Nhận thấy từng số hạng có dạng: $\sqrt{1 + \dfrac{1}{n^2} + \dfrac{1}{(n+1)^2}}\ \ (n\geqslant 2;\ n\in\Bbb N)$
Ta có:
$\quad \sqrt{1 + \dfrac{1}{n^2} + \dfrac{1}{(n+1)^2}}$
$= \sqrt{\left(1 + \dfrac{1}{n}\right)^2 - 2\cdot \dfrac{1}{n} + \dfrac{1}{(n+1)^2}}$
$= \sqrt{\left(1 + \dfrac{1}{n}\right)^2 - 2\cdot \dfrac{n+1}{n}\cdot \dfrac{1}{n+1} + \dfrac{1}{(n+1)^2}}$
$= \sqrt{\left(1 + \dfrac{1}{n}\right)^2 - 2\cdot \left(1 + \dfrac{1}{n}\right)\cdot \dfrac{1}{n+1} + \dfrac{1}{(n+1)^2}}$
$= \sqrt{\left(1 + \dfrac{1}{n} - \dfrac{1}{n+1}\right)^2}$
$= \left|1 + \dfrac{1}{n} - \dfrac{1}{n+1}\right|$
$= 1 + \dfrac{1}{n} - \dfrac{1}{n+1}$
Áp dụng công thức vừa chứng minh, ta được:
$ A = \left(1 + \dfrac{1}{2} - \dfrac13\right) + \left(1 + \dfrac{1}{3} - \dfrac14\right) + \cdots+ \left(1 + \dfrac{1}{1999} - \dfrac{1}{2000}\right)$
$\quad =1998 + \dfrac12 - \dfrac{1}{2000}$
$\quad = \dfrac{3\ 996\ 999}{2000}$
