Saudi Arabia Mathematical Competitions 2012

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