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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 $a_j:=\max\{k\in\mathbb Z\colon ke^{-j}+S_{j-1}\le1\}$ 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.

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.

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 $a_j:=\max\{k\in\mathbb Z\colon ke^{-j}+S_{j-1}\le1$ recurrently for $j=1,2,\dots$, where $S_{j-1}:=\sum_{i=1}^{j-1}a_ie^{-i}$. 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 $1-e^{-j}<S_j\le1$.

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 $a_j:=\max\{k\in\mathbb Z\colon ke^{-j}+S_{j-1}\le1\}$ 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.

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: $a_0:=0$, $a_j:=\max\{k\in\mathbb Z\colon ke^{-j}+S_{j-1}\le1$ for $j=1,2,\dots$, where $S_{j-1}:=\sum_{i=1}^{j-1}a_ie^{-i}$.

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 $a_j:=\max\{k\in\mathbb Z\colon ke^{-j}+S_{j-1}\le1$ recurrently for $j=1,2,\dots$, where $S_{j-1}:=\sum_{i=1}^{j-1}a_ie^{-i}$. 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 $1-e^{-j}<S_j\le1$.