Để ý rằng $\dfrac{1}{k(1999-k)}=\dfrac{1}{1999}\left(\dfrac{1}{k}+\dfrac{1}{1999-k}\right)$ với mọi $1\leq k\leq 1998$
Khi đó $S=\dfrac{1}{1999}\left[ \left(\dfrac{1}{1}+\dfrac{1}{1998}\right) +\left( \dfrac{1}{2}+\dfrac{1}{1997}\right) +...+\left(\dfrac{1}{1998}+\dfrac{1}{1}\right)\right]$
$S=\dfrac{1}{1999}.2\left[\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1998}\right]$
Dễ dàng chứng minh được $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}<2$ với mọi $n\in N$
Do đó $S<2.\dfrac{1}{1999}.2<2.\dfrac{1998}{1999}$