In finding the rotational partition function for a $C_{\infty v}$ molecule, one comes across the function\begin{equation}Z(x)=\sum_{j=0}^\infty(2j+1)e^{-j(j+1)x}\end{equation}Surprisingly, this admits the Laurent series\begin{equation}Z(x)=\frac{1}{x}+\frac{1}{3}+\frac{1}{15}x+\frac{4}{315}x^2+\frac{1}{315}x^3+\ldots\end{equation}The first term is generally found as the approximation\begin{equation}Z(x)\approx \int_0^\infty(2j+1)e^{-j(j+1)x}dj=\frac{1}{x}\quad\quad\quad\text{for }x\ll1,\end{equation}and the rest of the terms can be found by the Euler-Maclaurin formula. However, the Euler Maclaurin formula is a bit complicated, and I haven't figured out how to use it correctly to find the expansion of $Z(x)$. I am interested primarily in proving the value of the constant term\begin{equation}\lim_{x\to 0}\left(Z(x)-\frac{1}{x}\right)=\frac{1}{3}.\end{equation}I've tried expanding the exponentials as a Taylor series, but the constant terms end up adding to infinity, and I can't figure out how to annihilate them with an expansion of $\frac{1}{x}.$ Considering the $\frac{1}{x}$ term in the Laurent series for $Z(x)$, I figure that this problem might be able to be solved with Pade approximants, or Cauchy's integral formula.
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