Originally Posted by

**shawsend** I can perhaps start it:

$\displaystyle \frac{q^n}{1-q^n}=\sum_{k=1}^{\infty} \left(q^{n}\right)^k$

So that

$\displaystyle \sum_{n=1}^{\infty}\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\left(q ^n\right)^k$

$\displaystyle =q+q^2+q^3+q^4+\cdots$

$\displaystyle +q^2+q^4+q^6+q^8+\cdots$

$\displaystyle +q^3+q^6+q^9+q^{12}+\cdots$

$\displaystyle +q^4+q^8+q^{12}+q^{16}+\cdots$

$\displaystyle +q^5+q^{10}+q^{15}+q^{20}+\cdots$

$\displaystyle +\vdots$

Grouping like-powers, it becomes obvious that the number of terms for each power is the divisor function $\displaystyle d(nk)$ so that that sum can be written as:

$\displaystyle \sum_{n=1}^{\infty}\frac{q^n}{1-q^n}=\sum_{n=1}^{\infty}d(n)q^n$

The rest I assume may be approached from the general theory of Lacunary series and the Hadamard Gap Theorem although I wouldn't know how.