Ta để ý rằng
$\dfrac{1}{[x-(n-1)](x-n)} = \dfrac{1}{x-(n-1)} - \dfrac{1}{x-n}$
với mọi số tự nhiên $n$.
Khi đó ta có
$A = \dfrac{x}{x+1} + \dfrac{1}{(x+1)(x+2)} + \dfrac{1}{(x+2)(x+3)} + \cdots + \dfrac{1}{(x+2013)(x+2014)}$
$= \dfrac{x}{x+1} + \dfrac{1}{x+1} - \dfrac{1}{x+2} + \dfrac{1}{x+2} - \dfrac{1}{x+3} + \cdots + \dfrac{1}{x+2013} - \dfrac{1}{x+2014}$
$= \dfrac{x}{x+1} + \dfrac{1}{x+1} - \dfrac{1}{x+2014}$
$= \dfrac{x+1}{x+1} - \dfrac{1}{x+2014} = -\dfrac{1}{x+2014}$
