Don't beat around the bush...

Each of the sum can be expressed as $\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}$ for a = 1,2,3,...,1999

Now let me simplify that expression, and hope it can be simplified in a good way, because this problem may involve telescoping series: $\sqrt{1+\frac{1}{a^2}+\frac{1}{(a+1)^2}}$

$= \sqrt{ \frac{a^2 . (a+1)^2 + a^2 + (a+1)^2}{a^2 . (a+1)^2}}$

$= \sqrt{\frac{a^4 + 2a^3 + 3a^2 + 2a + 1}{(a^2+a)^2}}$

$= \sqrt{\frac{(a^4 + a^3 + a^2) + (a^3 + a^2 + a) + (a^2 + a + 1)}{(a^2+a)^2}}$

$= \sqrt{\frac{a^2(a^2 + a + 1) + a(a^2 + a + 1) + (a^2 + a + 1)}{(a^2+a)^2}}$

$= \sqrt{\frac{(a^2+a+1)^2}{(a^2+a)^2}}$

$= \frac{a^2+a+1}{a^2+a}$

$= 1 + \frac{1}{a(a+1)}$

Now the sum is become: $= 1 + \frac{1}{1.2} + 1 + \frac{1}{2.3} + 1 + \frac{1}{3.4} + ... + 1 + \frac{1}{1999.2000}$

$=1.1999 + \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + ... + \frac{1}{1999.2000}$

$=1999 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ... + (\frac{1}{1999} - \frac{1}{2000})$

Every terms in fraction cancel each out except $1$ and $\frac{1}{2000}$

$=1999 + 1 - \frac{1}{2000}$

Clearly the integer part is 1999, while the fractional part is $\frac{1999}{2000}$