A Möbius inversion sum

Let $q \in \mathbb{R}$ such that $|q|<1$ . Prove that

$\sum_{n=1}^{\infty} \mu(n) \cdot \frac{q^n}{1-q^n} = q$

Solution

We have successively:

$\begin{align*} \sum_{n=1}^{\infty} \mu(n) \cdot \frac{q^n}{1-q^n} &= \sum_{n=1}^{\infty} \mu(n) \sum_{m=0}^{\infty} q^{nm} \\ &= \sum_{n=1}^{\infty} \sum_{d \mid n} \mu(d)q^n \\ &= q + \sum_{n=2}^{\infty} \sum_{d \mid n} \mu(d)q^n \end{align*}$

since $\sum \limits_{d \mid n} \mu(d) =0$ whenever $n \geq 2.$

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A Möbius inversion sum

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