Lambert summation

Remarks

Example: let $\mu$ be the Möbius function and $a_n = \mu(n)/n$. Expanding $\frac{x^n}{1 - x^n} = \sum_{m \ge 1} x^{nm}$ and grouping by $N = nm$ gives $$\sum_{n \ge 1} \mu(n) \frac{x^n}{1 - x^n} = \sum_{N \ge 1} x^N \sum_{d \mid N} \mu(d) = x,$$ since $\sum_{d \mid N} \mu(d)$ is $1$ for $N = 1$ and $0$ for $N > 1$. The rearrangement is valid because the double series converges absolutely for $0 < x < 1$. So $(1 - x) \cdot x \to 0$, and $\sum \mu(n)/n$ is Lambert summable to $0$.

Non-example: for $1 + 1 + 1 + \cdots$, the inequality $1 - x^n \le n (1 - x)$ gives $n \frac{x^n}{1 - x^n} \ge \frac{x^n}{1 - x}$. Hence $(1 - x) \sum_{n \ge 1} n \frac{x^n}{1 - x^n} \ge \sum_{n \ge 1} x^n = \frac{x}{1 - x} \to +\infty$, and condition 2 fails.