Sum of Square Root Series

General term $T_n = \sqrt{1 + \frac{1}{n^2} + \frac{1}{(n+1)^2}}$.

Simplifying inside the square root: $1 + \frac{1}{n^2} + \frac{1}{(n+1)^2} = \frac{n^2(n+1)^2 + (n+1)^2 + n^2}{n^2(n+1)^2} = \frac{n^4 + 2n^3 + 3n^2 + 2n + 1}{n^2(n+1)^2} = \frac{(n^2 + n + 1)^2}{n^2(n+1)^2}$

So $T_n = \frac{n^2 + n + 1}{n(n+1)} = 1 + \frac{1}{n(n+1)} = 1 + \frac{1}{n} - \frac{1}{n+1}$.

Sum $S = \sum_{n=1}^{2007} T_n = \sum_{n=1}^{2007} \left(1 + \frac{1}{n} - \frac{1}{n+1}\right) = 2007 + \left(1 - \frac{1}{2008}\right) = 2008 - \frac{1}{2008}$.