I am wondering is there one way to extract the asymptotic behavior of $x$ in the following expression near $x=0$?

$$\sum_{n=1}^{\infty} n\log(1-\exp(-n x))$$ where $x $ is real.

  • $\begingroup$ At least the divergent parts of the asymptotic expansion seem like they can be extracted from the leading terms in the Euler-Maclaurin formula, see specifically the formula under "asymptotic expansion of sums." $\endgroup$ – Seth Whitsitt Oct 11 at 6:16

First, expand the logarithm into its Taylor series:

\begin{align} S&\equiv \sum_{n\ge 1}n\log(1-e^{-nx}) \\ &= -\sum_{n\ge 1} n \sum_{k\ge 1} \frac{e^{-knx}}{k} \\ &= -\sum_{k \ge 1} \frac{1}{k}\sum_{n\ge 1}n\,e^{-nkx} \end{align} To sum the inner series, differentiate the following identity with respect to $\beta$, $$\frac{1}{1-e^{-\beta}}-1=\sum_{m\ge 1}e^{-\beta m},$$ obtaining $$\sum_{m\ge 1}m\,e^{-\beta m}=\frac{e^{-\beta}}{(1-e^{-\beta})^2}.$$ Identifying $\beta$ with $kx$ in our original series, we see that \begin{align} S&=-\sum_{k \ge 1}\frac{1}{k} \left(\frac{e^{-kx}}{(1-e^{-kx})^2}\right) \end{align} Using $e^{-kx}=1-kx+\mathcal{O}(x^2)$, \begin{align} S&=\sim -\sum_{k\ge 1}\frac{1}{k^3}\frac{1-kx}{x^2} \\ &=\sum_{n\ge 1}\frac{k^{-2}}{x}-\frac{k^{-3}}{x^2} \end{align} Thus, $$S\sim -\frac{\zeta(3)}{x^2}+\frac{\zeta(2)}{x}\qquad \text{as } x\to 0$$


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