Q:

Power series equation with solution $1/e$

As $e$ is transcendental, there is no polynomial equation with integer coefficients having $e$ as a root.
Are there classical equations of the form
$$\sum_{i=0}^{\infty} a_ix^i =1$$
that have $e$ or $1/e$ as root, with $a_i\in \mathbb{Z}$ for each $i$?
For $1/e$, is it possible to have $a_i\ge 0$ for each $i$?

A:

If $a_i=0$ for all large enough $i$, then $e$, being transcendental, cannot be a solution to your equation, unless $a_i=0$ for all $i\ge1$. Otherwise, if $a_i\ne0$ for infinitely many $i$, then the series for $e$ cannot converge, given that the $a_i$'s are integers, because then $|a_ie^i|\ge e^i\not\to0$ if $a_i\ne0$. 
However, this can be done for $1/e$, as follows: let $a_0:=0$, and then let $a_j:=\max\big\{k\in\{0,1,\dots\}\colon ke^{-j}+S_{j-1}\le1\big\}$ recurrently for $j=1,2,\dots$, where $S_{j-1}:=\sum_{i=0}^{j-1}a_ie^{-i}$. 
Indeed, then for $j=1,2,\dots$ we have $S_j=a_je^{-j}+S_{j-1}\le1$, whereas $S_j+e^{-j}=(a_j+1)e^{-j}+S_{j-1}>1$, so that $1-e^{-j}<S_j\le1$, whence $S_j\to1$ as $j\to\infty$, as desired.  

