q^{nk}$, and $\sum_{n, k} \abs{a_n} \abs{q}^{nk} \le \frac{1}{1 - \abs{q}} \sum_n \abs{a_n} \abs{q}^n < \infty$, so the double series may be regrouped by $m = nk$. Example: for $a_n = 1$, the coefficient of $q^m$ is the number of divisors $d(m)$, so $\sum_n \frac{q^n}{1 - q^n} = \sum_m d(m) q^m$. Example: for $a_n = \mu(n)$, the Möbius function, the divisor sums vanish for $m > 1$, so $\sum_n \mu(n) \frac{q^n}{1 - q^n} = q$. Example: for $a_n = \varphi(n)$, Euler's totient function, the divisor sum of the totient is $m$, so the Lambert series is $\sum_m m q^m = q/(1-q)^2$. Non-example of convergence: for $a_n = n!$ and $0 < \abs{q} < 1$, 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.