TITLE: An infinite exponential sum
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.