Saudi Arabian Mathematical Competitions, 2012

Assume that there are polynomials $p, q \in \mathbb{R}[X]$ such that
$$\frac{p(n)}{q(n)} = 1 + \frac{1}{2!} + \dots + \frac{1}{n!}, \quad n \ge 1. \quad (1)$$
Then
$$\frac{p(n+1)}{q(n+1)} - \frac{p(n)}{q(n)} = \frac{1}{(n+1)!}, \quad n \ge 1,$$
so
$$\frac{p(n+1)q(n) - p(n)q(n+1)}{q(n)q(n+1)} = \frac{1}{(n+1)!}, \quad n \ge 1. \quad (2)$$
Define the polynomials $u, v \in \mathbb{R}[X]$ by
$$u(x) = p(x+1)q(x) - p(x)q(x+1), \quad v(x) = q(x)q(x+1).$$
From (2) it follows that $u$ is not the 0-constant polynomial and we have
$$\frac{u(n)}{v(n)} = \frac{1}{(n+1)!}, \quad n \ge 1 \quad (3)$$
and
$$\frac{u(n+1)}{v(n+1)} = \frac{1}{(n+2)!}, \quad n \ge 1.$$
It follows
$$\frac{u(n+1)}{u(n)} \cdot \frac{v(n)}{v(n+1)} = \frac{1}{n+2}, \quad n \ge 1. \quad (4)$$
We have
$$\lim_{n \to \infty} \frac{u(n+1)}{u(n)} = \lim_{n \to \infty} \frac{v(n+1)}{v(n)} = 1,$$
and from (4) we obtain the contradiction $1 = 0$.