An infinite exponential sum

Question

I was trying to create a problem for a test I'm writing, and I ended up attempting to evaluate

$$\sum_{n=1}^{\infty} \dfrac{1}{e^n-1}.$$

This definitely converges, but I have no idea how to go about finding an exact value. Anyone have any ideas?

Answer 1

Mathematica's calculation involves the q-digamma function:

$$ -\psi_{1/e}(1) + 1 - \log(e-1) $$

Answer 2

Goos's answer was close. Write $$\frac{1}{e^n-1} = \frac{e^{-n}}{1-e^{-n}} = \sum_{k=1}^\infty e^{-nk}$$ Then your sum is, after rearranging:

$$\sum_{m=1}^\infty \tau(m) e^{-m}$$

Where $\tau(m)$ is the number of positive divisors of $m$.

Probably not much more you can do with it.