As the euler number $e$ is transcendant, there is no polynomial equation with integer coefficients having $e$ as a root. Are they 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 $i$). Is it possible to have $a_i\ge 0$ for each $i$ (for $1/e$ )?

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 $i$?