Solution.

First, let’s rewrite the sum \(S\) using the sum notation. \[ S = \sum _{i=1}^{1999} \sqrt { 1 + \frac {1}{i^2} + \frac {1}{(i + 1)^2} } \] Now we see the terms \(1 + \frac {1}{i^2} + \frac {1}{(i + 1)^2}\) inside a square root. To get rid of the annoying square root, we can try writing the terms in a perfect square. Consider the following equation. \begin{align*} &\quad \ \left ( 1 + \frac {1}{n} + \frac {1}{n + 1} \right )^2 \\ &= 1 + \frac {1}{n^2} + \frac {1}{(n + 1)^2} + 2 \left ( \frac {1}{n} + \frac {1}{n + 1} + \frac {1}{n(n + 1)} \right ) \end{align*}

Notice that because \(\frac {1}{n} - \frac {1}{n+1} = \frac {1}{n(n+1)}\), it would be much preferable to have \(\frac {1}{n} - \frac {1}{n} - \frac {1}{n(n+1)}\) instead of \(\frac {1}{n} + \frac {1}{n + 1} + \frac {1}{n(n + 1)}\). Therefore, the following equations are obtained. \begin{align*} &\quad \ \left ( 1 + \frac {1}{n} - \frac {1}{n + 1} \right )^2 \\ &= 1 + \frac {1}{n^2} + \frac {1}{(n + 1)^2} + 2 \left ( \frac {1}{n} - \frac {1}{n + 1} - \frac {1}{n(n + 1)} \right ) \\ &= 1 + \frac {1}{n^2} + \frac {1}{(n + 1)^2} \end{align*}

Substituting the equation to the series, we obtain the following equations. \begin{align*} S &= \sum _{i=1}^{1999} \sqrt {1 + \frac {1}{i^2} + \frac {1}{(i + 1)^2}} = \sum _{i=1}^{1999} \sqrt {\left ( \frac {1}{i} - \frac {1}{i + 1} + 1 \right )^2} \\ &= \sum _{i=1}^{1999} \left ( \frac {1}{i} - \frac {1}{i + 1} + 1 \right ) = 1999 + \frac {1}{1} - \frac {1}{2000} \\ &= \frac {2000^2 - 1}{2000} = \frac {3999999}{2000} \end{align*}

Thus, \(S = \frac {3999999}{2000}\).