2 Transcribed robjohn's anwer posted at Mathematics SE

The version of this question posted on Mathematics Stack Exchange got a new answer yesterday by robjohn, which I've already approved.

Writing $g_1(x)=f(1/x)$ gives $$g_1(x)\equiv\sum_{k\ge1}\frac{c_kx^k}{(1+a_1x)(1+a_2x)\dots(1+a_kx)}\tag{1}$$ which vanishes at $x=0$.
Recursively define $$g_{n+1}(x)=\frac{(1+a_nx)g_n(x)}{x}-c_n\tag{2}$$ where $$c_n=\lim_{x\to0}\frac{g_n(x)}{x}\tag{3}$$ Then $$g_n(x)\equiv\sum_{k\ge n}\frac{c_kx^{k-n+1}}{(1+a_nx)(1+a_{n+1}x)\dots(1+a_kx)}\tag{4}$$ is another series like $(1)$ (which vanishes at $x=0$).
The series in $(1)$ may or may not converge, as with the Euler-Maclaurin Sum Series. As with most asymptotic series, we are only interested in the first several terms; the remainder (not the remaining terms) can be bounded by something smaller than the preceding terms. Therefore, convergence is not an issue.