I'm currently currently reading through a paper where the function
$$R(t) = 1 -\sqrt{\dfrac{2t}{\beta}} - \dfrac{1}{\beta} \displaystyle{\left(\dfrac{2t}{3} + V\sum_{n=1}^\infty\left[\dfrac{2e^{-n^2\pi^2kt}}{n^2\pi^2} - \frac{\text{erf}(\pi n\sqrt{kt})}{n^3\pi^{5/2}\sqrt{kt}} \right] \right)}, \tag 1$$
is given, and fast convergence for the involved infinite series (in the case of $t \ll 1$) is illustrated by using the identity:
$$\displaystyle{\sum_{n=1}^\infty e^{-n^2\pi^2kt} = -\frac{1}{2} + \frac{1}{2\sqrt{\pi kt}}}\left[1 + 2 \sum_{n=1}^\infty e^{-n^2/kt}\right] \tag 2$$
to derive
$$R(t) \sim 1 -\sqrt{\dfrac{2t}{\beta}} - \dfrac{1}{\beta} \left(\dfrac{2}{3}(1+kV)t - \dfrac{V\sqrt{kt}}{\sqrt\pi} - \displaystyle{\frac{2V}{\sqrt\pi}\sum_{n=1}^\infty\!\left[ \sqrt{kt}e^{-n^2/kt} \!- \frac{n^2}{\sqrt{kt}}E_1\left(\frac{n^2}{kt}\right) \right] }\right)$$
as $\beta \to \infty$, where $E_1$ is the exponential integral defined by
$$\displaystyle{E_1(z) = \int_{z}^\infty \frac{e^{-\xi}}{\xi}d\xi}.$$
I've tried to derive this asymptotic myself, however I have made little progress. It seems the writers may have modified (1) before substituting in (2) multiplied with $kt$, however I have no idea what this possible modification of (1) could be. I would be greatly appreciative if someone could give an outline of its derivation.
