Thật đơn giản: $\dfrac{a}{n.(n+a)}=\dfrac{1}{n}-\dfrac{1}{n+a}$
$\Rightarrow S=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{1016064}-\dfrac{1}{1018081}$
$S=1-\dfrac{1}{1018081}=\dfrac{1018080}{1018081} \approx 0,999999$