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 $i$?



Power series equation with solution $1/e$
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. 


