How's this:
Take the times $t_{pre}$ to be of the form $t_{pre} = T_{0} + iT$, where $i$ ranges over some set of integers of the form $\{0, 1, 2, . . . , N\}$, and we allow the possibility that $N$ is infinite and place no restriction on the sign of $T$, the interval size, though we take $T \ne\ 0$; since the case $T = 0$ boils down to there being only one value of $t_{pre}$, $T_{0}$, in which case $V(t)$ trivially becomes $(\epsilon_{0}/\tau)e^{-(t - T_{0})/\tau}$. (I'm dropping your subscripts to $\tau$ for convenience, i.e. to make typing slightly easier.) Then in general we have $V(t) = \sum_{i = 0}^{N}(\epsilon_{0}/\tau)e^{-(t-T_{0} - iT)/\tau}$; factoring out $(\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}$ yields $V(t) = (\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}\sum_{i = 0}^{N}e^{iT/\tau}$; in this form, $V(t)$ is a single exponential times a constant, as per your request. The sum can be cleaned up using standard formulae for geometric series (cf. http://en.wikipedia.org/wiki/Geometric_series), viz $\sum_{i = 0}^{N}e^{iT/\tau} = ((1 - e^{(N + 1)T/\tau})/(1- e^{T/\tau}))$, so finally $V(t) = (\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}((1 - e^{(N + 1)T/\tau})/(1- e^{T/\tau}))$. In the event $T > 0$, the sum diverges unless $N$ is finite; for $T < 0$, however, letting $N \to \infty$ we obtain $V(t) = (\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}(1/(1- e^{T/\tau}))$ or $V(t) = (\epsilon_{0}/\tau)(1/(1- e^{T/\tau}))e^{-(t-T_{0})/\tau}$.
It is interesting to note that, as $T < 0$ becomes smaller in magnitude, i.e. $T \to 0^{-}$, $V(t)$ explodes, as if it were the sum of an infinite number of identical terms $(\epsilon_{0}/\tau)e^{-(t-T_{0})/\tau}$; I leave a similar analysis when $T > 0$ to you.
I think you will find this stuff is generally covered under Laplace transforms, rather than Fourier series.
To the boys at http://mathoverflow.tqft.net/discussion/784/question-being-bumped-to-the-front-page/: I did it again! Accidentally hit "Save" instead of "Preview"! Only lasted an hour this time. Whew! Gotta be more careful with those buttons!
