Với $k\ge 2$ ta có:
$k^2\ge k+1$
$\Leftrightarrow \dfrac{1}{k}<\dfrac{k}{k+1}$
$\Leftrightarrow \dfrac{1}{k!k}<\dfrac{k}{(k+1)!}$
$\Leftrightarrow \dfrac{1}{k!k}<\dfrac{1}{k!}-\dfrac{1}{(k+1)!}$
Từ đó, suy ra:
$A<1+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+\ldots+\dfrac{1}{2013!}-\dfrac{1}{2014!}$
$=1+\dfrac{1}{2}-\dfrac{1}{2014!}<\dfrac{3}{2}$