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?


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.


Mathematica's calculation involves the q-digamma function:

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

  • $\begingroup$ And, of course, the $q$-digamma function is defined in terms of the derivative of the $q$-gamma function, which is itself defined as an even more complicated infinite sum. Not clear how useful that answer is :) $\endgroup$ – Thomas Andrews Nov 6 '14 at 1:09

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