Question

determine the value of the sum​ along with the steps

Answer 1

The n-th term in the sum (where n is a natural number 1 ≤ n ≤ 1999) is

$\sqrt{1 + \dfrac1{n^2} + \dfrac1{(n+1)^2}}$

and can be simplified as

$\sqrt{\dfrac{n^2(n+1)^2 + (n+1)^2 + n^2}{n^2(n+1)^2}} = \dfrac{n^2+n+1}{n(n+1)}$

and further expanded into partial fractions as

$\dfrac{n^2+n+1}{n(n+1)} = 1 + \dfrac1{n(n+1)} = 1 + \dfrac1n - \dfrac1{n+1}$

Then the sum telescopes:

$S = \left(1 + \dfrac11 - \dfrac12\right) + \left(1 + \dfrac12 - \dfrac13\right) + \left(1 + \dfrac13 - \dfrac14\right) + \cdots + \left(1 + \dfrac1{1999} - \dfrac1{2000}\right)$

$S = 1999 + \dfrac11 - \dfrac1{2000} = 2000 - \dfrac1{2000} = \boxed{\frac{3,999,999}{2000}}$